Question

The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calculate the Kspvalue, take the negative log and enter the (pKsp) value with 2 decimal places.

Answer 1

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$  

The equation for the ionization of magnesium phosphate is given as:

$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

 When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Thus S = 0.173 g/L or $\frac{0.173g/L}{262.8g/mol}=0.00065mol/L$

$K_{sp}=(3S)^3\times (2S)^2$

$K_{sp}=108S^5$

$K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}$

$pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90$