Question

1. what is the molar solubility in water of pbi₂? (the ( k_{sp} ) for ( \text{pbi}_2 ) is ( 3.2 \times 10^{-8} ))

a. ( 3.2 \times 10^{-8} , \text{m} )
b. ( 8.0 \times 10^{-8} , \text{m} )
c. ( (1.6)^{1/3} \times 10^{-8} , \text{m} )
d. ( (1.6)^{1/5} \times 10^{-8} , \text{m} )
e. ( 2 \times 10^{-3} , \text{m} )

Explanation:

Step1: Write the dissolution equation

The dissolution of $\ce{PbI2}$ in water is: $\ce{PbI2(s) <=> Pb^{2+}(aq) + 2I^-(aq)}$. Let the molar solubility of $\ce{PbI2}$ be $s$. Then, $[\ce{Pb^{2+}}] = s$ and $[\ce{I^-}] = 2s$.

Step2: Write the $K_{sp}$ expression

The solubility product constant expression for $\ce{PbI2}$ is $K_{sp} = [\ce{Pb^{2+}}][\ce{I^-}]^2$. Substituting the concentrations in terms of $s$, we get $K_{sp} = (s)(2s)^2 = 4s^3$.

Step3: Solve for $s$

We know that $K_{sp} = 3.2 \times 10^{-8}$. So, $4s^3 = 3.2 \times 10^{-8}$. First, divide both sides by 4: $s^3 = \frac{3.2 \times 10^{-8}}{4} = 8.0 \times 10^{-9}$. Then, take the cube root of both sides: $s = \sqrt[3]{8.0 \times 10^{-9}}$. Since $\sqrt[3]{8.0 \times 10^{-9}} = \sqrt[3]{8} \times \sqrt[3]{10^{-9}} = 2 \times 10^{-3}$? Wait, no, wait. Wait, $8.0 \times 10^{-9}$: $\sqrt[3]{8} = 2$, $\sqrt[3]{10^{-9}} = 10^{-3}$, so $s = 2 \times 10^{-3}$? Wait, no, wait, let's recalculate. Wait, $K_{sp} = 3.2 \times 10^{-8}$, so $4s^3 = 3.2 \times 10^{-8}$, so $s^3 = \frac{3.2 \times 10^{-8}}{4} = 8.0 \times 10^{-9}$? Wait, $3.2 \div 4 = 0.8$, so $0.8 \times 10^{-8} = 8.0 \times 10^{-9}$. Then $\sqrt[3]{8.0 \times 10^{-9}} = \sqrt[3]{8} \times \sqrt[3]{10^{-9}} = 2 \times 10^{-3}$? Wait, no, wait, $8.0 \times 10^{-9} = 8 \times 10^{-9}$, so cube root of 8 is 2, cube root of $10^{-9}$ is $10^{-3}$, so $s = 2 \times 10^{-3}$? Wait, but let's check the options. Option e is $2 \times 10^{-3}\ \text{M}$. Wait, but let's recheck the calculation. Wait, $K_{sp} = 3.2 \times 10^{-8}$, so $4s^3 = 3.2 \times 10^{-8}$, so $s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} = 8.0 \times 10^{-9}$. Then $s = \sqrt[3]{8.0 \times 10^{-9}} = 2.0 \times 10^{-3}\ \text{M}$. Yes, that's correct. Wait, but let's check the options again. Option e is $2 \times 10^{-3}\ \text{M}$. So that's the answer. Wait, but let's go through the steps again.

Wait, dissolution of $\ce{PbI2}$: $\ce{PbI2 -> Pb^{2+} + 2I^-}$. Let $s$ be the molar solubility (moles of $\ce{PbI2}$ dissolved per liter). Then $[\ce{Pb^{2+}}] = s$, $[\ce{I^-}] = 2s$. $K_{sp} = [\ce{Pb^{2+}}][\ce{I^-}]^2 = s(2s)^2 = 4s^3$. Given $K_{sp} = 3.2 \times 10^{-8}$, so $4s^3 = 3.2 \times 10^{-8}$. Divide both sides by 4: $s^3 = 0.8 \times 10^{-8} = 8.0 \times 10^{-9}$. Take cube root: $s = \sqrt[3]{8.0 \times 10^{-9}} = 2.0 \times 10^{-3}\ \text{M}$. So the molar solubility is $2 \times 10^{-3}\ \text{M}$, which is option e.

Answer:

e. $2 \times 10^{-3}\ \text{M}$