If the concentration of $ \left[\mathrm{NH}_{4}^{+}\right] $in a solution having $ 0.02 \mathrm{M} \mathrm{NH}_{3} $ and $ 0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} $ is $ \mathrm{a} \times 10^{-6} \mathrm{M} $, determine a . $$ \left[\mathrm{k}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\right] $$

Question

If the concentration of $ \left[\mathrm{NH}_{4}^{+}\right] $in a solution having $ 0.02 \mathrm{M} \mathrm{NH}_{3} $ and $ 0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} $ is $ \mathrm{a} \times 10^{-6} \mathrm{M} $, determine a . $$ \left[\mathrm{k}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\right] $$,

MARKS App · Accepted Answer

N H 4 O H ⇋ N H 4 + + O H - k b = N H 4 + O H - N H 4 O H ⇒ 1.8 × 10 - 5 = N H 4 + 0.01 0.02 {As O H - will mainly come from C a O H 2 only} ⇒ N H 4 + = 36 × 10 - 6 = a × 10 - 6 ⇒ a = 36

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