Explanation

(a)
(i)$Ag^{+} + NH_{3} \rightleftharpoons \left\lbrack Ag\left( NH_{3} \right) \right\rbrack^{+};$
$\ K_{1} = 3.5 \times 10^{- 3}$
(ii)$\left\lbrack Ag\left( NH_{3} \right) \right\rbrack^{+} + NH_{3} \rightarrow \left\lbrack Ag\left( {NH}_{3} \right)_{2} \right\rbrack^{+};\$
$\ K_{2} = 1.7 \times 10^{- 3}$
On the basis of above reaction,
$K_{1} = \frac{\left\lbrack Ag\left( NH_{3} \right) \right\rbrack^{+}\ }{\left\lbrack Ag^{+} \right\rbrack\left\lbrack NH_{3} \right\rbrack}$ …(i)
$K_{2} = \frac{\left\lbrack Ag\left( NH_{3} \right)_{2} \right\rbrack^{+}}{\left\lbrack Ag\left( NH_{3} \right) \right\rbrack^{+}\left\lbrack NH_{3} \right\rbrack}$ …(ii)
For the formation of $\left\lbrack Ag\left( NH_{3} \right)_{2} \right\rbrack^{+}$
$Ag^{+} + 2NH_{3} \rightleftharpoons \left\lbrack Ag\left( NH_{3} \right)_{2} \right\rbrack^{+}$
Formation constant $(K)$
$= \frac{\left\lbrack Ag\left( NH_{3} \right)_{2} \right\rbrack^{+}}{\left\lbrack Ag^{+} \right\rbrack\left\lbrack NH_{3} \right\rbrack^{2}}$ …(iii)
From Eqs. (i) and (ii)
$K = K_{1} \times K_{2}$
$= 3.5 \times 10^{- 3} \times 1.7 \times 10^{- 3}$
$= 5.95 \times 10^{- 6} \approx 6.08 \times 10^{- 6}$