Question:
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
\[ HCl \rightleftharpoons H^+ + Cl^- \qquad pK=-3 \]
\[ H_2SO_4 \rightleftharpoons 2H^+ + SO_4^{2-} \qquad pK=-3 \]
\[ H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \qquad pK=6.3 \]
\[ HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \qquad pK=10.3 \]
\[ HOCl \rightleftharpoons H^+ + OCl^- \qquad pK=7.5 \]
\(HOCl\) (in %) present in the total free chlorine in the water is (rounded off to the nearest integer).
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.
\[ HCl \rightleftharpoons H^+ + Cl^- \qquad pK=-3 \]
\[ H_2SO_4 \rightleftharpoons 2H^+ + SO_4^{2-} \qquad pK=-3 \]
\[ H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \qquad pK=6.3 \]
\[ HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \qquad pK=10.3 \]
\[ HOCl \rightleftharpoons H^+ + OCl^- \qquad pK=7.5 \]
\(HOCl\) (in %) present in the total free chlorine in the water is (rounded off to the nearest integer).
| 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 |
\[ HCl \rightleftharpoons H^+ + Cl^- \qquad pK=-3 \]
\[ H_2SO_4 \rightleftharpoons 2H^+ + SO_4^{2-} \qquad pK=-3 \]
\[ H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \qquad pK=6.3 \]
\[ HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \qquad pK=10.3 \]
\[ HOCl \rightleftharpoons H^+ + OCl^- \qquad pK=7.5 \]
\(HOCl\) (in %) present in the total free chlorine in the water is (rounded off to the nearest integer).
Show Hint
Find pH from the \(HCO_3^-/CO_3^{2-}\) pair using \(pK_2=10.3\), then use \(pH-pK_a\) with \(pK_a=7.5\) for HOCl/OCl- to get the HOCl fraction of total free chlorine.
Updated On: Jul 22, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 14
Solution and Explanation
Step 1: Understand what 'HOCl in the total free chlorine' means.
Free available chlorine in water exists as two species in an acid-base equilibrium: hypochlorous acid \(HOCl\) (the strong, effective disinfectant) and hypochlorite ion \(OCl^-\) (a much weaker disinfectant). Their ratio depends only on the pH of the water and the equilibrium constant of that reaction, \(pK=7.5\). So the first job is to find the pH of this water sample.
Step 2: Get the pH from the carbonate system data given.
The table gives both \(HCO_3^-\) and \(CO_3^{2-}\) concentrations, linked by the equilibrium \(HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}\) with \(pK_2=10.3\). Using the Henderson-Hasselbalch form of this equilibrium:
\[ pH = pK_2 + \log_{10}\left(\frac{[CO_3^{2-}]}{[HCO_3^-]}\right) \]
Substituting \([CO_3^{2-}]=0.01\) mM and \([HCO_3^-]=1.0\) mM:
\[ pH = 10.3 + \log_{10}\left(\frac{0.01}{1.0}\right) = 10.3 + (-2) = 8.3 \]
Step 3: Apply the same Henderson-Hasselbalch logic to the HOCl/OCl- pair.
For \(HOCl \rightleftharpoons H^+ + OCl^-\) with \(pK_a=7.5\):
\[ pH = pK_a + \log_{10}\left(\frac{[OCl^-]}{[HOCl]}\right) \]
Rearranging for the ratio:
\[ \frac{[OCl^-]}{[HOCl]} = 10^{(pH-pK_a)} = 10^{(8.3-7.5)} = 10^{0.8} \approx 6.31 \]
Step 4: Convert the ratio into a fraction of the total free chlorine.
Total free chlorine is \([HOCl]+[OCl^-]\). If \([OCl^-]/[HOCl]=6.31\), then for every 1 part \(HOCl\) there are 6.31 parts \(OCl^-\), so total parts \(=1+6.31=7.31\). The fraction that is \(HOCl\) is
\[ \frac{[HOCl]}{[HOCl]+[OCl^-]} = \frac{1}{1+10^{(pH-pK_a)}} = \frac{1}{1+6.31} = \frac{1}{7.31} = 0.1368 \]
Step 5: Convert to percentage.
\[ \%HOCl = 0.1368 \times 100 = 13.68\% \]
This makes sense: since the water's pH (8.3) is above the \(pK_a\) of hypochlorous acid (7.5), the equilibrium favors the dissociated form \(OCl^-\), so \(HOCl\) should be well under 50%, matching the result.
Final Answer:
Rounded to the nearest integer,
\[ \boxed{\%HOCl \approx 14\%} \]
Free available chlorine in water exists as two species in an acid-base equilibrium: hypochlorous acid \(HOCl\) (the strong, effective disinfectant) and hypochlorite ion \(OCl^-\) (a much weaker disinfectant). Their ratio depends only on the pH of the water and the equilibrium constant of that reaction, \(pK=7.5\). So the first job is to find the pH of this water sample.
Step 2: Get the pH from the carbonate system data given.
The table gives both \(HCO_3^-\) and \(CO_3^{2-}\) concentrations, linked by the equilibrium \(HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}\) with \(pK_2=10.3\). Using the Henderson-Hasselbalch form of this equilibrium:
\[ pH = pK_2 + \log_{10}\left(\frac{[CO_3^{2-}]}{[HCO_3^-]}\right) \]
Substituting \([CO_3^{2-}]=0.01\) mM and \([HCO_3^-]=1.0\) mM:
\[ pH = 10.3 + \log_{10}\left(\frac{0.01}{1.0}\right) = 10.3 + (-2) = 8.3 \]
Step 3: Apply the same Henderson-Hasselbalch logic to the HOCl/OCl- pair.
For \(HOCl \rightleftharpoons H^+ + OCl^-\) with \(pK_a=7.5\):
\[ pH = pK_a + \log_{10}\left(\frac{[OCl^-]}{[HOCl]}\right) \]
Rearranging for the ratio:
\[ \frac{[OCl^-]}{[HOCl]} = 10^{(pH-pK_a)} = 10^{(8.3-7.5)} = 10^{0.8} \approx 6.31 \]
Step 4: Convert the ratio into a fraction of the total free chlorine.
Total free chlorine is \([HOCl]+[OCl^-]\). If \([OCl^-]/[HOCl]=6.31\), then for every 1 part \(HOCl\) there are 6.31 parts \(OCl^-\), so total parts \(=1+6.31=7.31\). The fraction that is \(HOCl\) is
\[ \frac{[HOCl]}{[HOCl]+[OCl^-]} = \frac{1}{1+10^{(pH-pK_a)}} = \frac{1}{1+6.31} = \frac{1}{7.31} = 0.1368 \]
Step 5: Convert to percentage.
\[ \%HOCl = 0.1368 \times 100 = 13.68\% \]
This makes sense: since the water's pH (8.3) is above the \(pK_a\) of hypochlorous acid (7.5), the equilibrium favors the dissociated form \(OCl^-\), so \(HOCl\) should be well under 50%, matching the result.
Final Answer:
Rounded to the nearest integer,
\[ \boxed{\%HOCl \approx 14\%} \]
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