Solubility Product

Buffer solution

The solution which resists the change in the pH value when a small amount of acids or bases are added to them is known as a buffer solution. Buffer solutions are of two types:

1. Acidic buffer

It is a mixture of a weak acid and its salt with a strong base. Its pH is less than 7. Eg. A mixture of CH₃COOH and CH₃COONa.

2. Basic buffer

It is a mixture of a weak base and its salt with strong acid. Its pH is greater than 7. eg. A mixture of NH₄OH and NH₄Cl.

Buffer action

The capacity of a buffer solution to resist the change in pH is called buffer action. The buffer action of buffer solution can be explained as follows:

1. Consider an acidic buffer (CH₃COOH + CH₃COONa) to explain the buffer action.
CH₃COOH ⇌ CH₃COO^– + H⁺
CH₃COONa ⇌ CH₃COO^– + Na⁺

• Addition of H+ (from HCl): When small quantity of HCl is added to the buffer solution, the concentration of H+ ions in the solution increases. These H+ ions combine with CH3COO- ions to form unionized CH3COOH.
  CH3COO- (from buffer) + H+ (from acid) → CH3COOH
  In this way, pH of the solution remains constant.

• Addition of OH^– (from base): When small quantity of NaOH is added to the buffer solution, OH^– is neutralized by H⁺ of the base to give H₂O. So concentration of H⁺ is decreased which causes the disturbance of equilibrium state of ionization of CH₃COOH. However, according to Le-Chatlier’s principle, this disturbance is minimized by shifting the equilibrium towards forward direction i.e. more CH₃COOH will be ionized to fulfill the amount of H⁺ ion and therefore the pH remains constant.
  H⁺ (from buffer) + OH^– (from base) → H₂O

2. Consider a basic buffer (NH₄OH + NH₄Cl) to explain the buffer action.
NH₄OH ⇌ NH₄⁺ + OH^–
NH₄Cl → NH4⁺ + Cl^–

• Addition of H⁺ (from acid): When a small quantity of HCl is added to the buffer solution, H⁺ combines with OH^– of the buffer to give H₂O. So the concentration of OH^– is decreased which causes the disturbance of the equilibrium state of ionization of NH₄OH. However, according to Le-Chatlier’s principle, this disturbance is minimized by shifting the equilibrium towards the forward direction i.e. more NH₄OH will be ionized to fulfil the amount of OH^– ion and therefore the pH remains constant.
  OH^– (from buffer) + H⁺ (from acid) → H₂O

• Addition of OH^– (from base): When a small quantity of NaOH is added, OH^– combines with NH₄ of buffer to give NH₄OH and hence there is no change in pH.
  NH₄⁺(from buffer) + OH^– (from base) → NH₄OH

The pH of buffer solution can be calculated by Henderson equation, which is:

For\ acidic\ buffer,\ pH=pKa + log\frac{[salt]}{[acid]}\\ For\ basic\ buffer,\ pOH=pKa + log\frac{[salt]}{[base]}

Salt hydrolysis

The interaction of ions of the salt with water to produce acidic and basic solutions is called salt hydrolysis. There are four types of salt hydrolysis:

1. Salt of strong acid and weak base [NH₄Cl, (NH₄)₂SO₄, AlCl₃]:

The cations produced by this type of salt being the conjugate acid of a weak base is relatively strong. So the salt undergoes cationic hydrolysis to give an excess of free H⁺ ions and its aqueous solution is acidic in nature.

NH_{4}Cl\ \rightleftharpoons\ NH_{4}^{+} + Cl^{-}\\ \frac{NH_{4}^{+} + H_{2}O \rightleftharpoons\ NH_{4}OH + H^{+}}{NH_{4}Cl + H_{2}O \rightleftharpoons\ NH_{4}OH + HCl}

2. Salt of weak acid and strong base[NaCN, CH₃COONa, Na₂CO₃]:

The anions produced by this type of salt being the conjugate base of the weak acid is relatively strong. So the salt undergoes ionic hydrolysis to produce an excess of free OH^– ions and its aqueous solution is basic in nature.

CH_{3}COONa \rightleftharpoons\ CH_{3}COO^{-} + Na^{+}\\ \frac{CH_{3}COO^{-} + H_{2}O \rightleftharpoons\ CH_{3}COOH + OH^{-}}{CH_{3}COONa + H_{2}O \rightleftharpoons\ CH_{3}COOH + NaOH}

3. Salt of weak acid and weak base[NH₄CN, CH₃COONH₄]

The cations and anions produced by this type of salt being the conjugate acid and conjugate base pair of a weak base and a weak acid, both are strong. so both ions interact with water as:

NH_{4}CN \rightleftharpoons\ NH_{4}^{+} + CN^{-}\\ \frac{NH_{4}^{+}+CN^{-} + H_{2}O \rightleftharpoons\ NH_{4}OH + HCN}{NH_{4}CN + H_{2}O \rightleftharpoons\ NH_{4}OH + HCN}

4. Salt of strong acid and strong base (NaCl, NaNO₃, Na₂SO₄)

The cations and anions given by this type of salt are weak as they come from strong acids and strong bases. So no ions interact with water as:
NaCl ⇌ Na⁺ + Cl^–
Na⁺ + Cl^– + H₂O → No reaction
The aqueous solution of it is neutral.

