JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Sulphurous acid \(\left( H _{2} SO _{3}\right)\) has \(Ka _{1}=1.7 \times 10^{-2}\) and \(Ka _{2}=6.4 \times 10^{-8} .\) The \(pH\) of \(0.588 \,M\, H _{2} SO _{3}\) is ..... . (Round off to the Nearest Integer)
- A \(2\)
- B \(1\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(H _{2} SO _{3}\) [Dibasic acid] \(c=0.588 \,M\) \(\Rightarrow \quad pH\) of solution \(P\) due to First dissociation only since \(K _{ a },>> Ka _{2}\) \(\Rightarrow\) First dissociation of \(H _{2} SO _{3}\)…
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