Ionic Equilibrium

Ionic Equilibrium

The aqueous solution  of potash alum is acidic due to hydrolysis of

Q: The aqueous solution  of potash alum is acidic due to hydrolysis of

(A) K+

(B) Al3+

(C) SO4– –-

(D) presence of acid in its crystal as impurity

Sol: Al3+ having high charge density will associate strongly with water through its fractional negatively charged oxygen pole and in such interaction there will be hydrolysis to some extent:

Al3+ + H2O ⇔  [Al(OH)]2++ H+

The product cation may hydrolyse further

 Ans:  (B)

 The pH of a buffer is 6.745. When 0.01 mole of NaOH is added to 1 litre of it, the pH changes to 6.832. Its buffer capacity is

Q: The pH of a buffer is 6.745. When 0.01 mole of NaOH is added to 1 litre of it, the pH changes to 6.832. Its buffer capacity is

(A) 0.187

(B) 0.0115

(C) 0.076

(D) 0.896

Sol: $Buffer Capacity = \frac{No. of moles of base added /litre of buffer}{Change in pH} $

$= \frac{0.01}{6.832 – 6.745} = 0.11  $

Ans: (B)

pH value of pure water at 0°C will be

Q: pH value of pure water at 0°C will be

(A) Greater than 7

(C) Less than 7

(C) 7

(D) All the three

Sol: At 15°C, Kw = 10–14. So at temperature below 25°C Kw < 10–14

Ans:  pH > 7 (for neutral solution or pure water)

Dissociation constant of two acids HA & HB are respectively 4 × 10-10 & 1.8 × 10-5 whose pH value will be higher for a given molarity

Q: Dissociation constant of two acids HA & HB are respectively 4 × 10-10 & 1.8 × 10-5 whose pH value will be higher for a given molarity:

(A) HA

(B) HB

(C) Both same

(D) Can’t say

Sol: HA ⇔ H+ + A

HB ⇔H++ B

for HA

$K_{a1} = \alpha^2 C $

$\alpha_1^2 = \sqrt{\frac{K_{a1}}{C}} = \sqrt{\frac{4\times 10^{-10}}{C}}$

α­1 = 2 × 10–5 for C = 1 M

Similarly  , [H+] = C α= 2 × 10–5  ⇒ pH = 4.7

for HB

$\alpha_2^2 = \sqrt{\frac{K_{a2}}{C}} = \sqrt{\frac{1.8\times 10^{-5}}{C}}$ for C = 1 M

α2 = 4.2 × 10–3  ⇒ [H+] = Ca = 4.2 × 10–3

pH = – log (4.2 × 10–3) = 2.37

Ans: (A)

Let Kw at 100°C be 5.5 × 10^-13 M2. If an aqueous solution at this temperature has pH = 6.2. Its nature will be

Q: Let Kw at 100°C be 5.5 × 10-13 M2. If an aqueous solution at this temperature has pH = 6.2. Its nature will be

(A) acidic

(B) alkaline

(C) neutral

(D) can’t say

Sol: pKw = 12.26

Condition of neutrality pH = 6.13

Thus, pH = 6.2 > 6.13

Means alkaline

Ans:  (B)