Rozrakhunok pH in the range of hydrolyzing salts. A watery exponent. Hydrolysis of salts Table ph of different aqueous solutions of salts
Pure water є even with a weak electrolyte. The process of dissociation of water can be used to turn the lines: HOH ⇆ H + + OH -. As a result of dissociation of water, there is a place for any kind of water distribution and they H + and they OH -. Concentration of cich ions can be developed for additional help Rivnyannya Ionny Dobutku Vodi
C (H +) × C (OH -) = K w,
de K w - constant of ionic dobutku vodi ; at 25 ° C K w = 10 -14.
Razchini, in some concentrations of ions H + and OH - however, are called neutral razchini. Neutral range C (H +) = C (OH -) = 10 –7 mol / l.
In sour solution, C (H +)> C (OH -) і, as a distillation і from the level of ionic addition of water, C (H +)> 10 –7 mol / l, and C (OH -)< 10 –7 моль/л.
In the case of puddles, C (OH -)> C (H +); at tsom C (OH -)> 10 –7 mol / L, and C (H +)< 10 –7 моль/л.
pH is a value that characterizes acidity and water purity; the quantity to be called a watery exponent that insurance for the formula:
pH = -lg C (H +)
Acid pH<7; в нейтральном растворе pH=7; в щелочном растворе pH>7.
For the analogy with the "water indicator" (pH), the "hydroxyl" indicator (pOH) is introduced:
pOH = –lg C (OH -)
Water and hydroxyl indicators of tying
The hydroxyl index is used to determine the pH value in puddles.
Sirchana acid- a strong electrolyte, which is dissociated at the raznichnogo razvorno that I will follow the scheme: H 2 SO 4 ® 2 H + + SO 4 2–. It is clear from the dissociation process that C (H +) = 2 · C (H 2 SO 4) = 2 × 0.005 mol / l = 0.01 mol / l.
pH = -lg C (H +) = -lg 0.01 = 2.
Sodium hydroxide is a strong electrolyte, which can be dissociated without turning back according to the scheme: NaOH ® Na + + OH -. It is clear from the dissociation process that C (OH -) = C (NaOH) = 0.1 mol / l.
pOH = -lg C (H +) = -lg 0.1 = 1; pH = 14 - pOH = 14 - 1 = 13.
Dissociation of weak electrolyte is a very important process. Rivnovagi constant, written for the process of dissociation of weak electrolyte, called dissociation constant ... For example, for the process of dissociation of ocetic acid
CH 3 COOH ⇆ CH 3 COO - + H +.
The dermal stage of dissociation of bagato-basic acid is characterized by a constant of dissociation. Dissociation constant - previdkova value; div.
The level of concentration of ions (and pH) in the range of weak electrolytes should be set up until the task is set for a chemical solution for this purpose, if the constant is necessary and it is necessary to know the necessary ...
At 0.35% of the range of NH 4 OH, the molar concentration of ammonium hydroxide is 0.1 mol / l (butt of the converted percentage concentration in molar - div. Stock 5.1). The qiu value is often referred to as C0. C0 - the whole concentration of electrolyte in the distribution (concentration of electrolyte before dissociation).
NH 4 OH is taken with a weak electrolyte, which is reversely dissociated in water: NH 4 OH ⇆ NH 4 + + OH - (div. Also note 2 on side 5). Dissociation constant K = 1.8 · 10 -5 (previdkova value). Oskіlki weak electrolyte dissociation is not perfect, easily stewed, but produced x mol / l NH 4 OH, so the concentration of ions in ammonia and hydroxide-ions is also important (also expensive / l) = C x / l. Equally important is the concentration of unproductive NH 4 OH road: C (NH 4 OH) = (C 0 -x) = (0.1-x) mol / l.
Pidstavlyaєmo turn through x equally important concentration of all parts of the same name:
.
Even weaker electricity dissociation is insignificant (x ® 0) and the ixom at the denominator yak beforehand can be zehtuvati:
.
Call at the staff of the foreign chemistry and the bannerman doesn’t care for that kind of person, if (for the whole type of x - the concentration of electrolyte, for the production of food, - in 10 or less times when it’s about to concentrate)
С (OH -) = x = 1.34 ∙ 10 -3 mol / l; pOH = –lg C (OH -) = –lg 1.34 ∙ 10 –3 = 2.87.
pH = 14 - pOH = 14 - 2.87 = 11.13.
Dissociation stage Electrolyte can be used as a basis for the use of electrolyte concentration (x), for production, up to the over-the-counter concentration of electrolyte (C 0):
(1,34%).
The list of slides should be converted from the percentage concentration to molar (div. Butt 5.1). In this case, C0 (H3PO4) = 3.6 mol / L.
The concentration of ions in water in the range of high-base weak acids is carried out only at the first stage of dissociation. Strictly apparent, the overwhelming concentration of ions in water in the range of weak high-rich acid and high concentrations of ions H +, was established at the dermal stage of dissociation. For example, for phosphoric acid C (H +) zagalnaya = C (H +) in 1 stage + C (H +) in 2 stages + C (H +) in 3 stages. However, the dissociation of weak electrolytes against the first stage, and on other and advancing stages - an insignificant world, that
C (H +) in 2 stages ≈ 0, C (H +) in 3 stages ≈ 0 and C (H +) early ≈ C (H +) in 1 stage.
Do not have phosphoric acid and produced at the first stage x mol / l, so as to dissociation H 3 PO 4 ⇆ H + + H 2 PO 4 - next, for the most important concentration of the road, H + and H 2 in 4 - so and the concentration of nonproductive H 3 PO 4 is equally important (3.6-x) mol / l. The rate of change through x concentration of ions H + and H 2 PO 4 - and molecules of H 3 PO 4 changes the dissociation constants at the first stage (K 1 = 7.5 · 10 -3 is the pre-setting value):
K 1 / C 0 = 7.5 · 10 -3 / 3.6 = 2.1 · 10 -3< 10 –2 ; следовательно, иксом как слагаемым в знаменателе можно пренебречь (см. также пример 7.3) и упростить полученное выражение.
;
mol / l;
З (H +) = x = 0.217 mol / l; pH = -lg C (H +) = -lg 0.217 = 0.66.
(3,44%)
Zavdannya number 8
Ensure a) the pH of the range of strong acids and bases; b) the range of weak electrolyte and the steps of dissociation of electrolyte in a whole range (table 8). The strength of the solution is 1 g / ml.
Table 8 - Umovi zavdannya No. 8
| Option No. | a | b | Option No. | a | b |
| 0.01M H 2 SO 4; 1% NaOH | 0.35% NH 4 OH | ||||
| 0.01 MCa (OH) 2; 2% HNO 3 | 1% CH 3 COOH | 0.04M H 2 SO 4; 4% NaOH | 1% NH 4 OH | ||
| 0.5M HClO 4; 1% Ba (OH) 2 | 0.98% H 3 PO 4 | 0.7M HClO 4; 4% Ba (OH) 2 | 3% H 3 PO 4 | ||
| 0.02M LiOH; 0.3% HNO 3 | 0.34% H 2 S | 0.06M LiOH; 0.1% HNO 3 | 1.36% H 2 S | ||
| 0.1M HMnO 4; 0.1% KOH | 0.031% H 2 CO 3 | 0.2M HMnO 4; 0.2% KOH | 0.124% H 2 CO 3 | ||
| 0.4M HCl; 0.08% Ca (OH) 2 | 0.47% HNO 2 | 0.8M HCl; 0.03% Ca (OH) 2 | 1.4% HNO 2 | ||
| 0.05M NaOH; 0.81% HBr | 0.4% H 2 SO 3 | 0.07M NaOH; 3.24% HBr | 1.23% H 2 SO 3 | ||
| 0.02M Ba (OH) 2; 0.13% HI | 0.2% HF | 0.05M Ba (OH) 2; 2.5% HI | 2% HF | ||
| 0.02M H 2 SO 4; 2% NaOH | 0.7% NH 4 OH | 0.06MH 2 SO 4; 0.8% NaOH | 5% CH 3 COOH | ||
| 0.7M HClO 4; 2% Ba (OH) 2 | 1.96% H 3 PO 4 | 0.08M H 2 SO 4; 3% NaOH | 4% H 3 PO 4 | ||
| 0.04MLiOH; 0.63% HNO 3 | 0.68% H 2 S | 0.008M HI; 1.7% Ba (OH) 2 | 3.4% H 2 S | ||
| 0.3MHMnO 4; 0.56% KOH | 0.062% H 2 CO 3 | 0.08M LiOH; 1.3% HNO 3 | 0.2% H 2 CO 3 | ||
| 0.6M HCl; 0.05% Ca (OH) 2 | 0.94% HNO 2 | 0.01M HMnO 4; 1% KOH | 2.35% HNO 2 | ||
| 0.03M NaOH; 1.62% HBr | 0.82% H 2 SO 3 | 0.9M HCl; 0.01% Ca (OH) 2 | 2% H 2 SO 3 | ||
| 0.03M Ba (OH) 2; 1.26% HI | 0.5% HF | 0.09M NaOH; 6.5% HBr | 5% HF | ||
| 0.03M H 2 SO 4; 0.4% NaOH | 3% CH 3 COOH | 0.1M Ba (OH) 2; 6.4% HI | 6% CH 3 COOH | ||
| 0.002M HI; 3% Ba (OH) 2 | 1% HF | 0.04MH 2 SO 4; 1.6% NaOH | 3.5% NH 4 OH | ||
| 0.005MHBr; 0.24% LiOH | 1.64% H 2 SO 3 | 0.001M HI; 0.4% Ba (OH) 2 | 5% H 3 PO 4 |
Butt 7.5 They mixed 200 ml of 0.2M solution of H 2 SO 4 and 300 ml of 0.1M solution of NaOH. Protect the pH range by making sure that the concentration of ions Na + and SO 4 2 is at a wide range.
