Question: Calculate the minimum concentration of ammonia needed in \(1.0L\)of solution to dissolve\(3.0 \times 1{0^{ - 3}}mol\) silver bromide.

The product of the reactant concentration divided by the product concentration yields the dissociation constant.

\({K_d} = \)Product of concentration of reactants\( \div \)concentration of products

Consider the given values,

\(3.0 \cdot {10^{ - 3}}{\rm{\;mol of\;AgBr}}\)

Volume of a solution\( = 1L\)

Let the reaction of dissolution of \(AgBr\)is,

Thus, the dissociation constant is,

\({K_{sp}} = 5 \cdot {10^{ - 13}}\)

Since,\(0.003{\rm{\;mol of\;AgBr}}\)will dissolve into\(0.003{\rm{\;mol of\;A}}{{\rm{g}}^ + }\)and\(0.003{\rm{mol\;of\;B}}{{\rm{r}}^ - }\)2

Hence, the concentration of\({\rm{A}}{{\rm{g}}^ + }{\rm{and\;B}}{{\rm{r}}^ - }{\rm{are\;}}\frac{{0.003{\rm{mol}}}}{{1L}} = 0.003{\rm{M}}\)

Determine the maximum possible concentration of\(A{g^ + }\)that can be present without causing reprecipitation.

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {A{g^ + }} \right) \cdot \left( {B{r^ - }} \right)}\\{\left( {A{g^ + }} \right) = \frac{{{K_{sp}}}}{{\left( {B{r^ - }} \right)}}}\end{array}\)

\(\begin{array}{*{20}{c}}{ = \frac{{5 \cdot {{10}^{ - 13}}}}{{0.003}}}\\{ = 1.67 \cdot {{10}^{ - 10}}{\rm{M}}}\end{array}\)

Consider the reaction of formation of\({\left( {{\rm{Ag}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_2}} \right)^ + }\),

Hence, dissociation constant is\({K_f} = 1.7 \cdot {10^7}\)

Find the \(\left( {{\rm{N}}{{\rm{H}}_3}} \right)\)needed to make\(({\rm{A}}{{\rm{g}}^ + }){\rm{\;be\;}}1.67 \cdot {10^{ - 10}}{\rm{M}}\)

Solve for the value of\(x,\)

\({K_f} = \frac{{\left( {{{\left( {{\rm{Ag}}{{({\rm{CN}})}_2}} \right)}^ - }} \right)}}{{({\rm{A}}{{\rm{g}}^ + }) \cdot {{\left( {N{{\rm{H}}_3}} \right)}^2}}}\)

\(1.7 \cdot {10^7} = \frac{{0.003}}{{1.67 \cdot {{10}^{ - 10}} \cdot {{(x - 0.006)}^2}}}\)

\({(x - 0.006)^2} = \frac{{0.003}}{{1.7 \cdot {{10}^7} \cdot 1.67 \cdot {{10}^{ - 10}}}}\)

\({(x - 0.006)^2} = 1.0567\)

\(\begin{array}{*{20}{c}}{x - 0.006 = 1.028}\\{x = 1.034{\rm{M}}}\end{array}\)

Therefore, the minimum concentration of\(N{H_3}\)is \(1.034M\).