Question

For the reaction $Ag{\left(CN\right)}_2^{-}\rightleftharpoons A{\left(g\right)}^{+}+2{CN}^{-}$, the $K_c$ is $4\times {10}^{-19}$. Calculate $\left[A{\left(g\right)}^{+}\right]$ in solution which was originally 0.1 M in $KCN$ and 0.03 M in $AgN{O}_3$.

Answer 1

$2KCN\left(aq\right)+AgN{O}_3\left(aq\right)\to K\left[Ag\right(CN{)}_2\left]\right(aq)+KN{O}_3(aq)$

Here come of $KCN$ and $AgN{O}_3$ is given $0.1m\&0.03m$

$AgN{O}_3$ acts as a limiting reagent we need

twice as much KCN, so we use $0.03\times 2=0.06m$

of kCN to cannot all 0.03 mol the $A{g}^{+}$ to $Ag(CN{)}_2^{-}$

at first, and are left with $(0.1-0.06)=0.04M$ KCN

$Ag\left(CN{)}_2^{-}\right(aq)\rightleftharpoons A{g}^{+}(aq)+2C{N}^{-}(aq)$

initial $0.03m$ $0.04$

$0.03-x$ $x$ $0.04+2x$

$K_c=\frac{\left[A{g}^{+}\right][C{N}^{-}{]}^2}{\left[Ag\right(CN{)}_2^{-}]}$

$4\times {10}^{-19}=\frac{x\times (0.04+2x{)}^2}{(0.03-x)}$

$4\times {10}^{-19}=\frac{x\times 0.04}{0.03}$

$x=3\times {10}^{-19}$

concentration of $\left[A{g}^{+}\right]$ is $3\times {10}^{-19}M$