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$K_{c}$ for the the reaction, $\left[\mathrm{Ag}(\mathrm{CN})_{2}\right] \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-},$
the equillibrium constant at $25^{\circ} \mathrm{C}$ is $4.0 \times 10^{-19}$ then the silver ion concentration in a solution which was originally 0.1 molar in $\mathrm{KCN}$ and 0.03 molar in $\mathrm{AgNO}_{3}$ is:
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the equillibrium constant at $25^{\circ} \mathrm{C}$ is $4.0 \times 10^{-19}$ then the silver ion concentration in a solution which was originally 0.1 molar in $\mathrm{KCN}$ and 0.03 molar in $\mathrm{AgNO}_{3}$ is:
Solution:
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Verified Answer
The correct answer is:
$7.5 \times 10^{-18}$
$\left.\begin{array}{c}2 \mathrm{KCN}+\mathrm{AgNO}=\left[\underset{0.1}{0.03}=\left[\begin{array}{c}\mathrm{Ag}(\mathrm{CN})_{2} \\ 0\end{array}\right]^{-}\right. \\ (0.1-0.06) & 0\end{array}\right] \begin{array}{c}\mathrm{KNO}_{3}+\mathrm{K}_{0}^{+} \\ (0.03)\end{array}$
$\therefore\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}=\mathrm{Ag}^{+}+\underset{\mathrm{a}}{0.03} \mathrm{a} \quad 2 \mathrm{CN}$
$(0.03-\mathrm{a}) \quad \mathrm{a} .04(\mathrm{left} \mathrm{from} \mathrm{KCN}$
$\quad 0.04 \approx 0.04$
$\therefore \quad \mathrm{K}_{\mathrm{c}}=4 \times 10^{-19}=\frac{(0.04)^{2} \times \mathrm{a}}{0.03}$
$\mathrm{a}=7.5 \times 10^{-18}$
$\therefore\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}=\mathrm{Ag}^{+}+\underset{\mathrm{a}}{0.03} \mathrm{a} \quad 2 \mathrm{CN}$
$(0.03-\mathrm{a}) \quad \mathrm{a} .04(\mathrm{left} \mathrm{from} \mathrm{KCN}$
$\quad 0.04 \approx 0.04$
$\therefore \quad \mathrm{K}_{\mathrm{c}}=4 \times 10^{-19}=\frac{(0.04)^{2} \times \mathrm{a}}{0.03}$
$\mathrm{a}=7.5 \times 10^{-18}$
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