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Mobile App Chapter 15: Q15.3-95E (page 878)
What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for HSO4 − = 1.2×10–2.
Short Answer
The molar solubility of BaSO4 is\(4.68 \cdot {10^{ - 7}}{\rm{M}}\).
Step by step solution
Calculate the concentration of [SO42-]:
We have a 0.250 M solution of NaHSO4 \(\left( {{K_a}{\rm{\;for\;HSO}}_4^ - {\rm{is\;}}1.2 \cdot {{10}^{ - 2}}} \right)\).
The reaction of dissociation of HSO4-
Calculate the concentration of [SO42-]:
\(\begin{array}{*{20}{c}}{{K_a} = \frac{{\left[ {{H^ + }} \right] \cdot \left[ {SO_4^{2 - }} \right]}}{{\left[ {HSO_4^ - } \right]}}}\\{1.2 \cdot {{10}^{ - 2}} = \frac{{x \cdot x}}{{0.250 - x}}}\\{{x^2} = 0.003 - 0.012x}\\{0 = {x^2} + 0.012x - 0.003}\\{x = 0.0491}\\{\left[ {SO_4^{2 - }} \right] = x = 0.0491}\end{array}\)
Calculate the molar solubility of BaSO4
\({\rm{BaS}}{{\rm{O}}_4}({\rm{s}}) \to {\rm{B}}{{\rm{a}}^{2 + }}({\rm{aq}}) + {\rm{S}}{{\rm{O}}_4}^{2 - }({\rm{aq}})\)
- \({K_{sp}}{\rm{\;of\;BaS}}{{\rm{O}}_4}{\rm{\;is\;}}2.3 \cdot {10^{ - 8}}\)
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {B{a^{2 + }}} \right] \cdot \left[ {S{O_4}^{2 - }} \right]}\\{\left[ {B{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{\left[ {S{O_4}^{2 - }} \right]}}}\\{ = \frac{{2.3 \cdot {{10}^{ - 8}}}}{{0.0491}}}\\{ = 4.68 \cdot {{10}^{ - 7}}{\rm{M}}}\end{array}\)
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