Question

What is the molar solubility (s) of ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$ in terms of ${\mathrm{K}}_{\mathrm{sp}}$?

Answer 1

Step 1: Solubility product:

• ${\mathrm{K}}_{\mathrm{sp}}$ can be called a solubility product or the solubility constant.
• Consider, “the equilibrium between the undissolved solid and the ions in a saturated solution”. This can be represented by the equation:

${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)\stackrel{\mathrm{Saturated}\mathrm{solution}\mathrm{in}\mathrm{water}}{\rightleftharpoons }3{\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+{2{\mathrm{PO}}_4}^{3-}\left(\mathrm{aq}\right)$

• The equilibrium constant is given by equation:

${\mathrm{K}}_{\mathrm{sp}}=\frac{{\left[{\mathrm{Ba}}^{2+}\right]}^3{\left[{{\mathrm{PO}}_4}^{3-}\right]}^2}{\left[{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2\right]}$

• For a pure solid substance, the concentration remains constant and it can be written as:

${\mathrm{K}}_{\mathrm{sp}}=\mathrm{K}\left[{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2\right]={\left[{\mathrm{Ba}}^{2+}\right]}^3{\left[{{\mathrm{PO}}_4}^{3-}\right]}^2$

Step 2: Calculation of Solubility product:

• For solid barium phosphate in equilibrium with its saturated solution, the product of the concentrations of barium and phosphate ions is equal to its solubility product constant.
• The concentrations of the two ions will be equal to the molar solubility of the barium phosphate.
• If molar solubility is $\mathrm{S}$, then the solubility product will be:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ba}}^{2+}\right]}^3{\left[{{\mathrm{PO}}_4}^{3-}\right]}^2 \\ {\mathrm{K}}_{\mathrm{sp}}={\left[3\mathrm{S}\right]}^3{\left[2\mathrm{S}\right]}^2 \end{gathered}$$