GATE 2026 CE (CE2) – Question 63
The analysis of major cations and anions in a water sample collected from a city's water supply is given below. Ions present in minor concentrations are not given.
| Anions | $Cl^-$ | $SO_4^{2-}$ | $HCO_3^-$ | $CO_3^{2-}$ |
|---|---|---|---|---|
| Concentration (mM) | 1.5 | 0.5 | 1.0 | 0.01 |
| Cations | $Na^+$ | $Ca^{2+}$ | $Mg^{2+}$ | $K^+$ |
|---|---|---|---|---|
| Concentration (mM) | 2 | 0.5 | 0.25 | 0.02 |
$$HCl \rightleftharpoons H^+ + Cl^- \qquad pK = -3$$
$$H_2SO_4^{2-} \rightleftharpoons 2H^+ + SO_4^{2-} \qquad pK = -3$$
$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \qquad pK = 6.3$$
$$HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \qquad pK = 10.3$$
$$HOCl \rightleftharpoons H^+ + OCl^- \qquad pK = 7.5$$
$HOCl$ (in %) present in the total free chlorine in the water is _____ (*rounded off to the nearest integer*).
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Correct answer: 14
Explanation
The pH follows from the carbonate pair: $HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}$ gives $pH = pK_2 + \log\frac{[CO_3^{2-}]}{[HCO_3^-]} = 10.3 + \log\frac{0.01}{1.0} = 8.3$. For hypochlorous acid, $\frac{[OCl^-]}{[HOCl]} = 10^{pH - pK} = 10^{8.3 - 7.5} = 10^{0.8} = 6.31$. So the share of free chlorine present as $HOCl$ is $\frac{1}{1 + 6.31} = 0.137$, which is about 14 %.