Indication of the reaction of H 2 SO 4 + 2 NaOH → Na 2 SO 4 + 2 H 2 O to a high-speed ion-molecular vigle: H + + OH - → H 2 O
From the ionic-molecular reaction of the distillation, the reaction does not take place without the ion H + and OH - and the molecule of water. Ioni Na + and SO 4 2– do not take part in the reaction, that is, the majority of the reaction is the same, as before the reaction.
Rozrakhunok number of speeches before the reaction:
n (H 2 SO 4) = 0.2 mol / L × 0.1 L = 0.02 mol = n (SO 4 2-);
n (H +) = 2 × n (H 2 SO 4) = 2 × 0.02 mol = 0.04 mol;
n (NaOH) = 0.1 mol / L 0.3 L = 0.03 mol = n (Na +) = n (OH -).
Ioni OH - - for non-stable; the stench will reactivate. Together with them, the styli (tobto 0.03 mol) of ions H + was reacted.
Role of a number of ions from the reaction:
n (H +) = n (H +) before the reaction - n (H +), but reacted = 0.04 mol - 0.03 mol = 0.01 mol;
n (Na +) = 0.03 mol; n (SO 4 2–) = 0.02 mol.
Because there is a lack of breeding, then
V zag. "V range of H 2 SO 4 + V range of NaOH" 200 ml + 300 ml = 500 ml = 0.5 l.
C (Na +) = n (Na +) / V zag. = 0.03 mol: 0.5 L = 0.06 mol / L;
C (SO 4 2-) = n (SO 4 2-) / V zag. = 0.02 mol: 0.5 L = 0.04 mol / L;
C (H +) = n (H +) / V zag. = 0.01 mol: 0.5 L = 0.02 mol / L;
pH = -lg C (H +) = -lg 2 · 10 -2 = 1.699.
Zavdannya number 9
Determine the pH and molar concentration of metal cations and anionic acid surplus in the range, so that you can determine the result of the reduction of strong acid in the range of the meadow (Table 9).
Table 9 - Umovi zavdannya No. 9
| Option No. | Option No. | Ob'єmi that warehouse for the distribution of acids and meadows | |
| 300 ml 0.1 M NaOH and 200 ml 0.2 M H 2 SO 4 | |||
| 2 l 0.05 M Ca (OH) 2 and 300 ml 0.2 M HNO 3 | 0.5 l 0.1 M KOH and 200 ml 0.25 M H 2 SO 4 | ||
| 700 ml 0.1 M KOH and 300 ml 0.1 M H 2 SO 4 | 1 l 0.05 M Ba (OH) 2 and 200 ml 0.8 M HCl | ||
| 80 ml 0.15M KOH and 20 ml 0.2M H 2 SO 4 | 400ml 0.05M NaOH and 600ml 0.02M H 2 SO 4 | ||
| 100 ml 0.1 M Ba (OH) 2 and 20 ml 0.5 M HCl | 250 ml 0.4M KOH and 250 ml 0.1M H 2 SO 4 | ||
| 700ml 0.05M NaOH and 300ml 0.1M H 2 SO 4 | 200ml 0.05M Ca (OH) 2 and 200ml 0.04M HCl | ||
| 50 ml 0.2M Ba (OH) 2 and 150 ml 0.1M HCl | 150ml 0.08M NaOH and 350ml 0.02M H 2 SO 4 | ||
| 900ml 0.01M KOH and 100ml 0.05M H 2 SO 4 | 600ml 0.01M Ca (OH) 2 and 150ml 0.12M HCl | ||
| 250 ml 0.1 M NaOH and 150 ml 0.1 M H 2 SO 4 | 100 ml 0.2M Ba (OH) 2 and 50 ml 1M HCl | ||
| 1 l 0.05 M Ca (OH) 2 and 500 ml 0.1 M HNO 3 | 100 ml 0.5M NaOH and 100 ml 0.4M H 2 SO 4 | ||
| 100 ml 1M NaOH and 1900 ml 0.1M H 2 SO 4 | 25 ml 0.1 M KOH and 75 ml 0.01 M H 2 SO 4 | ||
| 300 ml 0.1 M Ba (OH) 2 and 200 ml 0.2 M HCl | 100ml 0.02M Ba (OH) 2 and 150ml 0.04M HI | ||
| 200 ml 0.05M KOH and 50 ml 0.2M H 2 SO 4 | 1 l 0.01 M Ca (OH) 2 and 500 ml 0.05 M HNO 3 | ||
| 500ml 0.05M Ba (OH) 2 and 500ml 0.15M HI | 250ml 0.04M Ba (OH) 2 and 500ml 0.1M HCl | ||
| 1 l 0.1 M KOH and 2 l 0.05 M H 2 SO 4 | 500 ml 1M NaOH and 1500 ml 0.1M H 2 SO 4 | ||
| 250ml 0.4M Ba (OH) 2 and 250ml 0.4M HNO 3 | 200 ml 0.1 M Ba (OH) 2 and 300 ml 0.2 M HCl | ||
| 80 ml 0.05M KOH and 20 ml 0.2M H 2 SO 4 | 50 ml 0.2M KOH and 200 ml 0.05M H 2 SO 4 | ||
| 300 ml 0.25 M Ba (OH) 2 and 200 ml 0.3 M HCl | 1 l 0.03M Ca (OH) 2 and 500 ml 0.1M HNO 3 |
Hydrolysis of salts
When deciding in water, be it salt, the dissociation of the salt in the cation and anion is displayed. If the strength is fixed by a strong base cation and an anion of a weak acid (for example, potassium nitrite KNO 2), then the nitrite-ion will cling to the H + ions, all of which are contained in water molecules, as a result of which a weak acid is formed. As a result of the process of interaction in the context of the establishment of a rivnovaga:
NO 2 - + HOH ⇆ HNO 2 + OH -
KNO 2 + HOH ⇆ HNO 2 + KOH.
Such a rank, at the level of salt, which is hydrolyzed according to anion, is an excess of ions OH - (the reaction of the middle is lush; pH> 7).
If the strength is fixed with a weak base cation and an anion of a strong acid (for example, ammonium chloride NH 4 Cl), then the NH 4 + cation of a weak base can absorb OH - form of water molecules and accept weakly dissociating electrolitone - hydroxy 1 hydroxide.
NH 4 + + HOH ⇆ NH 4 OH + H +.
NH 4 Cl + HOH ⇆ NH 4 OH + HCl.
The solution of the salt hydrolizes by the cation there is an excess of ions H + (the reaction of the middle is acid pH< 7).
In the case of hydrolysis of a salt, established by a cation of a weak base and an anion of a weak acid (for example, ammonium fluoride NH 4 F), cations of a weak base NH 4 + bond with ions OH -, which are present as water molecules, and an acid F is weak with H + ions, for which there is a weak base NH 4 OH and a weak acid HF: 2
NH 4 + + F - + HOH ⇆ NH 4 OH + HF
NH 4 F + HOH ⇆ NH 4 OH + HF.
The reaction of the middle in the range of salt, which is going to go by and according to the cation, and according to anion of the age, who from low-energy electroliths start to work out as a result of the hydrolysis, is strong (the decision can be made In case of hydrolysis of NH 4 F, the middle food will be acidic (pH<7), поскольку HF – более сильный электролит, чем NH 4 OH: KNH 4 OH = 1,8·10 –5 < K H F = 6,6·10 –4 .
In such a rank, the hydrolysis (to be spread out with water) is supplied with salt, approved:
- a strong base cation and a weak acid anion (KNO 2, Na 2 CO 3, K 3 PO 4);
- a cation of a weak base and an anion of a strong acid (NH 4 NO 3, AlCl 3, ZnSO 4);
- a weak base cation and a weak acid anion (Mg (CH 3 COO) 2, NH 4 F).
With water molecules, the cation of weak bases and (i) anion of weak acids; salts are fixed with cations of strong bases and with anions of strong acids hydrolysis is not susceptible to hydrolysis.
Hydrolysis of salts, approved by high-charge cations and anions, is often counterproductive; Below, on specific butts, the last of the mirkuvan is shown, which it is recommended to pre-trim when folding the hydrolysis of such salts.
Notes
1. Yak already indicated earlier (div. Note 2 on side 5) is an alternative point of view, but with a strong base of hydroxide ammonia. Acid reaction of the middle in the range of ammonium salts, approved by strong acids, for example, NH 4 Cl, NH 4 NO 3, (NH 4) 2 SO 4, can be explained in such a way by a reverse process of ammonium dissociation NH 4 + ⇄ NH 3 + or, more precisely, NH 4 + + H 2 O ⇄ NH 3 + H 3 O +.
