First slide

Solubility product(KSP)

Question

The solubility of AgCl in 0.1M NaCl is (K_sp of AgCl = 1.2 × 10^–10)

Easy

A

0.1M
B

1.2 x 10^-5
C

1.095 x 10^-5
D

1.2 x 10^-9

Solution

Solubility of AgCl will be suppressed in presence of NaCl due to commonion effect and the following equilibria gets established.
$AgCl (s) ⇌ \underset{S^{'}}{AgCl} (aq) \to \underset{S^{'}}{{Ag}^{+}} (aq) + \underset{S + 0.1}{C_{^{'}}^{-}} (aq)$
$NaCl \to {Na}^{+} (aq) + \underset{0.1}{{Cl}^{-}} (aq)$
$K_{sp} = (S^{'} + 0.1) \times S^{'}$
$\Rightarrow 1.2 \times 10^{- 10} = 10^{- 1} \times S^{'}$
$\Rightarrow S^{'} = 1.2 \times 10^{- 9}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App