JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 22)
Sulphurous acid (H2SO3) has Ka1 = 1.7 $$\times$$ 10$$-$$2 and Ka2 = 6.4 $$\times$$ 10$$-$$8. The pH of 0.588 M H2SO3 is __________. (Round off to the Nearest Integer).
Answer
1
Explanation
$$K{a_1}$$ of $${H_2}S{O_3} > > K{a_2}$$ of $${H_2}S{O_3}$$
$$ \therefore $$ The contribution of H+ from 2nd dissociation of H2SO3 can be neglected.
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$$ \Rightarrow $$ $${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$$
$$ \Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$$
$$ \Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha $$
$$ \Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$$
$$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$$
$$[{H^ + }] = c\alpha = 0.092$$
$$pH = - \log [{H^ + }]$$
$$ = 1.036$$
$$ \approx 1$$
$$ \therefore $$ The contribution of H+ from 2nd dissociation of H2SO3 can be neglected.
$$ \Rightarrow $$ $${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$$
$$ \Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$$
$$ \Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha $$
$$ \Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$$
$$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$$
$$[{H^ + }] = c\alpha = 0.092$$
$$pH = - \log [{H^ + }]$$
$$ = 1.036$$
$$ \approx 1$$
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