2. If ammonium hydroxide is used in a strong base, then in the range of ammonium salts, made up of weak acids, for example, NH 4 F should look like NH 4 + + F - ⇆ NH 3 + HF, in which ammonium molecules compete for and anions of weak acid.
Butt 8.1 To write down in the molecular and ionic-molecular view the specific reactions of hydrolysis of sodium carbonate. Set pH range (pH> 7, pH<7 или pH=7).
1. Equivalent dissociation of salts: Na 2 CO 3 ® 2Na + + CO 3 2–
2. Сіl is fixed with cations (Na +) of the strong base NaOH and anion (CO 3 2–) weak acid H2CO3. Otzhe, sіl guided by anіon:
CO 3 2– + HOH ⇆….
Hydrolysis at large vipadks opposite to the reverse (sign ⇄); for 1 ion, to take care of the fate of the process of hydrolysis, 1 molecule of HOH will be recorded .
3. Negatively charged ionic carbonate CO 3 2– is linked with positively charged ions H +, which are released from HOH molecules, and form hydrocarbonate ion HCO 3 -; the solution is filled with OH ions - (middle soil; pH> 7):
CO 3 2– + HOH ⇆ HCO 3 - + OH -.
The first stage of hydrolysis of Na2CO3.
4. The first stage of hydrolysis in the molecular view can be eliminated by removing all the evidence in the usual CO 3 2– + HOH ⇆ HCO 3 - + OH - anioni (CO 3 2–, HCO 3 - і OH -) with cations Na + , having approved salts Na 2 CO 3, NaHCO 3 and base NaOH:
Na 2 CO 3 + HOH ⇆ NaHCO 3 + NaOH.
5. As a result of hydrolysis at the first stage, hydrocarbonate was established, which takes part in the other stage of hydrolysis:
HCO 3 - + HOH ⇆ H 2 CO 3 + OH -
(negatively charged hydrocarbonate ions HCO 3 - to be linked with positively charged ions H + from volatile HOH molecules).
6. The adjustment of the other stage of hydrolysis in the molecular view can be eliminated by calling explicitly in the ordinary HCO 3 - + HOH ⇆ H 2 CO 3 + OH - anioni (HCO 3 - і OH -) with cations Na +, making 3 sіl NaHCO NaOH base:
NaHCO 3 + HOH ⇆ H 2 CO 3 + NaOH
CO 3 2– + HOH ⇆ HCO 3 - + OH - Na 2 CO 3 + HOH ⇆ NaHCO 3 + NaOH
HCO 3 - + HOH ⇆ H 2 CO 3 + OH - NaHCO 3 + HOH ⇆ H 2 CO 3 + NaOH.
Butt 8.2 Write down in the molecular and ionic-molecular view the specific reactions of hydrolysis of aluminum sulfate. Set pH range (pH> 7, pH<7 или pH=7).
1. Equivalent dissociation of salts: Al 2 (SO 4) 3 ® 2Al 3+ + 3SO 4 2–
2. Sile approved cations (Al 3+) of a weak base Al (OH) 3 and anions (SO 4 2–) of a strong acid H 2 SO 4. Otzhe, with the help of the cation; for 1 ion Al 3+ 1 HOH molecule will be recorded: Al 3+ + HOH ⇆….
3. Positively charged ions Al 3+ are linked with negatively charged ions OH -, which are added to the molecules of HOH, and form hydroxoaluminium AlOH 2+; growth with ions H + (sour; pH<7):
Al 3+ + HOH ⇆ AlOH 2+ + H +.
The first stage of hydrolysis of Al2 (SO4) 3.
4. The level of the first stage of hydrolysis in the molecular view can be eliminated by linking all manifestations to the level of Al 3+ + HOH ⇆ AlOH 2+ + H + cations (Al 3+, AlOH 2+ and H +) with anions SO 4 2– , having approved salts Al 2 (SO 4) 3 AlOHSO 4 and acid H 2 SO 4:
Al 2 (SO 4) 3 + 2HOH ⇆ 2AlOHSO 4 + H 2 SO 4.
5. As a result of hydrolysis at the first stage, cations of hydroxoaluminium AlOH 2+ were established, as they take part in the other stage of hydrolysis:
AlOH 2+ + HOH ⇆ Al (OH) 2 + + H +
(positively charged ions AlOH 2+ are linked with negatively charged ions OH - which are volatile from HOH molecules).
6. The adjustment of the other stage of hydrolysis in the molecular view can be eliminated by linking all apparent in the normal AlOH 2+ + HOH ⇆ Al (OH) 2 + + H + cations (AlOH 2+, Al (OH) 2 +, і H + ) with anions SO 4 2 - having made salts AlOHSO 4, (Al (OH) 2) 2 SO 4 і acid H 2 SO 4:
2AlOHSO 4 + 2HOH ⇆ (Al (OH) 2) 2 SO 4 + H 2 SO 4.
7. As a result of the other stage of hydrolysis, cations of dihydroxoaluminium Al (OH) 2 + were established, which take part in the third stage of hydrolysis:
Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H +
(positively charged ions Al (OH) 2 + are linked with negatively charged ions OH - which are volatile from HOH molecules).
8. The level of the third stage of hydrolysis in the molecular view can be eliminated by appearing in the form of Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H + cations (Al (OH) 2 + і H +) with anions SO 4 2–, having made the sil (Al (OH) 2) 2 SO 4 і acid H 2 SO 4:
(Al (OH) 2) 2 SO 4 + 2HOH ⇆ 2Al (OH) 3 + H 2 SO 4
As a result of the cycle of the world, there will be no more advances in the field of hydrolysis:
Al 3+ + HOH ⇆ AlOH 2+ + H + Al 2 (SO 4) 3 + 2HOH ⇆ 2AlOHSO 4 + H 2 SO 4
AlOH 2+ + HOH ⇆ Al (OH) 2 + + H + 2AlOHSO 4 + 2HOH ⇆ (Al (OH) 2) 2 SO 4 + H 2 SO 4
Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H + (Al (OH) 2) 2 SO 4 + 2HOH ⇆ 2Al (OH) 3 + H 2 SO 4.
Butt 8.3 Write down in the molecular and ionic-molecular view the specific reactions of ammonium orthophosphate hydrolysis. Set pH range (pH> 7, pH<7 или pH=7).
1. Equivalent dissociation of salts: (NH 4) 3 PO 4 ® 3NH 4 + + PO 4 3–
2. Sile approved cations (NH 4 +) weak base NH 4 OH ta anions
(PO 4 3–) weak acid H 3 PO 4. Otzhe, sіl gіdrolіzutsya і by cation, і by anіon : NH 4 + + PO 4 3– + HOH ⇆…; ( for one pair of ions NH 4 + and PO 4 3– in this vypadku enroll 1 molecule HOH ). Positively charged ions NH 4 + are linked with negatively charged ions OH -, which are absorbed by HOH molecules, and a weak base NH 4 OH is created, while negatively charged ions PO 4 3– are linked with ions H +, which are phosphoric
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2–.
The first stage of hydrolysis (NH 4) PO 4.
4. The first stage of hydrolysis in the molecular view can be eliminated by appearing in the original NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2– anioni (PO 4 3–, HPO 4 2–) s cations NH 4 + having made salts (NH 4) 3 PO 4, (NH 4) 2 HPO 4:
(NH 4) 3 PO 4 + HOH ⇆ NH 4 OH + (NH 4) 2 HPO 4.
5. Anioni hydrogen phosphate HPO 4 2– was established in the results of hydrolysis at the first stage, as well as NH 4 + cations take part in the other stage of hydrolysis:
NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 -
(The ions NH 4 + bond with the ions OH -, the ions HPO 4 2– - with the ions H +, which are absorbing from the HOH molecules, and the weak base NH 4 OH and the dihydrogen phosphate of the ions H 2 PO 4 -).
6. Even the other stage of hydrolysis in the molecular view can be eliminated by appearing in the original NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 - anioni (HPO 4 2– and H 2 PO 4 - ) with cations NH 4 + having made salts (NH 4) 2 HPO 4 і NH 4 H 2 PO 4:
(NH 4) 2 HPO 4 + HOH ⇆ NH 4 OH + NH 4 H 2 PO 4.
7. As a result of another stage of hydrolysis, the dihydrolysis of anioni H 2 PO 4 - was established, and together with cations NH 4 + take part in the third stage of hydrolysis:
NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4
(The NH 4 + ions bond with the OH - ions, the H 2 PO 4 - - ions with the H + ions, which are absorbed from the HOH molecules, and form the weak electrolytes NH 4 OH and H 3 PO 4).
8. The level of the third stage of hydrolysis in the molecular view can be eliminated by the presence of NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4 anion H 2 PO 4 - and cations NH 4 + having approved the sil NH 4 H 2 PO 4:
NH 4 H 2 PO 4 + HOH ⇆ NH 4 OH + H 3 PO 4.
