WBJEE · Chemistry · Ionic Equilibrium
The molar solubility (in mol \(\mathrm{L}^{-1}\) ) of a sparingly soluble salt \(M X_{4}\) is \({ }^{\prime} S\). The corresponding solubility product is \(K_{\mathrm{sp}} . S\) in terms of \('K_{\text {ip }}\)' s given by the relation
- A \(S=\left(\frac{K_{\mathrm{s} \rho}}{128}\right)^{1 / 4}\)
- B \(S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}\)
- C \(S=\left(256 K_{s p}\right)^{1 / 5}\)
- D \(S=\left(128 K_{\mathrm{sp}}\right)^{1 / 4}\)
Answer & Solution
Correct Answer
(B) \(S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}\)
Step-by-step Solution
Detailed explanation
\(\because\) For a compound \(A_{x}B_{y}\) \[ K_{s p}=x^{x} \cdot y^{y} \cdot S^{x+y} . \text { (where, } S=\text { solubility }) \] For \(M x_{4}\)…
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