GATE 2025 CE (CE2) – Question 34
Free residual chlorine concentration in water was measured to be 2 mg/l (as Cl$_2$). The pH of water is 8.5. By using the chemical equation given below, the HOCl concentration (in µmoles/l) in water is ________________ (round off to one decimal place).
HOCl ⇌ H$^+$ + OCl$^-$, pK = 7.50
Atomic weight: Cl(35.5)
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Correct answer: 2.59 to 2.61
Explanation
The free chlorine is the sum of HOCl and OCl⁻. In moles per litre, $\frac{2 \times 10^{-3}}{71} = 28.17$ µmol/l. At pH 8.5, $\frac{[OCl^-]}{[HOCl]} = 10^{pH - pK} = 10^{1} = 10$, so the HOCl is $\frac{1}{11}$ of the total: $\frac{28.17}{11} = 2.56 \approx 2.6$ µmol/l.