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  • Sulphurous acid   has <img alt="Ka_1 = 1.7 \times 10^{-2 }\text{and} Ka_2 = 6.4 \times 10 ^{-8}" src="/latex-image/?Ka_1%20%3D%201.7%20%5Ctimes%2010%5E%7B-2%20%7D%5Ctext%7Band%7D

Sulphurous acid  (H_2SO_3) has Ka_1 = 1.7 \times 10^{-2 }\text{and} Ka_2 = 6.4 \times 10 ^{-8}. The pH of 0.588 M (H_2SO_3) is ________ (Round off to the nearest integer)
 

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best_answer

\Rightarrow \left[\mathrm{H}^{\oplus}\right]=\mathrm{x}=\frac{-1.7+\sqrt{(1.7)^{2}+4 \times 100 \times 1}}{2 \times 100}=0.09186

\text { Therefore pH of sol. is : } \mathrm{pH}=-\log \left[\mathrm{H}^{\oplus}\right]

\Rightarrow \quad \mathrm{pH}=-\log (0.09186)=1.036 \simeq 1

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Kuldeep Maurya

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