JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The \(K _{ sp }\) for bismuth sulphide \(\left( Bi _{2} S _{3}\right)\) is \(1.08 \times 10^{-73}\). The solubility of \(Bi _{2} S _{3}\) in \(mol \,L ^{-1}\) at \(298 \,K\) is ......
- A \(1.0 \times 10^{-15}\)
- B \(2.7 \times 10^{-12}\)
- C \(3.2 \times 10^{-10}\)
- D \(4.2 \times 10^{-8}\)
Answer & Solution
Correct Answer
(A) \(1.0 \times 10^{-15}\)
Step-by-step Solution
Detailed explanation
\(Bi _{2} S _{3} \rightleftharpoons 2 Bi ^{3+}+3 S ^{2-}\) \(\,\quad\quad\quad\quad2s\quad\quad\quad3s\) \(k _{ sp } =(2 s )^{2}(3 s )^{3}\) \(=4 s ^{2} \times 27( s )^{3}\) \(=108( s )^{5}\) \(( s )^{5} =\frac{1.08 \times 10^{-73}}{108}\) \(\Rightarrow s =10^{-15}\)
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