Question

Concentration of $B{a}^{2+}$ ions in solution $A$ is:

• A. $3.5\times {10}^{-18}M$
• B. $4.7\times {10}^{-15}M$
• C. $2.5\times {10}^{-12}M$
• D. $4\times {10}^{-13}M$

Answer 1

The correct option is C $2.5\times {10}^{-12}M$

The mass of $K_{2}S{O}_4$ in the sample is 29.29 g.
The molecular weight of $K_{2}S{O}_4$ is 174 g/mol.
The number of moles of $K_{2}S{O}_4$ is equal to the number of moles of $S{O}_4^{2-}$ ions. It is equal to $\frac{29.29}{174}$
The mass of the solution is $900g+29.29g$ The density of the solution is $1.24g/ml$

Hence, the volume of the solution is $\frac{900+29.29}{1.24}\times \frac{1}{1000}$

Thus sulfate ion concentration is $\left[S{O}_4^{2-}\right]=\frac{\frac{29.29}{174}}{\frac{900+29.29}{1.24}\times \frac{1}{1000}}=0.224$

The expression for the solubility product is $K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
$\left[B{a}^{2+}\right]=\frac{K_{sp}}{\left[S{O}_4^{2-}\right]}$

Substitute values in the above expression.
$\left[B{a}^{2+}\right]=\frac{5.6\times {10}^{-13}}{0.244}=2.5\times {10}^{-12}M$