Question

If $A{g}^{+}+2N{H}_3\rightleftharpoons Ag(N{H}_3{)}_2^{+};K_1=1.7\times {10}^7$
$A{g}^{+}+C{l}^{-}\rightleftharpoons AgCl;K_2=5.4\times {10}^9$

For $AgCl+2N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}+C{l}^{-}$, the equilibrium constant is :

• A. $4.68\times {10}^{-3}$
• B. $5.2\times {10}^{-17}$
• C. $0.31\times {10}^{-2}$
• D. $3.1\times {10}^{-2}$

Answer 1

The correct option is C $0.31\times {10}^{-2}$

$A{g}^{+}+C{l}^{-}\rightleftharpoons AgCl;K_2=5.4\times {10}^9$
Reverse above reaction
$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-};K_2=\frac{1}{5.4\times {10}^9}=1.85\times {10}^{-10}$ .....(1)
$A{g}^{+}+2N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+};K_1=1.7\times {10}^7$ .....(2)
Add reactions (1) and (2), we get:

$AgCl+2N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}+C{l}^{-}K=K_1{K}_2=1.7\times {10}^7\times 1.85\times {10}^{-10}=0.31\times {10}^{-2}$