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Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

\(\begin{array}{l}(a)A{g_2}S{O_4}\\(b)PbB{r_2}\\(c)AgI\\(d)Ca{C_2}{O_4} \times {H_2}O\end{array}\)

Short Answer

Expert verified

The solubility product constant represented as Ksp can be defined as state in which a solid and its respective ions in given solution are in equilibrium. Its value indicates the extent to which a compound can undergo dissociation in water.

a)\(x = 1.5 \times 1{0^{ - 2}}M\)

b) \(x = 2 \times 1{0^{ - 2}}M\)

c)\(x = 9.2 \times 1{0^{ - 9}}M\)

d) \(x = 6.3 \times 1{0^{ - 5}}M\)

Step by step solution

01

Step 1: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\({\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)according to Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right){\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}{\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]^{\rm{2}}}\left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right]\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ \times x}}\;\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{3}}}\\{\rm{x = }}\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{\rm{1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = 1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)

02

Step 2: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\({\rm{PbB}}{{\rm{r}}_{\rm{2}}}{\rm{ = 4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)according to Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{PbB}}{{\rm{r}}_{\rm{2}}}} \right){\rm{ = 4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{P}}{{\rm{b}}^{{\rm{2 + }}}}} \right]{\left[ {{\rm{B}}{{\rm{r}}^{\rm{ - }}}} \right]^{\rm{2}}}\\{\rm{ = }}\;{\rm{x \times 4}}{{\rm{x}}^{\rm{2}}}{\rm{ = 4}}{{\rm{x}}^{\rm{3}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div }}}}{\rm{4}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt {{\rm{4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ \div 4}}} \\{\rm{x}}\;{\rm{ = }}\;{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)

03

Step 3: To  calculate the molar solubility of each of the following from its solubility product:

The solubility product of\(\;(AgI) = 8.5 \times 1{0^{ - 17}}\)according to KSP table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}{\rm{(AgI) = 8}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 17}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]\left[ {{{\rm{I}}^{\rm{ - }}}} \right]{\rm{ = }}\;{\rm{x \times x}}\;{\rm{ = }}\;{{\rm{x}}^{\rm{2}}}\\{\rm{x = }}\sqrt {{{\rm{K}}_{{\rm{sp}}}}} {\rm{ = }}\;\sqrt {{\rm{8}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 17}}}}} \\{\rm{x}}\;{\rm{ = }}\;{\rm{9}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{M}}\end{array}\)

04

Step 4: To calculate the molar solubility of each of the following from its solubility product:

The solubility product of\(Ca{C_2}{O_4} = 4 \times 1{0^{ - 9}}\)according to KSP table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{Ca}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}} \right){\rm{ = 4 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\left[ {{{\rm{C}}_{\rm{2}}}{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}} \right]{\rm{ = x \times x = }}{{\rm{x}}^{\rm{2}}}\\{\rm{x}}\;{\rm{ = }}\sqrt {{{\rm{K}}_{{\rm{sp}}}}} {\rm{ = }}\sqrt {{\rm{4 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}} \\x = 6.3 \times 1{0^{ - 5}}M\end{array}\)

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