As a result of the cycle of the world, there will be no more advances in the field of hydrolysis:
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2– (NH 4) 3 PO 4 + HOH ⇆ NH 4 OH + (NH 4) 2 HPO 4
NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 - (NH 4) 2 HPO 4 + HOH ⇆ NH 4 OH + NH 4 H 2 PO 4
NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4 NH 4 H 2 PO 4 + HOH ⇆ NH 4 OH + H 3 PO 4.
The process of hydrolysis against the first stage is important, so the reaction of the middle stage in the range of salt, which is going on and on the cation, and according to the announcement of the date, starts a person who is from a small number of early stages of electrolysis. U vipadku
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2–
the reaction of the middle will be normal (pH> 7), fragments of ion HPO 4 2– - weak electrolyte, low NH 4 OH: KNH 4 OH = 1.8 · 10 –5> KHPO 4 2– = K III H 3 PO 4 = 1 , 3 × 10 –12 (dissociation of ion HPO 4 2– - dissociation of H 3 PO 4 at the third stage, thus KHPO 4 2– = K III H 3 PO 4).
Zavdannya number 10
Record the molecular and ionic-molecular view of the specific reactions of salt hydrolysis (Table 10). Set pH range (pH> 7, pH<7 или pH=7).
Table 10 - Umovi zavdannya No. 10
| Option No. | List of salts | Option No. | List of salts |
| a) Na 2 CO 3 b) Al 2 (SO 4) 3 c) (NH 4) 3 PO 4 | a) Al (NO 3) 3, b) Na 2 SeO 3, c) (NH 4) 2 Te | ||
| a) Na 3 PO 4 b) CuCl 2 c) Al (CH 3 COO) 3 | a) MgSO 4, b) Na 3 PO 4, c) (NH 4) 2 CO 3 | ||
| a) ZnSO 4 b) K 2 CO 3 c) (NH 4) 2 S | a) CrCl 3 b) Na 2 SiO 3 c) Ni (CH 3 COO) 2 | ||
| a) Cr (NO 3) 3, b) Na 2 S, c) (NH 4) 2 Se | a) Fe 2 (SO 4) 3, b) K 2 S, c) (NH 4) 2 SO 3 |
Table extension 10
| Option No. | List of salts | Option No. | List of salts |
| a) Fe (NO 3) 3 b) Na 2 SO 3 c) Mg (NO 2) 2 | |||
| a) K 2 CO 3 b) Cr 2 (SO 4) 3 c) Be (NO 2) 2 | a) MgSO 4 b) K 3 PO 4 c) Cr (CH 3 COO) 3 | ||
| a) K 3 PO 4 b) MgCl 2 c) Fe (CH 3 COO) 3 | a) CrCl 3 b) Na 2 SO 3 c) Fe (CH 3 COO) 3 | ||
| a) ZnCl 2 b) K 2 SiO 3 c) Cr (CH 3 COO) 3 | a) Fe 2 (SO 4) 3 b) K 2 S c) Mg (CH 3 COO) 2 | ||
| a) AlCl 3 b) Na 2 Se, c) Mg (CH 3 COO) 2 | a) Fe (NO 3) 3, b) Na 2 SiO 3, (NH 4) 2 CO 3 | ||
| a) FeCl 3 b) K 2 SO 3 c) Zn (NO 2) 2 | a) K 2 CO 3 b) Al (NO 3) 3 c) Ni (NO 2) 2 | ||
| a) CuSO 4, b) Na 3 AsO 4, c) (NH 4) 2 SeO 3 | a) K 3 PO 4 b) Mg (NO 3) 2 c) (NH 4) 2 SeO 3 | ||
| a) BeSO 4 b) K 3 PO 4 c) Ni (NO 2) 2 | a) ZnCl 2, Na 3 PO 4, c) Ni (CH 3 COO) 2 | ||
| a) Bi (NO 3) 3 b) K 2 CO 3 c) (NH 4) 2 S | a) AlCl 3 b) K 2 CO 3 c) (NH 4) 2 SO 3 | ||
| a) Na 2 CO 3 b) AlCl 3 c) (NH 4) 3 PO 4 | a) FeCl 3, b) Na 2 S, c) (NH 4) 2 Te | ||
| a) K 3 PO 4 b) MgCl 2 c) Al (CH 3 COO) 3 | a) CuSO 4, b) Na 3 PO 4, c) (NH 4) 2 Se | ||
| a) ZnSO 4 b) Na 3 AsO 4 c) Mg (NO 2) 2 | a) BeSO 4, b) b) Na 2 SeO 3, c) (NH 4) 3 PO 4 | ||
| a) Cr (NO 3) 3 b) K 2 SO 3 c) (NH 4) 2 SO 3 | a) BiCl 3 b) K 2 SO 3 c) Al (CH 3 COO) 3 | ||
| a) Al (NO 3) 3, b) Na 2 Se, c) (NH 4) 2 CO 3 | a) Fe (NO 3) 2, b) Na 3 AsO 4, c) (NH 4) 2 S |
List of Literature
1. Lur'є, Yu.Yu. Dovidnik of analytical chemistry / Yu. Lur'є. - M .: Khimiya, 1989 .-- 448 p.
2. Rabinovich, V.A. A short him_chniy dovidnik / V.A. Rabinovich, Z. Ya. Khavin - L .: Khimiya, 1991 .-- 432 p.
3. Glinka, N.L. Zagalna chemistry / N.L. Glinka; for ed. V.A. Rabinovich. - 26th view. - L .: Khimiya, 1987 .-- 704 p.
4. Glinka, N.L. The head of state is right from the background chemistry: a master book for universities / N.L. Glinka; for ed. V.A.Rabinovich and H.M. Ruby - 22nd species. - L .: Khimiya, 1984 .-- 264 p.
5. General and non-organic chemistry: lecture notes for students of technological specialties: about 2 years. / Mogiliv State University of Food; author-ordered V.A. Ogorodnik_v. - Mogilov, 2002. - Part 1: Homes of food chemistry. - 96 p.
First seen
ZAGALNA CHIMIA
Methodical instructions and control tests
for students of technological specialties by correspondence form of education
Teacher: Ogorodnikov Valeriy Anatoliyovich
Editor T.L. Mateusz
Technical editor O.O. Shcherbakova
Signed up to the friend. Format 60'84 1/16
Druk is offset. Garnitura Times. Druck stencil
Mind. p_ch. arch. Uch. view. l. 3.
Circulation ekz. Substitution.
Supervised on the risograph of the editorial-viddilu
mortgage
"Mogiliv State University of Food"
1.Razrakhunok pH in the range of strong acids and bases.
The pH value in the range of strong monobasic acids and bases is carried out according to the formulas:
pH = - log C to і pH = 14 + log C
De C to, C about molar concentration of acid and abo base, mol / l
2.Reduce the pH in the range of weak acids and bases
Use the following formulas: pH = 1/2 (pK to - lgC to) і pH = 14 - 1/2 (pK - lg C O)
3.Razrakhunok pH in the range of hydrolysing salts
Razrіznyayut 3 types of salt hydrolysis:
a) hydrolysis of salt according to anion (the salt is fixed in a weak acid and strong base, for example, CH 3 COO Na). The value of pH should be adjusted according to the formula: pH = 7 + 1/2 pK to + 1/2 lg C
b) hydrolysis of salt by cation (the salt is fixed with a weak base and a strong acid, for example NH 4 Cl).
c) hydrolysis of salt by cation and anion (with a weak acid and weak base, for example CH 3 COO NH 4). In most cases, the pH value is reduced by the following formula:
pH = 7 + 1/2 pK to - 1/2 pK o
If the oil is taken as a weak, richly basic acid or weakly rich in protic base, then in the formula (7-9), the pH value is given pK to і pK about the residual dissociation gap
4.Razrakhunok pH in the range of small sums of acids and bases
When poured acid and base, the pH is removed from the sum of the amount of taken acid and base and strength.
4.Buffer systems
To the buffer systems, sums must be carried:
a) weak acidity and salts, for example CH 3 COO H + CH 3 COO Na
b) weak base and її salts, for example NH 4 OH + NH 4 Cl
c) the sum of acidic salts of acidity, for example NaH 2 PO 4 + Na 2 HPO 4
d) sum of acidic and middle salts, for example NaHCO 3 + Na 2 CO 3
e) the sum of basic salts of different basicity, for example, Al (OH) 2 Cl + Al (OH) Cl 2, etc.
The pH value in buffer systems follows the formulas: pH = pK to - log C to / C і pH = 14 - pK o + log C pro / C s
Acid-base buffer solutions, Henderson-Haselbach family. Zagalny characteristic. The principle of dії. Rozrahunok pH buffered roschin. Buffer Umnist.
Buffer solutions - Systems that adapt to the same value of any parameter (pH, potential of the system, etc.) when changing the warehouse of the system.
