To calculate the solubility product (Ksp) of A\(_{1/2}\)CrO\(_4\), given its solubility and molecular weight, we can follow these steps:
1. **Understand the Dissociation in Water:**
Suppose A\(_{1/2}\)CrO\(_4\) dissociates in water as follows:
\[ A_{1/2}CrO_4 \rightarrow \frac{1}{2}A^{2+} + CrO_4^{2-} \]
Given this stoichiometry, we can set up the dissociation where \(s\) is the molar solubility of A\(_{1/2}\)CrO\(_4\).
2. **Calculate the Molar Solubility:**
Convert the given solubility in mg/L to mol/L.
\[ \text{Solubility in mg/L} = 25 \text{ mg/L} \]
\[ \text{Molecular weight (MW)} = 331.7 \text{ g/mol} \]
Convert mg to g and then use the molecular weight to find moles:
\[ \text{Solubility in g/L} = 25 \text{ mg/L} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.025 \text{ g/L} \]
\[ \text{Molar solubility (s) in mol/L} = \frac{0.025 \text{ g/L}}{331.7 \text{ g/mol}} \approx 7.54 \times 10^{-5} \text{ mol/L} \]
3. **Set Up the Expression for Ksp:**
Based on the dissociation stoichiometry:
\[ A_{1/2}CrO_4 \rightarrow \frac{1}{2}A^{2+} + CrO_4^{2-} \]
For every 1 mole of A\(_{1/2}\)CrO\(_4\) that dissolves, 0.5 moles of \(A^{2+}\) and 1 mole of \(CrO_4^{2-}\) are produced. Therefore, if the molar solubility is \(s\):
\[
\begin{cases}
[A^{2+}] &= \frac{1}{2} \times s \\
[CrO_4^{2-}] &= s \\
\end{cases}
\]
Hence,
\[
K_{sp} = [A^{2+}] \cdot [CrO_4^{2-}]
\]
Using the values:
\[
[A^{2+}] = \frac{1}{2} \times 7.54 \times 10^{-5} \text{ mol/L} = 3.77 \times 10^{-5} \text{ mol/L}
\]
\[
[CrO_4^{2-}] = 7.54 \times 10^{-5} \text{ mol/L}
\]
Therefore:
\[
K_{sp} = (3.77 \times 10^{-5}) \times (7.54 \times 10^{-5})
\]
\[
K_{sp} \approx 2.84 \times 10^{-9}
\]
So, the solubility product (Ksp) of A\(_{1/2}\)CrO\(_4\) is approximately \(2.84 \times 10^{-9}\).
The solubility of A_{1/2}*Cr*O_{4} is 25 mg/lit. Calculate solubility product, if molecular weight in 331.7 g/mol.
1 answer