Question

If $A{g}^{+}+2N{H}_3\rightleftharpoons {\left[Ag\right(N{H}_3{)}_2]}^{+}$ ; $K_1=1.7\times {10}^7$ ; $A{g}^{+}+C{l}^{-}\rightleftharpoons AgCl$ ; $K_2=5.4\times {10}^9$ Then for $AgCl+2N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}C{l}^{-}$ ; equilibrium constant will be?

• A. $0.31\times {10}^{-2}$
• B. $3.2\times {10}^2$
• C. $9.18\times {10}^{16}$
• D. $1.00\times {10}^{-17}$

Answer 1

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

If we have different equilibrium constants for different reactions and the net reactions is obtained by multiplying those equilibrium constant.

$A{g}^{+}+2N{H}_3\leftrightharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+};K_1=1.7\times {10}^7-----\left(1\right)$

$A{g}^{+}+C{l}^{-}\leftrightharpoons AgCl;K_2=5.4\times {10}^9-------\left(2\right)$

$AgCl\leftrightharpoons A{g}^{+}+C{l}^{-};K_3=\frac{1}{K_2}=\frac{1}{5.4}\times {10}^{-9}------\left(3\right)$

Adding equation 1 and 3,

$AgCl+2N{H}_3\leftrightharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}C{l}^{-}$

$K=K_1\times K_3=\frac{1.7\times {10}^7}{5.4\times {10}^9}$

$K=0.31\times {10}^{-2}$

So, correct option is A.