The solubility of Zn3(PO4)2 in water is = 5.9 x 10-5 g/L, with this given information calculate the solubility product constant (Ksp) of this salt. (Molar Mass Zn3(PO4)2 = 386.11g/mol).
The Correct Answer and Explanation is:
Let’s work through the problem step-by-step.
Given:
- Solubility of Zn₃(PO₄)₂ in water = $5.9 \times 10^{-5}$ g/L
- Molar mass of Zn₃(PO₄)₂ = 386.11 g/mol
Goal:
Calculate the solubility product constant, $K_{sp}$, for Zn₃(PO₄)₂.
Step 1: Convert solubility from g/L to mol/L (molarity)
Solubility (S) in mol/L is:
$$
S = \frac{\text{solubility in g/L}}{\text{molar mass in g/mol}} = \frac{5.9 \times 10^{-5}}{386.11} \approx 1.53 \times 10^{-7} \, \text{mol/L}
$$
So, the molar solubility $S$ is approximately $1.53 \times 10^{-7}$ mol/L.
Step 2: Write the dissociation equation for Zn₃(PO₄)₂
$$
\text{Zn}_3(\text{PO}_4)_2 (s) \leftrightarrow 3 \text{Zn}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq)
$$
Step 3: Express ion concentrations in terms of $S$
From the dissociation:
- For every 1 mole of Zn₃(PO₄)₂ dissolved:
- $3$ moles of Zn²⁺ are produced.
- $2$ moles of PO₄³⁻ are produced.
Therefore,
$$
[\text{Zn}^{2+}] = 3S = 3 \times 1.53 \times 10^{-7} = 4.59 \times 10^{-7} \, \text{mol/L}
$$
$$
[\text{PO}_4^{3-}] = 2S = 2 \times 1.53 \times 10^{-7} = 3.06 \times 10^{-7} \, \text{mol/L}
$$
Step 4: Write the expression for $K_{sp}$
$$
K_{sp} = [\text{Zn}^{2+}]^3 \times [\text{PO}_4^{3-}]^2
$$
Substitute the values:
$$
K_{sp} = (4.59 \times 10^{-7})^3 \times (3.06 \times 10^{-7})^2
$$
Calculate powers:
- $(4.59 \times 10^{-7})^3 = 4.59^3 \times (10^{-7})^3 = 96.59 \times 10^{-21} = 9.66 \times 10^{-20}$
- $(3.06 \times 10^{-7})^2 = 3.06^2 \times (10^{-7})^2 = 9.36 \times 10^{-14}$
Multiply:
$$
K_{sp} = (9.66 \times 10^{-20}) \times (9.36 \times 10^{-14}) = 9.04 \times 10^{-33}
$$
Final answer:
$$
\boxed{
K_{sp} \approx 9.0 \times 10^{-33}
}
$$
Explanation:
The solubility product constant, $K_{sp}$, quantifies the extent to which a sparingly soluble ionic compound dissolves in water. For Zn₃(PO₄)₂, the solubility is given in grams per liter, but to calculate $K_{sp}$, we first need to convert this to molar solubility (mol/L) using the molar mass.
Once molar solubility $S$ is known, we use the dissociation stoichiometry to find the concentration of each ion in solution. Zn₃(PO₄)₂ dissociates into 3 zinc ions and 2 phosphate ions per formula unit dissolved, so the ion concentrations are multiples of $S$.
The $K_{sp}$ expression incorporates the ion concentrations raised to the power of their coefficients in the balanced dissociation equation. The very small solubility of Zn₃(PO₄)₂ leads to extremely low ion concentrations, so $K_{sp}$ is correspondingly tiny, indicating the salt is very insoluble.
This approach can be generalized to any ionic salt: find molar solubility from mass solubility, relate ion concentrations via dissociation coefficients, then calculate $K_{sp}$ by raising concentrations to their stoichiometric powers and multiplying.
In this case, the very low $K_{sp}$ of about $9 \times 10^{-33}$ reflects the near insolubility of Zn₃(PO₄)₂ in water, consistent with phosphate salts’ typical behavior. This knowledge helps predict precipitation and dissolution equilibria in chemical and environmental systems involving zinc and phosphate ions.
