Question
Calculate the solubility of Al(OH), in g/L. in water at 25°C if K 8.5×10 Given Mol wt. = 78 g mol.
Answers
GPT-4o mini
To find the solubility of Al(OH)₃ in g/L, we first need to understand the solubility product constant (\(K_{sp}\)) and the dissociation of Al(OH)₃ in water.
The compound Al(OH)₃ dissociates in water as follows:
\[
\text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^- (aq)
\]
Let \(s\) be the solubility of Al(OH)₃ in mol/L. According to the dissociation:
- The concentration of \(\text{Al}^{3+}\) ions will be \(s\).
- The concentration of \(\text{OH}^-\) ions will be \(3s\).
The solubility product constant (\(K_{sp}\)) expression for this reaction is:
\[
K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3
\]
Substituting the concentrations in terms of \(s\):
\[
K_{sp} = s (3s)^3 = s \cdot 27s^3 = 27s^4
\]
Given \(K_{sp} = 8.5 \times 10^{-6}\), we have:
\[
27s^4 = 8.5 \times 10^{-6}
\]
Now, solving for \(s\):
\[
s^4 = \frac{8.5 \times 10^{-6}}{27}
\]
Calculating the right side:
\[
s^4 = \frac{8.5 \times 10^{-6}}{27} \approx 3.1481 \times 10^{-7}
\]
Next, we take the fourth root of both sides to find \(s\):
\[
s = (3.1481 \times 10^{-7})^{1/4}
\]
Calculating the fourth root:
\[
s \approx 0.0178 \, \text{mol/L}
\]
Now, to convert this concentration from mol/L to g/L, we use the molar mass of Al(OH)₃, which is 78 g/mol:
\[
\text{solubility in g/L} = s \times \text{molar mass} = 0.0178 \, \text{mol/L} \times 78 \, \text{g/mol}
\]
\[
\text{solubility in g/L} \approx 1.39 \, \text{g/L}
\]
Therefore, the solubility of Al(OH)₃ in water at 25°C is approximately **1.39 g/L**.
The compound Al(OH)₃ dissociates in water as follows:
\[
\text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^- (aq)
\]
Let \(s\) be the solubility of Al(OH)₃ in mol/L. According to the dissociation:
- The concentration of \(\text{Al}^{3+}\) ions will be \(s\).
- The concentration of \(\text{OH}^-\) ions will be \(3s\).
The solubility product constant (\(K_{sp}\)) expression for this reaction is:
\[
K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3
\]
Substituting the concentrations in terms of \(s\):
\[
K_{sp} = s (3s)^3 = s \cdot 27s^3 = 27s^4
\]
Given \(K_{sp} = 8.5 \times 10^{-6}\), we have:
\[
27s^4 = 8.5 \times 10^{-6}
\]
Now, solving for \(s\):
\[
s^4 = \frac{8.5 \times 10^{-6}}{27}
\]
Calculating the right side:
\[
s^4 = \frac{8.5 \times 10^{-6}}{27} \approx 3.1481 \times 10^{-7}
\]
Next, we take the fourth root of both sides to find \(s\):
\[
s = (3.1481 \times 10^{-7})^{1/4}
\]
Calculating the fourth root:
\[
s \approx 0.0178 \, \text{mol/L}
\]
Now, to convert this concentration from mol/L to g/L, we use the molar mass of Al(OH)₃, which is 78 g/mol:
\[
\text{solubility in g/L} = s \times \text{molar mass} = 0.0178 \, \text{mol/L} \times 78 \, \text{g/mol}
\]
\[
\text{solubility in g/L} \approx 1.39 \, \text{g/L}
\]
Therefore, the solubility of Al(OH)₃ in water at 25°C is approximately **1.39 g/L**.