1. pH of 10-12 M HCl is[MOE Model]
(c) At very low concentrations (<10-7 M), water's autoionization dominates, making pH ≈ 7 (neutral). 2. The pH value of a solution of NaOH is 10. Assuming complete dissociation, the concentration of OH- ion is[MOE 2062]
- 10-10
- 10-1
- 10-4
- None of above
(c) pH = 10 ⇒ pOH = 4 ⇒ [OH-] = 10-4 M. 3. Why is precipitate of AgCl obtained when a drop of AgNO3 is added to aqueous NaCl solution?[MOE 2062]
- Ionic product of AgCl > solubility product
- Solubility product > ionic product of AgCl
- Solubility product = ionic product of AgCl
- Solubility product ≠ ionic product of AgCl
(a) Precipitation occurs when ionic product (Q) exceeds solubility product (Ksp). 4. Ionic product of water is[MOE 2060]
- 101 moles/litre
- 10-14 moles2/litre2
- 10-7 moles/litre
- 10-14 moles/litre
(b) Kw = [H+][OH-] = 10-14 at 25°C (units: mol2/L2). 5. pH of 0.02 M NaOH is[MOE 2063]
(b) [OH-] = 0.02 M ⇒ pOH = -log(0.02) ≈ 1.7 ⇒ pH = 14 - 1.7 = 12.3. 6. pH of 50 cc of 0.01 N HCl is decreased by
- Adding 50 cc of 0.01 N HCl
- Adding 50 cc of water
- Adding 10 cc of 0.0001 N HCl
- Adding 50 cc of 0.1 N HCl
(d) Adding 0.1 N HCl increases [H+] significantly, lowering pH. 7. 4 ml of 0.5 N HCl is mixed with 1 ml of 2 N KOH. The pH of resulting solution is[MOE 2056]
(b) HCl meq = 4 × 0.5 = 2; KOH meq = 1 × 2 = 2. Exact neutralization ⇒ pH = 7. 8. Solubility of AB2 is x. Then the solubility product will be[IOM 2007]
(a) For AB2 ⇌ A2+ + 2B-, Ksp = [A2+][B-]2 = x(2x)2 = 4x3. 9. The solubility product of a sparingly soluble salt AB2 is 1.08 × 10-23 at 25°C. Its molar solubility is[IOM 1996]
- 3 × 10-8 M
- 3 × 10-6 M
- 3 × 10-4 M
- 3 × 10-2 M
(a) Ksp = 4x3 ⇒ x = (1.08 × 10-23/4)1/3 ≈ 3 × 10-8 M. 10. pH of a solution containing 2 g of sodium hydroxide per litre of water will be[IOM 2005]
(d) [NaOH] = 2 g/L ÷ 40 g/mol = 0.05 M ⇒ pOH = -log(0.05) ≈ 1.3 ⇒ pH = 14 - 1.3 = 12.7. 11. The highest pH value is shown by 0.1 M[IOM 2008, MOE 2065]
(c) Ca(OH)2 is a strong base releasing 2 OH- per formula unit ⇒ highest [OH-] and pH. 12. pH of 0.1% solution of NaOH is
(d) 0.1% NaOH = 0.1 g/100 mL = 1 g/L ⇒ [OH-] = 1/40 = 0.025 M ⇒ pOH ≈ 1.6 ⇒ pH ≈ 12.4.