What [I] is required to just begin precipitation of a BiI3(s) from a solution of 0.62 M Bi3+(aq)? (Suggestion: Write the dissolving equation and Ksp expression for BiI3 before starting this problem.)
a. 1.3 x 10-18 M
b. 7.9 x 10-7 M
c. 2.2 x 10-6 M
d. 1.1 x 10-6 M
e. 1.1 x 10-9 M
The correct answer and explanation is :
To determine the minimum iodide ion concentration ([I^-]) required to begin the precipitation of (\text{BiI}_3 (s)), we need to analyze the solubility equilibrium of bismuth(III) iodide.
Step 1: Write the Dissociation Equation and (K_{sp}) Expression
The dissolution of (\text{BiI}_3) in water is represented by:
[
\text{BiI}_3 (s) \rightleftharpoons \text{Bi}^{3+} (aq) + 3\text{I}^- (aq)
]
The solubility product constant ((K_{sp})) expression for this equilibrium is:
[
K_{sp} = [\text{Bi}^{3+}][\text{I}^-]^3
]
Given that precipitation begins when the ionic product equals the solubility product ((K_{sp})), we set up the equilibrium condition:
[
K_{sp} = (0.62)[I^-]^3
]
Step 2: Solve for ([I^-])
From reference data, the solubility product constant for (\text{BiI}_3) is:
[
K_{sp} = 8.3 \times 10^{-19}
]
Setting up the equation:
[
8.3 \times 10^{-19} = (0.62)[I^-]^3
]
Solving for ([I^-]):
[
[I^-]^3 = \frac{8.3 \times 10^{-19}}{0.62}
]
[
[I^-]^3 = 1.34 \times 10^{-18}
]
[
[I^-] = \sqrt[3]{1.34 \times 10^{-18}}
]
[
[I^-] \approx 1.1 \times 10^{-6} \text{ M}
]
Step 3: Select the Correct Answer
From the given answer choices, the closest match is:
[
\mathbf{d. \ 1.1 \times 10^{-6} \ M}
]
Explanation
Precipitation begins when the ionic product ([\text{Bi}^{3+}][\text{I}^-]^3) reaches or exceeds the solubility product (K_{sp}). Since the solution contains (0.62 \ M) (\text{Bi}^{3+}), we calculated the minimum iodide concentration required for precipitation. Using the (K_{sp}) value and solving for ([I^-]), we found that the threshold concentration is approximately (1.1 \times 10^{-6} \ M), which corresponds to option d.
This means that as soon as the iodide ion concentration reaches this value, (\text{BiI}_3) will begin to precipitate out of solution. If the iodide concentration is lower, the solution remains unsaturated, and no precipitation occurs.