Question
What is the equilibrium concentration of Ca^2+ ions in a saturated solution of calcium sulfate in water at 25°C?
Answers
GPT-4o mini
To find the equilibrium concentration of Ca²⁺ ions in a saturated solution of calcium sulfate (CaSO₄) at 25°C, we first look at the solubility equilibrium of calcium sulfate:
\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
The solubility product constant, \( K_{sp} \), expression for this equilibrium is:
\[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \]
At saturation, the concentration of Ca²⁺ ions is equal to the concentration of sulfate ions (\( \text{SO}_4^{2-} \)). Thus, we can let:
\[ [\text{Ca}^{2+}] = [\text{SO}_4^{2-}] = s \]
where \( s \) is the solubility of calcium sulfate.
Substituting this into the \( K_{sp} \) expression gives us:
\[ K_{sp} = s \cdot s = s^2 \]
The \( K_{sp} \) for calcium sulfate at 25°C is typically around \( 2.4 \times 10^{-5} \). Setting this equal to \( s^2 \) gives:
\[ s^2 = 2.4 \times 10^{-5} \]
To find \( s \), we take the square root of both sides:
\[ s = \sqrt{2.4 \times 10^{-5}} \]
Calculating this gives:
\[ s \approx 4.9 \times 10^{-3} \]
Therefore, since the concentration of \( \text{Ca}^{2+} \) ions is equal to the solubility \( s \), we find that the equilibrium concentration of Ca²⁺ ions in a saturated solution of calcium sulfate at 25°C is approximately:
\[ \text{[Ca}^{2+}] \approx 4.9 \times 10^{-3} \, \text{M} \]
\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
The solubility product constant, \( K_{sp} \), expression for this equilibrium is:
\[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \]
At saturation, the concentration of Ca²⁺ ions is equal to the concentration of sulfate ions (\( \text{SO}_4^{2-} \)). Thus, we can let:
\[ [\text{Ca}^{2+}] = [\text{SO}_4^{2-}] = s \]
where \( s \) is the solubility of calcium sulfate.
Substituting this into the \( K_{sp} \) expression gives us:
\[ K_{sp} = s \cdot s = s^2 \]
The \( K_{sp} \) for calcium sulfate at 25°C is typically around \( 2.4 \times 10^{-5} \). Setting this equal to \( s^2 \) gives:
\[ s^2 = 2.4 \times 10^{-5} \]
To find \( s \), we take the square root of both sides:
\[ s = \sqrt{2.4 \times 10^{-5}} \]
Calculating this gives:
\[ s \approx 4.9 \times 10^{-3} \]
Therefore, since the concentration of \( \text{Ca}^{2+} \) ions is equal to the solubility \( s \), we find that the equilibrium concentration of Ca²⁺ ions in a saturated solution of calcium sulfate at 25°C is approximately:
\[ \text{[Ca}^{2+}] \approx 4.9 \times 10^{-3} \, \text{M} \]