Acid-base is called a buffer solution , so that the pH value is approximately constant when not added to large quantities of strong acid or strong base, as well as when diluted and concentrated. Acid-base buffers to replace weak acids and bases bound to them. A strong acid, when added to a buffer, “transforms” to a weak acid, and a strong base to a weak base. Formula for pH buffering solution: pH = pK about + lg C about /WITH s Tse rivnyannya Henderson - Hasselbach ... The third level of vaping, the pH of the buffer should be based on the ratio of the concentration of weak acid and the base bound with it. Oscillations during the breeding process do not change, then the pH value will become permanent. Breeding can not be bezidezhnaya. If the pH value is too high, the range will change, and the concentration of the components will probably become malimi, but it will not be possible to use the auto-test of water, but, in another way, the efficiency of the uncharged part of the charge or the charge
Buffer development of a constant pH value at the addition of a lack of small amounts of a strong acid and a strong base. The value of the buffer depends on the change in pH due to the ratio of the concentration of a weak acid and the base bound to it, as well as from the total concentration - to be characterized by a buffer mn_styu.
Buffer Umnist - the value of an infinitely small increase in the concentration of a strong acid and a strong base in the range (without a change in volume) in a cyclonic cyme with an increase in pH (line 239, 7.79)
In a strongly acidic and strong meadow middle ground, the buffer єmn_st is significantly increased. Razchini, in which to reach the top of the concentration of a strong acid and a strong base, also may buffer power.
Buffer mnіst is maximum at pH = RK. To adjust a certain pH value, there is such a buffer gap, for which the pKy value goes up to the weak acid store and is closer to the pH value. The buffer solution can be used to adjust the pH values to be in the pKa + _ 1 interval. Such an interval is called the working power of the buffer.
19. The main understanding, tied with complex spokes. Classification of complex spoluk. Konstanty Rivnovagi, scho vikoristovuyutsya for the har-ki complex spoluks: constants of education, constants of dissociation (zagalny, steps, thermodynamics, real and clever concentration)
Most often, a complex is called a particle, established as a result of the donor-acceptor interaction of the central atom (ion), called a complex, and charging of neutral particles, called ligands. Complex solutions and ligands are guilty of self-reliance in the middle, de-development of complex solutions.
It is complex to store from the inner and outer spheres. К3 (Fe (CN) 6) - К3-outer sphere, Fe-complex-solution, CN-ligand, complex-solution + ligand = inner sphere.
Dentistry is the number of donor centers in a ligand, but in a donor-acceptor interaction when a complex part is approved. Ligands are monodentate (Cl-, H2O, NH3), bidentate (C2O4 (2-), 1,10-phenanthroline) and polydentate.
Coordinate number is the number of donor centers in ligands, of which the central atom is interconnected. The designated vische has butt: 6-coordination number. (Ag (NH3) 2) + -coordinate number 2, so as amiak monodentate ligand, and (Ag (S2O3) 2) 3- - coordination number 4, so as thiosulfate ion-bidentate ligand.
Classification.
1) Relevant to its charge: anion ((Fe (CN) 6) 3-), cation ((Zn (NH3) 4) 2 +) і uncharged or non-electrolyte complex (HgCl2).
2) Regarding the number of atoms in the metal: mononuclear and polynuclear complexes. Before the storage of a mononuclear complex there is one metal atom, and before the storage of a polynuclear one, two or more. Polynuclear complex particles, which replace the same atoms with a metal, are called homonuclear (Fe2 (OH) 2) 4+ or Be3 (OH) 3) 3+), and the atoms of other metals are called heteronuclear (Zr2Al (OH) 5) 6+ ).
3) Regularly from the har-ra of ligands: single-family and multi-ligand (zmishanoligand) complexes.
Chelati-cyclic complex spoluks of ions of metals with polydentate ligands (call them organic), in which the central atom should be included in the warehouse of one or even deciduous cycles.
Constant... The technique of a complex ion is characterized by its dissociation constant, as it is called the constant of instability.
As a matter of fact, the clues about the part of the constant of non-performance in the daytime, to the backward constant of the non-performance of the complex ion:
Significant constant of non-stability of the road in addition to the number of stage constant of non-stability.
In the analytical chemistry, it will remain for an hour to replace the stiffness constants of the complex ion: 
The constant of stiffness is carried before the process of establishing a complex іon and the costly value of the constant of non-efficiency: Kusch = 1 / Knest.
The constant of stiffness characterizes the equilibrium of the complex.
Thermodynamic and concentration of constant div. side 313.
20. Infusion of new factors on the process of complexation and efficiency of complex processes. Concentration injections react to complex solutions. The growth of molar parts of strong metal ions and complexes at the most important sum.
1) The strength of complex spoluks to lie in the nature of the complex and ligands. The laws of the change in the stability of bagatoch complexes of metals with the help of other ligands can be explained with additional help. Theory of hard and mild acids and bases (ZhMKO): mild acids are used for more stiff bases, and hard for hard ones. Ligands (l. Bases), and Ag + or Hg2 + (m. C-ty) s S-sod Ligands (m. Basic) Complexes of metal cations with polydentate ligands yavl.
2) Ionna strength. At the same time, the increase in the efficiency of the complex will change.
3) temperature. When the complex is illuminated, the delta N is greater than 0, then when the temperature is adjusted, the complex's strength will increase, if the delta H is less than 0, then it will change.
4) secondary r-tsii. The injection of pH on the strength of the complex is due to the nature of the ligand and the central atom. As soon as the base of the ligand complex enters the warehouse, the base is strong, then with a decrease in pH, the protonation of such ligands and a decrease in the molar part of the ligand will occur, so that you take part in the established complex. The injected pH will be stronger, the greater the strength given to the base and the lesser the strength of the complex.
5) concentration. With an increase in concentration, the ligand grows in the place of complexes due to the great coordination number and the concentration of strong ions in metal decreases. With an excess of ionic metal in the solution, the monoligand complex dominates.
The molar part of the ions in the metal, not tied at the complex
Molar part of complex particles
The choice of salts, how to hydrate, to be found in medical practice. So, if acid breaks down on a range of acids, the slides are churned with water, and then with sodium carbonate Na 2 CO 3. This method allows the neutralization of acid surpluses and residues of water solution of Na 2 CO 3 to a small reaction. However, the solution of Na 2 CO 3 is unlikely to be used for lowering the acidity of the slurry juice through the addition of a high puddle. For cich tsіley zastosovuyut the difference in sodium hydrocarbonate NaHCO 3 yakі is characterized by the lowest pH values. At the same time, for an accurate test of the preparation, which is based on hydrolysis, it is necessary to estimate the pH value of the salts in order to hydrolyze.
1. In the range of salt types NH 4 Cl:
de , pC,- Negative tens of logarithms of similar values.
Oskilki at t 0 = 20-25 0 C = 14, then, even:
2. In the range of salt types CH 3 COONa:
3. In the range of salt types NH 4 CN:
At the time of equilibrium = the fraction of the formula will turn into zero and pH = 7.
As soon as it is hydrolized in a number of steps, then it is possible to increase the value of the pH value in the range of salt, which is deprived of the first step of hydrolysis.
Etaloni virіshennya zavdan
1. Calculate the constant і steps for the hydrolysis of salt NH 4 Cl in the range of (NH 4 Cl) = 0.1 mol / l, where (NH 3 × H 2 O) = 1.8 × 10 - 5.
NH 4 Cl + H-OH ⇄ NH 3 ∙ H 2 O + HCl2. Calculate the constant і steps of the hydrolysis of Na 2 CO 3 at the first step in the range (Na 2 CO 3) = 0.01 mol / l, for H 2 CO 3 = 4 × 10 - 7; = 5 × 10-11.
Hydrolysis of Na 2 CO 3 is often used:
Na 2 CO 3 + H-OH ⇄ NaHCO 3 + NaOH (1 step)
The speedy viglyad has a viglyad like this:
CO 3 2 - + H-OH ⇄ HCO 3 - + ВІН -NaHCO 3 + H-OH ⇄ H 2 CO 3 + NaOH (2 steps)
HCO 3 - + H-OH ⇄ H 2 CO 3 + ВІН -Hydrolysis of Na 2 CO 3 in the first step shall be brought up to the approval of hydrocarbonate-ion HCO 3 - which is a weak electrolyte:
HCO 3 - ⇄ H + + CO 3 2 -
The given definition of dissociation of H 2 CO 3 of another stage and is characterized by a constant (H 2 CO 3) = 5 × 10 - 11.
3. Measure the steps of NaNO 2 hydrolysis at the range of salts with concentrations of 0.1 and 0.001 mol / l, where (HNO 2) = 4 × 10 - 4.
Introduced value: 1 = 0.1 mol / l; З 2 = 0.001 mol / l.
Todi: 
; 
.
Rozdilimo one viraz on іnshe and і іtrimaєmo:
NaCN + H-OH ⇄ HCN + NaOH
pH> 7 The middle is weak.
6. Know the difference in the pH value of the Na 2 S and NaHS solutions with the same salt concentrations, when (H 2 S) = 7, (H 2 S) = 13.
From the first ryvnyannya other and otrimaєmo:
Power supply for self-control
1. What process is called salt hydrolysis?
2. What is the reason for the change in pH in the range of hydrolysis?
2. What type of salts are you aware of hydrolysis in the market? Hover butt.
3. Why does the salt type NaCl, KI, CaCl 2 not know the hydrolysis?
4. At any rate for the hydrolysis of salts, acid (basic) salts are established? Hover butt.
5. Do any types of drops have non-collar hydrolysis of salt? Hover butt.
6. What products are established by the interaction of chromium (III) chloride and ammonium sulfide (NH 4) 2 S in the water solution?