Some Important Numericals

1. Is an aqueous solution containing hydrogen ion concentration 3 x 10⁵ mol/lit acidic, basic or neutral ? (acidic)
2. Is an aqueous solution containing hydrogen ion concentration 3.33 x 10^-1 mol/lit acidic, basic or neutral ? (basic)
3. Calculate the pH of 0.1 N H₂SO₄. (1)
4. 49 gm of H₂SO₄ is present in 1000 ml of its solution. What is the pH of solution? (0)
5. Calculate the pH of an aqueous solution containing 10^-7 moles of NaOH per litre. (7.302)
6. Two litres of 1M HCl and one litres of 1M NaOH solution are mixed together. Calculate the strength of acid formed and pH of the resulting solution. (0.33M, 0.8)
7. 10^-2 mole of KOH is dissolved in 10 litres of water. What will be the pH of the solution? (11)
8. What will be the H⁺ ion concentration of a solution having pH 5.5? (3.16 x 10^-6 mol/lit)
9. 200 ml of an aqueous solution of HCl (pH = 2) is mixed with 300 ml of an aqueous solution of NaOH (pH = 12). What will be the pH of resulting mixture solution? (11.3)
10. Calculate the pH of 1M acetic acid solution. To what volume one litre of this solution be diluted so that the pH of the solution that is formed will be twice of original value (Ka = 1.8 x 10^-5) (2.37, 54975.26 lit)
11. If the volume of 25 cm³ of 0.05M Ba(NO₃)₂ are mixed with 25 cm³ of 0.02M NaF. Will any BaF₂ precipitate? (Ksp of BaF₂ at 298 K = 1.7 x 10^-6) (1.7 x 10^-8)
12. The solubility product of Ca(OH)₂ at 25°C is 4.42 x 10^-5. A 500 ml of a saturated solution of Ca(OH)₂ is mixed with an equal volume of 0.4M NaOH. How much Ca(OH)₂ is precipitated? (0.754 g)
13. What is the minimum volume of water required to dissolve 1gm of CaSO₄ at 298K? (Ksp for CuSO₄ is 9.1 x 10^-6) (2.45 x 10³ ml)
14. Calculate the pH of a saturated solution of Mg(OH)₂ , Ksp for Mg(OH)₂ is 8.9 x 10^-12. (10.35)
15. What mass of KOH should be dissolved in one litre of solution to prepare a solution having pH 12 at 25°C?(at. wt. of K =39) (0.56 gm)
16. 0.41 kg of NaOH is placed in 100 ml of 0.1N H₂SO₄. Find the pH of resulting solution. (11.4)
17. What is the pH of a solution of NaOH whose concentration is 0.4 gm/lit? (12)
18. Calculate the strength in gm/lit of NaOH whose pH is 11. (4 x 10^-2 gm/lit)
19. 0.00143 gm of AgCl dissolve in 1 lit of water at 25°C to form a saturated solution. What is the solubility product of the salt ? (Ag=108, Cl=35.5) (9.92 x 10^-11)
20. A sample of AgCl is treated with 5 ml of 2M Na₂CO₃ solution to produce Ag₂CO₃ . The remaining solution contained 0.003 gm of Cl^– per litre. Calculate the solubility product of AgCl. (Ksp of Ag₂CO₃ = 8.2 x 10^-12). (1.2 x 10^-10)
21. The pH of a solution of KOH is 10. Calculate the hydroxyl ion concentration. (10^-4 M)
22. The pH of 0.1M HCN solution is 5.2. What is the value of ionization constant(Ka) for the acid? (3.969 x 10^-10)
23. The solubility of AgCl in water at 298 K is 1.43 x 10^-3 gm/lit. Calculate its solubility in 0.5 M KCl solution ? 1.98 x 10^-10 mol/lit)
24. Calculate the hydroxyl ion concentration of a solution having pH 10.5. (3.16 x 10^-4 mol/lit)
25. The solubility of CaF₂ in water at 18°C is 2.05 x 10^-4 mol/lit. Calculate is solubility product. (3.446 x 10^-11)
26. The solubility product of BaSO₄ in water at 25°C is 10^-10 mol/lit. Calculate its solubility in gm/lit. (Ba=137) (2.33 x 10^-3 gm/lit)
27. The solubility product of chalk is 9.3 x 10^-8. Calculate its solubility in gm/lit. (3.49 x 10^-2 gm/lit)
28. The pH of HCl solution is 3. Calculate the strength of HCl in terms of molarity. (10^-3M)