Question

The equilibrium constants are given for the following reaction,
$A{g}^{+}+N{H}_3$ $\leftrightharpoons$ $\left[Ag\right(N{H}_3){]}^{+}$ ; $K_1=3.5\times {10}^{-3}$

$\left[Ag\right(N{H}_3){]}^{+}+N{H}_3$ $\leftrightharpoons$ $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ ; $K_2=1.7\times {10}^{-3}$

then the formation constant of $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ is:

• A. $5.95\times {10}^{-6}$
• B. $5.20\times {10}^{-3}$
• C. $2.06$
• D. $1.80\times {10}^{-3}$

Answer 1

The correct option is A $5.95\times {10}^{-6}$

Given:
$A{g}^{+}+N{H}_3$ $\leftrightharpoons$ $\left[Ag\right(N{H}_3){]}^{+}$ ; $k_1=3.5\times {10}^{-3}$
$\left[Ag\right(N{H}_3){]}^{+}+N{H}_3$ $\leftrightharpoons$ $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ ; $k_2=1.7\times {10}^{-3}$
on adding the above two equilibrium reactions we get:
$A{g}^{+}+2N{H}_3$ $\leftrightharpoons$ $\left[Ag\right(N{H}_3{)}_2{]}^{+}$
hence equilibrium constant or formation constant of above reaction becomes :
$K_{formation}=K_1\times K_2$
$K_{formation}=5.95\times {10}^{-6}$