7. What is called the constant of hydrolysis? What kind of bureaucrats do they have and which ones do not have a hydrolysis constant?
8. What is called a hydrolysis step? How is it tied to the constant of hydrolysis of different types of salts?
9. Do you pour in the factors for the step size for the hydrolysis of the salt?
10. Why should hydrolysis be increased when the temperature is increased?
11. For any kind of dilution salts, is it practically impossible to pour hydrolysis onto the steps?
12. In a bright way, it is possible to hydrolyze FeCl 3 until Fe (OH) 3 is approved?
13. With hydrolysis of any salts, the pH is close to 7?
14. Why is the range of NaHCO 3 a weak reaction, and the range of NaHSO 3 is weakly acidic? (H2CO3) = 4 × 10-7, (H2SO3) = 1.7 × 10-2.
15. It is necessary to prepare the range of FeSO 4 salt during hydrolysis, which will establish a low-rooted spoluca (turbidity). Do yakuy middle (sour abo lousy) have gotuvati razchin, what will uniknuti yogo dimness? For what?
Options for independent revision
Option number 1
1. Write an equal hydrolysis (in molecular and ionic view) and in the value of the reaction of the middle range of water solutions in perehovannyh salts: Na 2 SO 4, FeCl 2, Na 2 S.
3. Calculate the pH value of the range of CH 3 COOK s C (CH 3 COOK) = 0.005 mol / l, where (CH 3 COOH) = 1.8 × 10 - 5.
Option number 2
1. Write an equal hydrolysis (in molecular and ionic view) and the value of the reaction of the middle range of water solutions in perehovannyh salts: MnSO 4, KI, Na 2 SiO 3.
3. Calculate the pH value of the range NaNO 2 C (NaNO 2) = 0.01 mol / l, where (HNO 2) = 4 × 10 - 4.
4. Ratio of the values of the Pb (NO 3) 2 hydrolysis constants for the first and the other step, for Pb (OH) 2 = 9.6 × 10 - 4; = 3 × 10-8.
Option number 3
1. Write an adequate hydrolysis (in molecular and ionic viglyad) and the significance of the reaction of the middle range of water solutions in perehovannyh salts: Ca (NO 3) 2, Na 2 SO 3, Cu (NO 3) 2.
2. Calculate the constant of the steps for the hydrolysis of KClO in the range of C (KClO) = 0.1 mol / l, where (HClO) = 5.6 × 10 - 8.
3. Calculate the pH value of the salt KCN s C (KCN) = 0.05 mol / l, where (HCN) = 8 × 10 - 10.
Option number 4
1. Write an adequate hydrolysis (in molecular and ionic view) and the value of the reaction of the middle range of water solutions in perehovannyh salts: K 3 PO 4, CaCl 2, ZnCl 2.
2. Measure the steps of NaCN hydrolysis at the range of molar concentration of the salt equivalent of 0.1 and 0.001 mol / l, but (HCN) = 8 × 10 - 10.
3. Calculate the pH value of the range NH 4 NO 3 3 C (NH 4 NO 3) = 0.1 mol / l, if (NH 3 × H 2 O) = 1.8 × 10 - 5.
Option No. 5
1. Write a specific hydrolysis (in molecular and ionic viglyad) and in the value of the reaction of the middle range of water solutions in perekhovannyh salts: CuSO 4, Li 2 S, NaBr.
3. Calculate the pH value of the range of NH 4 I with a salt concentration of 0.02 mol / l, where (NH 3 × H 2 O) = 1.8 × 10 - 5.
4. Ratio of the values of the hydrolysis constants for Na 2 SiO 3 for the first and the other stage, for H 2 SiO 3 = 1.3 × 10 - 10; = 2 × 10-12.
Option number 6
1. Write an equal amount of hydrolysis (in molecular and ionic view) and the significance of the reaction of the middle range of water solutions in perehovannyh salts: SrCl 2, Fe (NO 3) 3, K 2 S.
2. Determination of the magnitude of the step of NaF hydrolysis at the range of molar concentration of the salt equivalent of 0.2 and 0.002 mol / l. (HF) = 6.6 10 - 4.
3. Calculate the pH value of the range of HCOOH with a molar salt concentration of 0.05 mol / l, where (HCOOH) = 2.2 × 10 - 4.
Option number 7
1. Write an adequate hydrolysis (in molecular and ionic view) and the importance of the reaction of the middle range of water solutions in perehovannyh salts: NaNO 3, ZnSO 4, Ca (OCl) 2.
3. Calculate the pH value of the range C 6 H 5 COONa with a salt concentration of 0.01 mol / l, where (C 6 H 5 COOH) = 6.3 × 10 - 5.
Option number 8
1. Write an adequate hydrolysis (in molecular and ionic view) and in the value of the reaction of the middle range of water solutions in perekhovannyh salts: Pb (NO 3) 2, CaS, KC1.
2. Ratio of the values of the constants of the steps in the hydrolysis of NaF and NaCN salts at the same concentrations, when (HF) = 6.6 × 10 - 4; (HCN) = 8 × 10 - 10.
3. Calculate the pH value of the range CH 3 COONH 4 with a molar salt concentration of 0.05 mol / l, where (CH 3 COOH) = 1.8 × 10 - 5; (NH 3 × H 2 O) = 1.8 × 10-5.
Option number 9
1. Write an adequate hydrolysis (in molecular and ionic viglyad) and in the main reaction of the middle range of water solutions in perehovannyh salts: Ba (NO 3) 2, NiCl 2, K 2 SO 3.
3. Calculate the pH value of the range of salt KF from a concentration of 0.001 mol / l, when (HF) = 6.6 × 10 - 4.
Option number 10
1. Write an adequate hydrolysis (in molecular and ionic viglyad) and the value of the reaction of the middle range of water solutions in perehovannyh salts: CoSO 4, Na 2 C 2 O 4, Sr (NO 3) 2.
2. Ratios of the values of the constants and steps in the NH 4 F hydrolysis at the ratios with concentrations of 0.02 mol / L and 0.002 mol / L, when (HF) = 6.6 × 10 - 4, (NH 3 × H 2 O) = 1 , 8 × 10-5.
3. Calculate the pH value of the range NH 4 CN with a concentration of 0.01 mol / l, when (HCN) = 8 × 10 - 10 (NH 3 × H 2 O) = 1.8 × 10 - 5.
4. Ratio of the value of the constants for the hydrolysis of Na 2 S for the first and the other step, when (H 2 S) = 1 × 10 - 7; (H 2 S) = 1 × 10 - 13.
Option number 11
1. Write an equal hydrolysis (in molecular and ionic viglyad) and the value of the reaction of the middle range of water solutions in perehovannyh salts: BaS, K 2 SO 4, CrCl 3.
2. Calculate the constant of the steps of HCOONa hydrolysis at the range with a molar concentration of a salt of 0.001 mol / l, where (HCOOH) = 2.2 × 10 - 4.
3. Calculate the pH value of the range of NH 4 F with a concentration of 0.02 mol / l, when (NH 3 × H 2 O) = 1.8 × 10 - 5 (HF) = 6.6 × 10 - 4.
Option number 12
1. Write a specific hydrolysis (in molecular and ionic view) and the importance of the reaction of the middle range of water solutions in perehovannyh salts: Ni (NO 3) 2, K 2 CO 3, BaCl 2.
2. Ratio of the value of the constants of the steps in the hydrolysis of NH 4 NO 3 at ranges with salt concentrations of 0.02 and 0.002 mol / l, where (NH 3 × H 2 O) = 1.8 × 10 - 5.
3. Calculate the pH value of the KClO range from a salt concentration of 0.04 mol / l, where (HClO) = 5.6 × 10 - 8.
Option No. 13
1. Write a specific hydrolysis (in molecular and ionic view) and the importance of the reaction of the middle range of aqueous solutions in perehovannyh salts: NaI, K 2 SiO 3, Fe 2 (SO 4) 3.
2. Calculate the constant і of the hydrolysis steps of C 2 H 5 COONa in the range of (C 2 H 5 COONa) = 0, l mol / l, where (C 2 H 5 COOH) = 1.3 × 10 - 5.
3. Calculate the pH value of the range of NaHCO 3 with a concentration of 0.1 mol / l, when (H 2 3) = 4 × 10 - 7, (H 2 CO 3) = 5 × 10 - 11.
Option number 14
1. Write an adequate hydrolysis (in molecular and ionic viglyad) and the importance of the reaction of the middle range of water solutions in perehovannyh salts: Na 2 HPO 4, KNO 3, Bi (NO 3) 3.
2. Count the steps of NH 4 F hydrolysis in the range (NH 4 F) = 0.02 mol / l, where (HF) = 6.6 × 10 - 4, (NH 3 × H 2 O) = 1.8 × 10-5.
Hydrolysis of salts is a reaction of interaction of ions of salt with water, as a result of which weak electricity is established. The difference between neutrality - soli - nabuv in its own, either sour or luscious, reaction. As a result, salts at their own heart are set in the result of neutralization reactions, with the interaction of acids and bases. Of the three types of salts, hydrolysis can be reduced to three types, which can be established during the interaction:
1) weak acid and strong base;
2) strong acid and weak base;
3) weak acid and weak base.
