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The Ksp of CaSO4 is 4.93× 10-5 M2. Calculate the solubility (in g/L) of CaSO4(s) in...

The Ksp of CaSO4 is 4.93× 10-5 M2. Calculate the solubility (in g/L) of CaSO4(s) in 0.350 M Na2SO4(aq) at 25 ℃.
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Answer #1
Concepts and reason

Solubility product a substance is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients.

First write the expression of solubility product for the substance CaSO4{\rm{CaS}}{{\rm{O}}_4} . Calculate the solubility of CaSO4{\rm{CaS}}{{\rm{O}}_4} by using the concentration of common ion present in Na2SO4{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} and CaSO4{\rm{CaS}}{{\rm{O}}_4} .

Fundamentals


‎The solubility product is a heterogeneous equilibrium constant, and it depends on temperature. The larger the solubility product of a substance, the higher is its solubility.

Expression of solubility product for CaSO4{\rm{CaS}}{{\rm{O}}_4} is as follows:

CaSO4Ca2++SO42SSSKsp=[Ca2+][SO42]Ksp=(S)(S)=S2\begin{array}{l}\\{\rm{CaS}}{{\rm{O}}_{\rm{4}}} \to {\rm{ C}}{{\rm{a}}^{2 + }}{\rm{ + S}}{{\rm{O}}_4}^{2 - }\\\\S{\rm{ }}S{\rm{ }}S\\\\{\rm{Ksp }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right]\\\\{\rm{Ksp }} = \left( S \right)\left( S \right) = {\rm{ }}{S^2}\\\end{array}

Ksp=[Ca2+][SO42]{\rm{Ksp }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right]

Concentration of Na2SO4{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} is 0.350 M.

So,

[Na2SO4]=[SO42]=0.350M\left[ {{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}} \right] = \left[ {{\rm{S}}{{\rm{O}}_4}^{2 - }} \right] = 0.350{\rm{ M}}

Substitute these values.

4.93×105=[Ca2+]×(0.350)[Ca2+]=1.41××104M\begin{array}{l}\\{\rm{4}}{\rm{.93}} \times {\rm{1}}{{\rm{0}}^{ - 5}}{\rm{ }} = \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }} \times \left( {0.350} \right)\\\\\left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]{\rm{ }}\, = 1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ M}}\\\end{array}

Solubility of is equal to solubility of Ca2+{\rm{C}}{{\rm{a}}^{2 + }} .

Thus, solubility of CaSO4{\rm{CaS}}{{\rm{O}}_4} in Na2SO4{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} is 1.41××104M1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ M}} .

SolubilityofCaSO4ing/mL=1.41××104(molL)×136(gmol)=0.019g/L\begin{array}{c}\\{\rm{Solubility of CaS}}{{\rm{O}}_4}{\rm{in g/mL = }}1.41 \times \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ }}\left( {\frac{{{\rm{mol}}}}{{\rm{L}}}} \right) \times 136\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right)\\\\ = 0.019{\rm{ g/L}}\\\end{array}

Ans:

Solubility of CaSO4{\rm{CaS}}{{\rm{O}}_4} in Na2SO4{\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} is 0.019g/L0.019{\rm{ g/L}} .

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