Sulphurous acid (\(\mathrm{H_2SO_3}\)) has \(\mathrm{Ka_1}\) = 1.7 \(\times\) \(10^{-2}\) and \(\mathrm{Ka_2}\) = 6.4 \(\times\) \(10^{-8}\). The pH of 0.588 M \(\mathrm{H_2SO_3}\) is __________. (Round off to the Nearest Integer).
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Correct Answer: 1
Solution and Explanation
\(\mathrm{Ka_1}=\frac{x^2}{0.588-x}=1.7\times 10^{-2}\Rightarrow x^2+0.017x-0.009996=0\).
\(x=\frac{-0.017+\sqrt{0.000289+0.039984}}{2}=\frac{-0.017+0.2007}{2}\approx 0.0918\ \mathrm{M}\).
pH = \(-\log(0.0918)\approx 1.04\approx 1\).
Answer: 1
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