If pK _ b for CN ^ - at 25 ^ o C is 4.7, the pH of 0.5 M aqueous NaCN solution is

Correct answer: B. 11.5 p k _ b ofCN^ - = 4.7 p k _ a ofHCN = 4.7 NaCN is salt of weak acid (HCN) and strong base (NaOH) So pH = 7 + (1)/(2)p k _ a + (1)/(2)logC = 7 + (1)/(2) × 9.3 + (1)/(2) (0.5) ∴ pH = 11.5

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

If pK _ b for CN ^ - at 25 ^ o C is 4.7, the pH of 0.5 M aqueous...

{pK}_{b}

for

{CN}^{-}

{25}^{o}C

is 4.7, the pH of 0.5 M aqueous NaCN solution is

Options

A10
B11.5
C11
D12

Correct answer

B. 11.5

Step-by-step solution

p{k}_{b}ofCN^{-} = 4.7

p{k}_{a}ofHCN = 4.7

NaCN

is salt of weak acid \left(HCN\right)

and strong base \left(NaOH\right)

pH = 7 + \dfrac{1}{2}p{k}_{a} + \dfrac{1}{2}logC

= 7 + \dfrac{1}{2} \times 9.3 + \dfrac{1}{2}\log \left(0.5\right)<br />

∴ pH = 11.5

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