The fourth type of salt, which is established by the interaction of a strong base and strong acid, for example NaOH and HCI, for the reaction
NaOH + HCI = NaCl + H2O
Hydrolysis is not susceptible to hydrolysis, as well as NaCl is also a strong electrolyte and the molecules are dissociated to hydration (to be cooled by water molecules); When the level of equalization is 2H 2 O ↔ N Z O + + VIN - it does not break down; therefore, the hydrolysis does not fail, the solution becomes neutral. The pH range of such a range is 7.
Apply hydrolysis to the skin type of okremo salts.
1. If the silt is set up with a weak acidic acid CH 3 COOH and a strong base NaOH, for example sodium acetate CH 3 COONa, then an equal amount of hydrolysis can be written as follows:
● in molecular form
СН З СООNa + Н 2 О CH 3 CCOH + NaOHl; (2.8 a)
● in the form
СН З СОО - + Na + + H 2 0 CH 3 COOH + Na + + ВІН -; (2.8 b)
● in the quickest form
СН З СОО - + Н 2 0 CH 3 COOH + + ВІН -. (2.8 st)
Yak can be seen from the induction, when CH 3 COONa is hydrolyzed through the reaction of acetate ions in water and in weak otsic acid in the range of concentration, they accumulate in BIN - and the pH range will be greater than 7.
The constant of the equal reaction (2.8.c) is written in the viewer:
. (2.9)
Accepting the concentration of water and the value of the constant, which combined with the constant K s, we can take the viraz for the constant hydrolysis:
. (2.10)
Viraziv through іonny dobutok Vodi, maєmo
. (2.11)
So yak in the rest of the way 
is the value, the rotary constant of the dissociation of ocetic acid 
,
viraz for a constant hydrolysis of salt, established by a weak acid and a strong base (2.10), is written in the offensive rank:
Yak is obviously from the last formula, less acid is weak, tobto. In less of a dissociation constant, in a larger world, it is strong to hydrolysis.
In addition, the process of hydrolysis can be characterized also by the step of hydrolysis "h", which is the ratio of the number of salt molecules, which is known to hydrolysis, to the cob number of molecules. The concentration of this part of the salt, which was fed to the hydrolysis numerically to the same concentration of the ions of the VIN - in the range, yak, at its own thief, as a matter of fact (2.8c), rivnu the concentration of acid, is available.
[CH 3 COOH] = [ВІН -] = h ∙ С,
de С is the primary concentration of СН 3 СООNa, g-mol / l. Concentration of acetate-ions [СН 3 СОО -]
[CH 3 COO -] = C - h ∙ C = C ∙ (1-h).
With the urahuvannya of the entered value of h, we can accept the viraz, which is the constant of the hydrolysis step:
. (2.13)
With the value of h, the denominator of the last viraz can be zehtuvati, and even formula (2.13) can be written as follows:
stars. (2.15)
The stage of hydrolysis is more demanding, which is more varied, as well as the temperature, which is the same as the temperature of growth K W. The addition of the іonіv ВІН -, based on the principle of substitution of Le Chatelier, will lead to the process of hydrolysis.
If the oil is fixed with a basic acidic acid, then hydrolysis is opposite to the first step. So, for example, the level of hydrolysis of sodium Na 2 CO 3 can be written as follows:
CO 3 2- + H 2 O ↔ HCO 3 - + OH -
і the constant of hydrolysis is the value of the constant of dissociation of high acidity at the first step:
H 2 CO 3 ↔ H ++ HCO 3 -
For rejection of the formula for the development of pH levels, which is established as a result of hydrolysis, re-conversion of viraz (2.10), for which it is acceptable, the value of the concentration of acetate-ions through two small steps from the concentration of hydrochloric acid is practically 2.16.
tobto. concentration of ions in hydroxyl [OH -], which was established as a result of hydrolysis, before [OH -] = C. (2.17)
As soon as possible with the operator p≡ -lg, tsei viraz will write yak
pOH = -lg =, (2.18)
abo, vrahoyuchi virazi (2.7. and 2.12)
pH = 14 - = 7 + 
. (2.19)
2. Iaksho sil is made with a strong acid and a weak base,
NH 4 0H + HCl, = NH 4 Cl + H 2 Pro,
then the ryvnyannya gidrolizu will be written like this:
● in molecular form
N H 4 Cl + H 2 0 = NH 4 0 H + HCl; (2.20 a)
● in the form
NH + 4 + Cl -, + 2H 2 0 = NH 4 0H + H 3 0 + + Cl -; (2.20 b)
● in the quickest form
NH + 4 + 2H 2 0 = NH 4 0H + H 3 0 +. (2.20 in)
Constant for hydrolysis in the whole vipadk of maє viglyad
.(2.21)
If you multiply the number and the standard of the rivnyannya by [OH -], then viraz for K G nabude viglyadu
. (2.22)
At once diluted, it is possible to accept the solution, the concentration of the hydrolyzed part of the salt, which is used [H 3 0 +], the additional concentration of the base, tobto. =, And the concentration of ions in the road salt concentration (C). Todi 
(2.23)
Otzhe, the concentration of ions in hydroxon, which was established during hydrolysis,
= . (2.24)
Having shrunk by the value of p = - lg; otrimaєmo
pH = = 7 - 
. (2.25)
Hydrolysis step
. (2.26)
Otzhe, the base is weak (less), then the greater the concentration of ions in the rose, tobto. more against salt hydrolysis, established with a strong acid and weak base. If they are adjusted, they can be weakened because they will help the process of hydrolysis, so it’s worthwhile for the children (2.20 c.) To be able to get along with it.
3. Hydrolysis of salt, approved by a weak base and a weak acid, for example, ammonium acetate СН З СООНН 4 according to the scheme
CH 3 COONH 4 + H 2 O ↔ CH 3 COOH + NH 4 OH,
the opposite will increase.
Hydrolysis constant
. (2.27)
The pH of such a range must be found only in terms of the values of the dissociation constants of the acid and the base, and not in terms of the concentration of salt:
= (2.28)
і 
. (2.29)
Such a rank, salt is given to hydrolysis, as a result of which a weak electrolyte is established, which does not occur well.
2.4. Buffer solutions
Buffers are water solutions of electrolytes, as they save practically unchanged pH values when diluted or supplied with small amounts of acid in a meadow. Buffer solutions є a sum of either a weak acid and a salt, set up with an acidic and a strong base, or a weak base and a strong base, set up with a strong base and a strong acid.
Yaksho, for example. add to the range of weak oztic acid CH 3 COOH sil, if you want to take the same anion (for example, sodium acetate CH 3 COONa), then, according to the principle of Le Chatel, the equal process of acid dissociation
CH 3 СOOH ↔ СН 3 СОО - + Н + (2.30)
As soon as it’s pushed, it’s practical to strangle the process of acid dissociation and the steps of dissociation α to zero (α = 0).
But it will be dissociated nationally for good reason
СН З СООNa ↔ СН 3 СОО - + Na + (2.31)
In general, the concentration of non-dissociated molecules of acid and high concentration of acid C acid in the sum of acids and salts, and the concentration of acetate ions CH 3 COO - is the concentration of salt C of the salt.
How to set the q values for the viraz for the acid dissociation constant
, (2.32)
then the concentration of ions [N Z O +] at the time of delivery
(2.33)
. (2.34)
In such a rank, in order to increase the pH of the buffer solution, folded from a weak acid and a salt, established by a strong acid and a strong base, the nobility of only cob con-
centering of cich components.
For the sum of the weak base NH 4 OH and the salts of NH 4 Cl, the anion of the anion of the strong hydrochloric acid, speeding up in front of the microorganisms, it can be shown that the acidity of such a range is variable
, (2.35)
and the pH of the buffer sum is equal to the
pH = p - lg. (2.36)
It can be seen at the bottom of the accumulated deposits that the pH of the buffer solutions does not accumulate due to the dilution, so the concentration of acid, and the concentration of salt (or the base and the salt), are too low. Tse persha specialty buffer solutions.
If the amount of acid is small enough to reach the buffer, or in the meadow, then the pH of the range will change even slightly. Tse їхnya friend vіdminna rice.
For example, even before the acetate buffer solution, to avenge the sum of CH C COOH and CH 3 COONa, until a small amount of HCI is given, then sodium acetate will be combined with hydrochloric acid, but I will add the dissociation to 3 -
CH 3 COO - + Na + + H 3 O - + Cl - ↔ CH 3 COOH + Na + + Cl -. + H 2 Pro
The change in the concentration of ions [H 3 0 +], as well as the pH range, is practically not readable from the standard (2.36). The change in pH value change at the addition of acid and base, because of the strong buffering power of the range. That area of concentration, in some pH buffer solutions, is practically invisible, is called the buffer mnistyu:
Thus, the buffer the amount of g-equivalents of acid and in the meadow, you can add up to 1 liter of buffer solution, to change the pH value by one. Buffer solutions are widely used for setting standard pH values when calibrating different adjustments, so that the acidity of the solutions is changed, for example, pH meters.
Lecture No. 12. Electrolytic dissociation of water.
