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Mobile App Chapter 15: Q52E (page 875)
Question: Hydrogen sulfide is bubbled into a solution that is 0.10 M in both \(P{b^{2 + }}\)and \(F{e^{2 + }}\)and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with \({H_2}S\) ((\({H_2}S\)) = 0.10 M). What concentrations of \(P{b^{2 + }}\)and \(F{e^{2 + }}\)remain in the solution? For a saturated solution of \({H_2}S\)we can use the equilibrium:
Short Answer
The concentration of \({\rm{P}}{{\rm{b}}^{2 + }}\)ions left in the solution is \(6.31 \cdot {10^{ - 3}}\).
The concentration of \({\rm{F}}{{\rm{e}}^{2 + }}\) ions left in the solution is \(3.33 \cdot {10^7}\).
Step by step solution
Find the Concentration of \(P{b^{2 + }}\):
Given,
\({{\rm{H}}_2}{\rm{S\;}} = {\rm{\;}}0.1{\rm{M}}{\rm{.\;}}\)
Calculate \(\left( {{{\rm{S}}^{2 - }}} \right)\) ,
\(\begin{array}{*{20}{c}}{K = \frac{{\left( {{{\rm{S}}^{2 - }}} \right){{\left( {{{\rm{H}}^ + }} \right)}^2}}}{{\left( {{{\rm{H}}_2}{\rm{S}}} \right)}}}\\{1.0 \cdot {{10}^{ - 26}} = \frac{{\left( {{{\rm{S}}^{2 - }}} \right){{(0.3{\rm{M}})}^2}}}{{0.1{\rm{M}}}}}\end{array}\)
\(\begin{array}{*{20}{c}}{\left( {{{\rm{S}}^{2 - }}} \right) = \frac{{1.0 \cdot {{10}^{ - 27}}}}{{0.09}}}\\{\left( {{{\rm{S}}^{2 - }}} \right) = 1.11 \cdot {{10}^{ - 26}}}\end{array}\)
\(\begin{array}{*{20}{c}}{{{\rm{K}}_{sp}} = 7 \cdot {{10}^{ - 29}}}\\{{{\rm{K}}_{sp}} = \left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right)\left( {{{\rm{S}}^{2 - }}} \right)}\end{array}\)
\(7 \cdot {10^{ - 29}} = \left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right) \cdot 1.11 \cdot {10^{ - 26}}\)
\(\begin{array}{*{20}{c}}{\left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right) = \frac{{7 \cdot {{10}^{ - 29}}}}{{1.11 \cdot {{10}^{ - 26}}}}}\\{\left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right) = 6.31 \cdot {{10}^{ - 3}}}\end{array}\)
The concentration of \({\rm{P}}{{\rm{b}}^{2 + }}\)ions left in the solution is \(6.31 \cdot {10^{ - 3}}\).
Find the Concentration of \(F{e^{2 + }}\):
\(\begin{array}{*{20}{c}}{{{\rm{K}}_{sp}} = 3.7 \cdot {{10}^{ - 19}}}\\{{{\rm{K}}_{sp}} = \left( {{\rm{F}}{{\rm{e}}^{2 + }}} \right)\left( {{{\rm{S}}^{2 - }}} \right)}\end{array}\)
\(3.7 \cdot {10^{ - 19}} = \left( {{\rm{F}}{{\rm{e}}^{2 + }}} \right) \cdot 1.11 \cdot {10^{ - 26}}\)
\(\begin{array}{*{20}{c}}{\left( {{\rm{F}}{{\rm{e}}^{2 + }}} \right) = \frac{{3.7 \cdot {{10}^{ - 19}}}}{{1.11 \cdot {{10}^{ - 26}}}}}\\{\left( {{\rm{F}}{{\rm{e}}^{2 + }}} \right) = 3.33 \cdot {{10}^7}}\end{array}\)
The concentration of \({\rm{F}}{{\rm{e}}^{2 + }}\) ions left in the solution is \(3.33 \cdot {10^7}\).
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