JEE Advanced 2022 Paper 1 · Q03 · Ionic Equilibrium & pH
A solution is prepared by mixing 0.01 mol each of H$_2$CO$_3$, NaHCO$_3$, Na$_2$CO$_3$, and NaOH in 100 mL of water. pH of the resulting solution is _______. [Given: pK$_{a_1}$ and pK$_{a_2}$ of H$_2$CO$_3$ are 6.37 and 10.32, respectively; log 2 = 0.30]
Reveal answer + step-by-step solution
Correct answer:10.02
Solution
First, neutralise the strong base NaOH (0.01 mol) with the strong acid H$_2$CO$_3$ (0.01 mol): H$_2$CO$_3$ + NaOH $\rightarrow$ NaHCO$_3$ + H$_2$O. After this step: H$_2$CO$_3 = 0$, NaOH = 0, NaHCO$_3$ = 0.01 + 0.01 = 0.02 mol, Na$_2$CO$_3$ = 0.01 mol.
The resulting mixture is a CO$_3^{2-}$/HCO$_3^-$ buffer. Using the Henderson–Hasselbalch equation with pK$_{a_2}$ = 10.32:
pH = pK$_{a_2}$ + log([CO$_3^{2-}$]/[HCO$_3^-$]) = 10.32 + log(0.01/0.02) = 10.32 + log(1/2) = 10.32 - 0.30 = 10.02.
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