Question: A roll of \(35 - mm\) black and white photographic film contains about \(0.27g\) of unexposed \(AgBr\) before developing. What mass of \(N{a_2}{S_2}{O_3} \cdot 5{H_2}O\) (sodium thiosulfate penta hydrate or hypo) in \(1.0L\)of developer is required to dissolve the \(AgBr\)as \(Ag\left( {{S_2}{O_3}} \right)_2^{3 - }\left( {{K_f} = 4.7 \times 1{0^{13}}} \right)?\)

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 mass of\({\rm{AgBr\;is\;}}0.27{\rm{g}}\)and the volume is\(1L\) .

Hence, the number of moles of\(AgBr\)is,

\({n_{AgBr}} = \frac{{0.27{\rm{g}}}}{{187.77{\rm{g}}/{\rm{mol}}}} = 1.438 \cdot {10^{ - 3}}{\rm{mol}}\)

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

Thus,

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

Since,\(1.438 \cdot {10^{ - 3}}{\rm{mol\;of\;AgBr}}\)will dissolve into\(1.438 \cdot {10^{ - 3}}{\rm{mol\;of\;A}}{{\rm{g}}^ + }\)and\(1.438 \cdot {10^{ - 3}}{\rm{mol\;of\;B}}{{\rm{r}}^ - }\)

Therefore, the concentration of\({\rm{A}}{{\rm{g}}^ + }{\rm{and\;B}}{{\rm{r}}^ - }{\rm{are\; }}1.438 \cdot {10^{ - 3}}{\rm{M}}\)

Determine the maximum possible concentration of\(A{g^ + }\),

\({K_{sp}} = \left( {A{g^ + }} \right) \cdot \left( {B{r^ - }} \right]\)

\(\left( {A{g^ + }} \right) = \frac{{{K_{sp}}}}{{\left( {B{r^ - }} \right)}}\)

\(\begin{array}{*{20}{c}}{ = \frac{{5 \cdot {{10}^{ - 13}}}}{{1.438 \cdot {{10}^{ - 3}}}}}\\{ = 3.48 \cdot {{10}^{ - 10}}}\end{array}\)

Consider, the reaction of formation of\({\rm{Ag}}\left( {{{\rm{S}}_2}{{\rm{O}}_3}} \right)_2^{3 - }\),

Thus, the dissociation constant is,

\({K_f} = 4.7 \cdot {10^{13}}\)

Find the concentration of\(\left( {{{\rm{S}}_2}{\rm{O}}_3^{2 - }} \right]\),

Solve the value of\(x,\)

\({K_f} = \frac{{\left( {Ag{{\left( {{S_2}{O_3}} \right)}_2}^{3 - }} \right)}}{{\left( {A{g^ + }} \right) \cdot {{\left( {{S_2}O_3^{2 - }} \right)}^2}}}\)

\(4.7 \cdot {10^{13}} = \frac{{1.438 \cdot {{10}^{ - 3}}}}{{3.48 \cdot {{10}^{ - 10}} \cdot {{\left( {x - 2.876 \cdot {{10}^{ - 3}}} \right)}^2}}}\)

\({\left( {x - 2.876 \cdot {{10}^{ - 3}}} \right)^2} = \frac{{1.438 \cdot {{10}^{ - 3}}}}{{3.48 \cdot {{10}^{ - 10}} \cdot 4.7 \cdot {{10}^{13}}}}\)

\(\begin{array}{*{20}{c}}{x - 2.876 \cdot {{10}^{ - 3}} = \sqrt {8.79 \cdot {{10}^{ - 8}}} }\\{ = 2.965 \cdot {{10}^{ - 4}}}\\{x = 3.17 \cdot {{10}^{ - 3}}{\rm{M}}}\end{array}\)

Hence, the number of moles of\({\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} \cdot 5{{\rm{H}}_2}{\rm{O}}\)is\(3.17 \cdot {10^{ - 3}}\).

Calculate the mass of\({\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} \cdot 5{{\rm{H}}_2}{\rm{O}}\),

\(\begin{array}{*{20}{c}}{m = 3.17 \cdot {{10}^{ - 3}}{\rm{mol}} \cdot 248.126{\rm{g}}/{\rm{mol}}}\\{ = 0.787{\rm{g}}}\end{array}\)

Therefore, the mass of \({\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} \cdot 5{{\rm{H}}_2}{\rm{O}}\)required is\(0.787g\).