JEE Advanced 2022 · Paper 1 · Chemistry Q3

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ChemistryNumerical+3Question 39 of 54

A solution is prepared by mixing 0.01 mol each of $\ce{H2CO3}$ , $\ce{NaHCO3}$ , $\ce{Na2CO3}$ , and NaOH in 100 mL of water. $p$ H of the resulting solution is _____.

[Given: $pK_{a1}$ and $pK_{a2}$ of $\ce{H2CO3}$ are 6.37 and 10.32, respectively; log 2 = 0.30]

Your answer

Answer and worked solution

Official answer: 10

Neutralisation first. NaOH neutralises the strongest acid present:
$\ce{H2CO3 + NaOH -> NaHCO3 + H2O}.$
That leaves 0.02 mol $\ce{HCO3-}$ and 0.01 mol $\ce{CO3^2-}$ .

Buffer. This is a $\ce{HCO3-}/\ce{CO3^2-}$ buffer:
$pH=pK_{a2}+\log\frac{0.01}{0.02}=10.32-0.30=10.02.$

Topics:Buffer solutions Acids and bases

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