Predict expression for \( \alpha \) in terms of \( K_{eq} \) and concentration
C: \( A_2B_3 \, (aq) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq) \)
C: \( A_2B_3 \, (aq) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq) \)
Show Hint
- \( \left( \frac{K_{eq}}{5C^4} \right)^{1/5} \)
- \( \left( \frac{K_{eq}}{108C^4} \right)^{1/5} \)
- \( \left( \frac{4K_{eq}}{5C^4} \right)^{1/5} \)
- \( \left( \frac{9K_{eq}}{5C^4} \right)^{1/5} \)
The Correct Option is B
Solution and Explanation
Step 1: Write the reaction and assume initial concentration of \(A_2B_3\) is \(C\), and degree of dissociation is \(\alpha\).
\[ A_2B_3(aq)\rightleftharpoons 2A^{3+}(aq)+3B^{2-}(aq) \]
At equilibrium:
\[ [A_2B_3]=C(1-\alpha) \] \[ [A^{3+}]=2C\alpha \] \[ [B^{2-}]=3C\alpha \]
Step 2: Write the expression for equilibrium constant \(K_{eq}\).
\[ K_{eq}=\frac{[A^{3+}]^2[B^{2-}]^3}{[A_2B_3]} \]
Substituting the equilibrium concentrations:
\[ K_{eq}=\frac{(2C\alpha)^2(3C\alpha)^3}{C(1-\alpha)} \]
Step 3: Simplify the expression.
\[ K_{eq}=\frac{4C^2\alpha^2 \cdot 27C^3\alpha^3}{C(1-\alpha)} \] \[ K_{eq}=\frac{108C^5\alpha^5}{C(1-\alpha)} \] \[ K_{eq}=\frac{108C^4\alpha^5}{1-\alpha} \]
Step 4: For a weak electrolyte, \(\alpha\) is very small, so \(1-\alpha \approx 1\).
Therefore, \[ K_{eq}=108C^4\alpha^5 \] \[ \alpha^5=\frac{K_{eq}}{108C^4} \] \[ \alpha=\left(\frac{K_{eq}}{108C^4}\right)^{1/5} \]
Final Answer:
\[ \boxed{\alpha=\left(\frac{K_{eq}}{108C^4}\right)^{1/5}} \]
So, the correct option is (B)
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