Question

$Ba{SO}_4\left(s\right)\rightleftharpoons {Ba}^{2+}\left(aq\right)+{SO}_4^{2-}\left(aq\right)$
Whai is $K_{sp}$ for above reaction?

• A. $\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$
• B. $\frac{\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]}{Ba{SO}_4}$
• C. $\frac{\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]}{\left[Ba{SO}_4\right]}$
• D. None

Answer 1

Answer: (A)

$\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$

For the reaction $Ba{SO}_4\left(s\right)\rightleftharpoons {Ba}^{2+}\left(aq\right)+{SO}_4^{2-}\left(aq\right)$, the expression for the solubility product $K_{sp}$ is $\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$. It is the product of the molar concentrations of cations and anions. The concentration of each ion is raised to the power of its coefficient in the balanced equilibrium equation.