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Step-by-Step Solution
Step 1: Identify the Known Data
ā¢ Solubility of CdSO4 in pure water, s = 8.0 \times 10^{-4}\,\text{mol L}^{-1} .
ā¢ Required: Solubility ( S_{1} ) of CdSO4 in 0.01\,\text{M} H2SO4.
ā¢ We assume S_{1} is much less than 0.01 , so the sulfate ion concentration from H2SO4 remains approximately 0.01\,\text{M} .
Step 2: Express the Solubility Product in Pure Water
If s is the molar solubility of CdSO4 in pure water, then
K_{sp} = [\text{Cd}^{2+}]\,[\text{SO}_{4}^{2-}] = s \times s = s^2.
Step 3: Calculate K_{sp} Using Data from Pure Water
Given s = 8.0 \times 10^{-4} mol Lā1,
K_{sp} = (8.0 \times 10^{-4})^2 = 64 \times 10^{-8}.
Step 4: Setup the Solubility Product in 0.01 M H2SO4
Let S_{1} be the solubility of CdSO4 in 0.01\,\text{M} H2SO4. Because S_{1} is much smaller than 0.01 , the total concentration of sulfate ions is approximately 0.01\,\text{M} . Hence:
K_{sp} = [\text{Cd}^{2+}]\,[\text{SO}_{4}^{2-}] = S_{1} \times 0.01.
Step 5: Solve for the New Solubility
We already know K_{sp} = 64 \times 10^{-8} . Thus,
64 \times 10^{-8} = S_{1} \times 0.01.
So,
S_{1} = \frac{64 \times 10^{-8}}{0.01} = 64 \times 10^{-6}\,\text{mol L}^{-1}.
Final Answer
The solubility of CdSO4 in 0.01\,\text{M} H2SO4 is 64 \times 10^{-6}\,\text{mol L}^{-1} .
