Question

at a particular temperature, the solubility of in₂(so₄)₃ in water is 0.0045 m.
what is the value of ksp?

Explanation:

Step1: Write the dissolution equation

$$\text{In}_2(\text{SO}_4)_3(s) ightleftharpoons 2\text{In}^{3 +}(aq)+3\text{SO}_4^{2-}(aq)$$

Let the solubility of $\text{In}_2(\text{SO}_4)_3$ be $s$. Given $s = 0.0045\ M$.

Step2: Determine the concentrations of ions

From the stoichiometry of the dissolution equation, $[\text{In}^{3+}]=2s$ and $[\text{SO}_4^{2 -}]=3s$.
Substitute $s = 0.0045\ M$, so $[\text{In}^{3+}]=2\times0.0045\ M = 0.009\ M$ and $[\text{SO}_4^{2 -}]=3\times0.0045\ M=0.0135\ M$.

Step3: Calculate $K_{sp}$

The solubility - product constant expression for $\text{In}_2(\text{SO}_4)_3$ is $K_{sp}=[\text{In}^{3+}]^2[\text{SO}_4^{2 -}]^3$.
Substitute the values of $[\text{In}^{3+}]$ and $[\text{SO}_4^{2 -}]$ into the $K_{sp}$ expression:
$$K_{sp}=(0.009)^2\times(0.0135)^3$$
$$K_{sp}=8.1\times10^{- 5}\times2.460375\times10^{-6}$$
$$K_{sp}=2.0\times10^{-10}$$

Answer:

$2.0\times 10^{-10}$