JEE MainChemistryModeratePhysical Chemistry

In a solution containing 0.01 M HCl and 0.1 M H_2CO_3, ratio
of produced from strong acid & weak acid respectively is x: 1.
Report your answer as x/500. Given: of H_2CO_3 are 4 × 10^-7 & 4 ×
10^-11 respectively.

Show answer and solution

4 \times 10^{- 7} = \frac{0.01 \times \left\lbrack {HCO}_{3}^{-} \right\rbrack}{0.1}

\Rightarrow

[H+]~from weak acid~ = [HCO_3^-] = 4 × 10^-6 M

\therefore

Ratio =

\frac{0.01}{4 \times 10^{- 6}} = \frac{2500}{1} = x

So, answer

= \frac{x}{500} = x

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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