Question

If $500mL$ of $0.4mAgN{O}_3$ is mixed with $500mL$ of $2MN{H}_3$ solution, then what is the concentration of $Ag(N{H}_3{)}^{+}$ in solution?
$\left[K_{f_1}\right[Ag\left(N{H}_3{)}^{+}\right]={10}^3;K_{f_2}\left[Ag\right(N{H}_2{)}_2^{+}]={10}^4]$

• A. $3.33\times {10}^{-7}M$
• B. $3.33\times {10}^{-5}M$
• C. $3\times {10}^{-4}M$
• D. ${10}^{-7}M$

Answer 1

Answer: (B)

$3.33\times {10}^{-5}M$

$A{g}^{+}+N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3){]}^{+}$
$\left[Ag\right(N{H}_3){]}^{+}+N{H}_3\rightleftharpoons [Ag(N{H}_3{)}_2{]}^{+}$
$K_{f1}=\frac{\left[Ag\right(N{H}_3){]}^{+}}{\left[A{g}^{+}\right]\left[N{H}_3\right]}$

$\frac{x-y}{(0.4-x)(2-x-y)}={10}^3$......(1)

$K_{f2}=\frac{\left[Ag\right(N{H}_3{)}_2{]}^{+}}{\left[Ag\right(N{H}_3\left){]}^{+}\right[N{H}_3]}$

$\frac{y}{(x-y)(2-x-y)}={10}^4$......(2)

On solving equations (1) and (2), we get $x-y=3.33\times {10}^{-5}M$

This is equal to the concentration of $\left[Ag\right(N{H}_3){]}^{+}$ in the solution.