Question

When 0.1 mole solid NaOH is added in 1lt of 0.1 M NH3(aq) then which statement is wrong? (Kb = 2 × 10–5, log 2 = 0.3)

• A. Degree of dissociation of NH3 approaches to zero.
• B. Change in pH by adding NaOH would be 1.85
• C. In solution, [Na+] = 0.1 M, [NH3] = 0.1 M, [OH–] = 0.2 M.
• D. On addition of OH–, Kb of NH3 does not changes.

Answer 1

Answer: (C)

In solution, $Na+$ = 0.1 M, $NH3$ = 0.1 M, $OH–$ = 0.2 M.

Answer 2

Initial pH = $\frac { 1 } { 2 }$ (pKb – log C) = $\frac { 1 } { 2 }$ (5 – log 2 – log 0.1) = 2.85

After adding NaOH, pOH of solution = 1

Change in pOH = 1.85