Question

Conductivity of a saturated solution of $Mg_3(PO_4)_2$ is $1.2 \times 10^{-5} S cm^{-1}$. Limiting equivalent conductivities of $MgCl_2, Na_3PO_4$ and $NaCl$ are 160, 140 and 100 $\frac{S cm^2}{eq}$ respectively. Find $K_{sp}$ of $Mg_3(PO_4)_2$

Answer 1

$1.08 \times 10^{-23}$

Answer 2

To find the $K_{sp}$ of $Mg_3(PO_4)_2$, we need to follow these steps:

1. Calculate the limiting equivalent conductivity ($\Lambda_{eq}^\circ$) of $Mg_3(PO_4)_2$.
2. Use the conductivity of the saturated solution and $\Lambda_{eq}^\circ$ to find the molar solubility (S) of $Mg_3(PO_4)_2$.
3. Calculate $K_{sp}$ from the molar solubility (S).

Step 1: Calculate $\Lambda_{eq}^\circ$ for $Mg_3(PO_4)_2$

According to Kohlrausch's Law of independent migration of ions, the limiting equivalent conductivity of an electrolyte is the sum of the limiting equivalent conductivities of its constituent ions. We are given:

$\Lambda_{eq}^\circ(MgCl_2) = \lambda_{eq}^\circ(Mg^{2+}) + \lambda_{eq}^\circ(Cl^-) = 160 \frac{S cm^2}{eq}$

$\Lambda_{eq}^\circ(Na_3PO_4) = \lambda_{eq}^\circ(Na^+) + \lambda_{eq}^\circ(PO_4^{3-}) = 140 \frac{S cm^2}{eq}$

$\Lambda_{eq}^\circ(NaCl) = \lambda_{eq}^\circ(Na^+) + \lambda_{eq}^\circ(Cl^-) = 100 \frac{S cm^2}{eq}$

We need $\Lambda_{eq}^\circ(Mg_3(PO_4)_2) = \lambda_{eq}^\circ(Mg^{2+}) + \lambda_{eq}^\circ(PO_4^{3-})$. This can be obtained by:

$\Lambda_{eq}^\circ(Mg_3(PO_4)_2) = \Lambda_{eq}^\circ(MgCl_2) + \Lambda_{eq}^\circ(Na_3PO_4) - \Lambda_{eq}^\circ(NaCl)$

$\Lambda_{eq}^\circ(Mg_3(PO_4)_2) = 160 + 140 - 100 = 200 \frac{S cm^2}{eq}$

Step 2: Calculate the molar solubility (S) of $Mg_3(PO_4)_2$

The relationship between conductivity ($\kappa$), equivalent conductivity ($\Lambda_{eq}$), and equivalent concentration ($C_{eq}$) is:

$\kappa = \frac{\Lambda_{eq} \times C_{eq}}{1000}$

where $\kappa$ is in $S cm^{-1}$, $\Lambda_{eq}$ is in $S cm^2 eq^{-1}$, and $C_{eq}$ is in $eq/L$. For a sparingly soluble salt like $Mg_3(PO_4)_2$ in a saturated solution, we can approximate $\Lambda_{eq} \approx \Lambda_{eq}^\circ$.

Given $\kappa = 1.2 \times 10^{-5} S cm^{-1}$.

$C_{eq} = \frac{\kappa \times 1000}{\Lambda_{eq}^\circ}$

$C_{eq} = \frac{1.2 \times 10^{-5} S cm^{-1} \times 1000}{200 S cm^2 eq^{-1}}$

$C_{eq} = \frac{1.2 \times 10^{-2}}{200} = 6 \times 10^{-5} eq/L$

Now, relate the equivalent concentration ($C_{eq}$) to the molar solubility (S). For $Mg_3(PO_4)_2$, the total positive charge is $3 \times (+2) = +6$, and the total negative charge is $2 \times (-3) = -6$. Therefore, 1 mole of $Mg_3(PO_4)_2$ corresponds to 6 equivalents. So, $C_{eq} = 6S$

$6 \times 10^{-5} eq/L = 6S$

$S = 1.0 \times 10^{-5} mol/L$

Step 3: Calculate $K_{sp}$ of $Mg_3(PO_4)_2$

The dissolution equilibrium for $Mg_3(PO_4)_2$ is:

$Mg_3(PO_4)_2(s) \rightleftharpoons 3Mg^{2+}(aq) + 2PO_4^{3-}(aq)$

If the molar solubility is S, then at equilibrium:

$[Mg^{2+}] = 3S$

$[PO_4^{3-}] = 2S$

The solubility product constant ($K_{sp}$) is given by:

$K_{sp} = [Mg^{2+}]^3 [PO_4^{3-}]^2$

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

$K_{sp} = (27S^3)(4S^2)$

$K_{sp} = 108S^5$

Substitute the value of S:

$K_{sp} = 108 \times (1.0 \times 10^{-5})^5$

$K_{sp} = 108 \times (1.0^5 \times (10^{-5})^5)$

$K_{sp} = 108 \times 1.0 \times 10^{-25}$

$K_{sp} = 1.08 \times 10^{-23}$

The final answer is $1.08 \times 10^{-23}$.

Explanation of the solution:

1. Kohlrausch's Law: $\Lambda_{eq}^\circ(Mg_3(PO_4)_2) = \Lambda_{eq}^\circ(MgCl_2) + \Lambda_{eq}^\circ(Na_3PO_4) - \Lambda_{eq}^\circ(NaCl) = 160 + 140 - 100 = 200 \frac{S cm^2}{eq}$.
2. Solubility from Conductivity: Use $\kappa = \frac{\Lambda_{eq}^\circ \times C_{eq}}{1000}$. Given $\kappa = 1.2 \times 10^{-5} S cm^{-1}$, $C_{eq} = \frac{1.2 \times 10^{-5} \times 1000}{200} = 6 \times 10^{-5} eq/L$. Since $1 \text{ mol of } Mg_3(PO_4)_2 = 6 \text{ eq}$, molar solubility $S = \frac{C_{eq}}{6} = \frac{6 \times 10^{-5}}{6} = 1.0 \times 10^{-5} mol/L$.
3. $K_{sp}$ Calculation: For $Mg_3(PO_4)_2 \rightleftharpoons 3Mg^{2+} + 2PO_4^{3-}$, $K_{sp} = [Mg^{2+}]^3 [PO_4^{3-}]^2 = (3S)^3 (2S)^2 = 108S^5$. Substitute $S = 1.0 \times 10^{-5} mol/L$: $K_{sp} = 108 \times (1.0 \times 10^{-5})^5 = 108 \times 10^{-25} = 1.08 \times 10^{-23}$.