JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The solubility of \(AgCN\) in a buffer solution of \(pH =3\) is \(x\). The value of \(x\) is [Assume : No cyano complex is formed; \(K _{ sp }( AgCN )\) \(=2.2 \times 10^{-16}\) and \(\left. K _{ a }( HCN )=6.2 \times 10^{-10}\right]\)
- A \(0.625 \times 10^{-6}\)
- B \(1.9 \times 10^{-5}\)
- C \(2.2 \times 10^{-16}\)
- D \(1.6 \times 10^{-6}\)
Answer & Solution
Correct Answer
(B) \(1.9 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(\frac{ K _{ sp }}{ Ka }=\frac{ s ^{2}}{\left( H ^{+}\right)} ; \quad s =\sqrt{\frac{ K _{ sp }}{ K _{ a }}\left( H ^{+}\right)}\) \(s=\sqrt{\frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}} \times 10^{-3}}\) \(s =1.9 \times 10^{-5}\)
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