[{H^ + }] of a solution that is 0.01,M in HCN and 0.02,M in NaCN ({Kₐ} for HCN = 6.2 × {10^{ - 10

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6-2.Equilibrium-II (Ionic Equilibrium)

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What is $[{H^ + }]$ of a solution that is $0.01\,M$ in $HCN$ and $0.02\,M$ in $NaCN$ $({K_a}$for $HCN = 6.2 \times {10^{ - 10}})$

A

$3.1 \times {10^{10}}$

B

$6.2 \times {10^5}$

C

$6.2 \times {10^{ - 10}}$

D

$3.1 \times {10^{ - 10}}$

Solution

(d) ${K_a} = \frac{{[{H^ + }]\,\,[C{N^ – }]}}{{[HC{N^ – }]}}$

$6.2 \times {10^{ – 10}} = \frac{{[{H^ + }]\,\,[0.02]}}{{[0.01]}}$

$[{H^ + }] = \frac{{6.2 \times {{10}^{ – 10}} \times 0.01}}{{0.02}} = 3.1 \times {10^{ – 10}}$

Standard 11

Chemistry

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