Uninvolved on those who do not use electrolyte in water, they are often dissociated from the approved cation hydroxon and hydroxide-anion:
H 2 O + H 2 O H 3 O + + OH -
Often it’s easy to write down the form given to the process:
H 2 O H + + OH -
Tsya rivnovaga is characterized by the following constant:
Oskіlki in clean water and dilution water razor = const, the whole viraz can be reconfigured to an offensive look:
K W =
Otriman constant is called іonny Vitvіr Vodi. For 25 ° С K W = 10 -14. Sounds like a drink, for clean water and neutral ratios = = Ö10 -14 = 10 -7. Obviously, sour razines> 10 -7, and lugs< 10 -7 . На практике часто пользуются an indicator of the concentration of cations in water- Negative tens logarithm (pH = -lg). Acid pH ranges< 7, в щелочных pH >7 in neutral center pH = 7. Similarly, you can enter the hydroxyl index pOH = -lg. The water and hydroxyl indicators of tying are simpler for the matings: pH + pOH = 14.
Clearly apply the pH value of water solutions of strong and weak acids.
Application No. 1. Centimolar break (0.01 mol / l) hydrochloric acid (strong monobasic acid).
HCl = H + + Cl -
C HCl = 0.01; pH = -lg 0.01 = 2
Application number 2. Centimolar difference (0.01 mol / l) sodium hydroxide (strong one-acid base).
NaOH = Na + + OH -
C NaOH = 0.01; pOH = -lg 0.01 = 2;
pH = 14 - pOH = 12
Supplement No. 3. Centimolar breakdown (0.01 mol / l) oztic acid (weak monobasic acid).
CH 3 COO - + H + CH 3 COOH
=. For weak electrolyte. "
= 1.75 x 10 -5; ; "
pH = - log = -1/2 (logK a + logC) = 1/2 (pK a - logC) = 1/2 (4.75 + 2) = 3.38
Application No. 4. Centimolar variation (0.01 mol / l) amicac (ammonium hydroxyde, weak one-acid base).
NH 3 + H 2 O NH 4 + + OH -
=. Oskilki ammonium hydroxide is a weak electrolyte, then "C. Submitting the formulas in the constant of ionization of ammonia yak base, we can say:
= 1.8 x 10 -5; ; =
pOH = -lg = 1/2 (pK b - lgC);
pH = 14 - pOH = 14 + 1/2 (logC - pK b) = 14 + 1/2 (-2 - 4.76) = 10.62
Hydrolysis of salts ... The change in the acidity of aqueous solutions of salts in terms of acidity clean water start with hydrolysis. Gidroliz - the price of exchange of communication with water... For quickness to hydrolysis salt to go to chotiri tipi:
1. Salt, established with a strong acid and strong base (for example, NaCl, Na 2 SO 4), does not undergo hydrolysis. Water solutions of such salts may be neutral (pH = 7).
2. Salt, established with a weak base and a weak acid, is hydrolyzed by a meaning world and is often irreversible, for example,
Al 2 S 3 + 6H 2 O = 2Al (OH) 3 + 3H 2 S
The acidity of these differences is based on a larger speech, and is close to neutral (pH "7).
3. Salts, set with a weak base and strong acid, hydrolyze reversely, sounding hydroxide-anioni, and sounding acidic reaction (pH< 7). Например, гидролиз хлорида аммония можно описать следующими уравнениями:
NH 4 Cl + H 2 O NH 3 × H 2 O + HCl
When you look at it, you can see that not all the power is needed, but only the cation. The cations of salts, approved with weak acidic acid bases, hydrolyze rapidly, afterwards, in water and hydroxide-anion:
Al 3+ + H 2 O Al (OH) 2+ + H +
Al (OH) 2+ + H 2 O Al (OH) 2 + + H +
Al (OH) 2 + H 2 O Al (OH) 3 H +
Summarizing the hydrolysis of the aluminum cation with such a viglyad:
Al 3+ + 3H 2 O Al (OH) 3 + 3H +
4. Salts, established with a strong base and a weak acid, are hydrolyzed according to anion, which is borne from water to water. Hydroxide-anioni, scho zvіlnyayuyutsya, nadayut razhnuyu reaction (pH> 7). For example, hydrolysis of sodium acetate is like this:
CH 3 COONa + H 2 O CH 3 COOH + NaOH
Obviously, hydrolysis of anionic salts of weak acid-rich acids often occurs, for example,
PO 4 3- + H 2 O HPO 4 2- + OH -
HPO 4 2- + H 2 O H 2 PO 4 - + OH -
H 2 PO 4 - + H 2 O H 3 PO 4 + OH -
Sumarne rіvnyannya hydrolizu phosphate-anionu maє takiy viglyad
PO 4 3- + 3H 2 O H 3 PO 4 + 3OH -
Hydrolysis is not only salt, but covalent inorganic organic spoluks... For example:
PCl 3 + 3H 2 O = H 3 PO 3 + 3HCl
An important role is played in the life of living organisms in the land of hydrolysis of biological molecules - proteins and polypeptides, fats, and also polysaccharides.
Glybin for hydrolysis is characterized by step to hydrolysis(h) - to the number of speeches, which have become proud of the hydrolysis, to the outright number of speeches in the range... The circulating hydrolysis can also be characterized as a constant. For example, for the process of hydrolysis of acetate-anion, the constant of hydrolysis is recorded by the next rank:
Equally important is the concentration of water from the constant, the hydrolysis does not enter, some parts of it will be automatically transferred to the left part of the balance.
The level of hydrolysis is constant, as well as the pH of water solutions of salts is discernible on specific butts.
Application No. 5. Centimolar difference (0.01 mol / l) to ammonium chloride (sil, formulated with a weak base and strong acid). We can write down the hydrolysis ratio in the form of a hydrolysis that can be stored for constant hydrolysis.
NH 4 + + H 2 O NH 3 × H 2 O + H +
Having multiplied the number and the standard of the right part of ryvnosti by the concentration of hydroxides-ions, the constant of hydrolysis can be reconverted by the next rank:
5.56 × 10 -10
, = = Ch, а = C - Ch = C (1-h). As a matter of fact,
Oskіlki h<< 1, а (1-h) ® 1, полученное выражение можно упростить:
; stars h "
2.36 × 10 -4 or 0.0236%
It can be seen from the ryvnyans that the constant and the level of hydrolysis of salt will grow from the changes in the constant of dissociation, so that. from changes in strength. In addition, the steps for the hydrolysis and the rate of reduction in the reduction of the concentration (increase in the dilution) of the salt. The constant of hydrolysis, as a constant, be it equal, because of the concentration, do not lie. Increase the temperature to increase the level of that constant hydrolysis, oscillations of hydrolysis - endothermic process.
When the pH value rises, the range is soli vrahumo, uh =, and the first one is close to C.
; stars "
pH = - log = -1/2 (logK w + logC + pK b) = 7 - 1/2 (pK b + logC) = 7 - 1/2 (4.76 - 2) = 5.62
Appendix No. 6. Centimolar difference (0.01 mol / l) to sodium acetate (sil, fixed with a strong base and a weak acid). We can write down the hydrolysis ratio in the form of a hydrolysis that can be stored for constant hydrolysis.
CH 3 COO - + H 2 O CH 3 COOH + OH -
Multiplying the numeral and the denominator of the right part of ryvnosti by the concentration of cation vodnya, it can be recast into the following form:
1 × 10 -14 / 1.75 × 10 -5 = 5.71 × 10 -10
, = = Ch, а = C - Ch = C (1-h).
As a matter of fact,
; ; stars h =
2.39 × 10 -4 or 0.0239%
When the pH value rises, the value of pH is vrahumo, uh =, a »C.
; stars ";
pOH = -lg = -1/2 (logK w + logC + pK a) = 7 - 1/2 (pK a + logC)
pH = 14 - pOH = 7 + 1/2 (pK a + logC) = 7 + 1/2 (4.75 - 2) = 9.75
Appendix No. 7. Centimolar razchin (0.01 mol / l) to ammonium acetate (sil, fixed with a weak base and weak acid). We can write down the hydrolysis ratio in the form of a hydrolysis that can be stored for constant hydrolysis.
NH 4 + + CH 3 COO - + H 2 O NH 3 × H 2 O + CH 3 COOH
Having multiplied the number and the standard of the right part of ryvnosti by the addition of the concentration of cation water and hydroxide-anion (ionic addition of water), it can be recaptured by the offensive rank:
= = 0.32 × 10 -4
, = = Ch
C - Ch = C (1-h), apparently,
0.0056 or 0.56%
The ammonia molecule is hydrated, so that it becomes established as a result of hydrolysis, dissociation, and the addition of hydroxide-anion:
NH 3 × H 2 O NH 4 + + OH -
; stars
Similarly, dissociation of ocetic acid and without prescribing the establishment of cations in water:
CH 3 COO - + H + CH 3 COOH
We know about the performance of the concentration of danikh ions:
As a matter of fact until the level of hydrolysis =, a =, todi
Oskilki = K w /, then 2 =; stars =
pH = - log = 1/2 (pK w + pK a - pK b) = 7 + 1/2 (pK a - pK b) = 7 + 1/2 (4.75 - 4.76) = 6.995
Literature: p. 243 - 255; with. 296 - 302