First slide

Common ion effect

Question

When 0.1 mole of solid NaOH is added in 1 L of 0.1 M ${NH}_{3} (aq)$ then which statement is going to be wrong ?
$(K_{b} = 2 \times 10^{- 5}, \log 2 = 0.3)$

Moderate

A

degree of dissociation of ${NH}_{3}$ approaches to zero.
B

change in pH would be 1.85
C

concentration of $[{Na}^{+}] = 0.1 M, [{NH}_{3}] = 0.1 M, [{OH}^{-}] = 0.2 M$
D

on addition of ${OH}^{-}, K_{b} of {NH}_{3}$ does not changes

Solution

Before adding 0.1 M NaOH $p^{H}$ of weak base 0.1M ${NH}_{3}$ is
$p^{H} = \frac{1}{2} [p^{k_{b}} - \log_{10} C] = \frac{1}{2} [4.7 + 1] = 2.85$
When 0.1 M NaOH is added due to common ion effect dissociation of ${NH}_{3}$ decreases.
${NH}_{3} + H_{2} O ⇌ {NH}_{4}^{+} + {OH}^{-} K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{3}]} = \frac{0.1 [{OH}^{-}]}{0.1 + x} = 2 \times 10^{- 5} x = [{OH}^{-}] combining with [{NH}_{4}^{+}] to form ammonia solution in backward direction [{OH}^{-}] = 2 \times 10^{- 5} [since 0.1 + x ≅ 0.1] p^{OH} = 4.7 Change in p^{OH} = 4.7 - 2.85 = 1.85 For a given basic solution change in p^{OH} is also equal to change in p^{H}$

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