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).

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.

Answer 1

Correct Answer: 14

Free chlorine in water splits between two forms, $HOCl$ and $OCl^-$, and the split is controlled entirely by pH through the equilibrium constant $pK=7.5$. To answer this question we need two things: the water's pH, and then the standard acid-fraction formula applied to the chlorine equilibrium.

The pH itself is not given directly, but the table hands us a matched pair of carbonate species, $HCO_3^-$ and $CO_3^{2-}$, whose equilibrium ($pK_2=10.3$) lets us back-calculate the pH the same way a buffer equation would:

\[ pH = pK_2 + \log\frac{[CO_3^{2-}]}{[HCO_3^-]} = 10.3 + \log\frac{0.01}{1.0} \]

Since $\log(0.01) = -2$:

\[ pH = 10.3 - 2 = 8.3 \]

Now switch to the chlorine system. A useful way to compute the un-dissociated fraction $\alpha_0$ of a weak acid directly, without first solving for the concentration ratio, is the standard alpha-fraction formula:

\[ \alpha_{HOCl} = \frac{[H^+]}{[H^+]+K_a} = \frac{1}{1+10^{(pH-pK_a)}} \]

Here $pH - pK_a = 8.3 - 7.5 = 0.8$, so:

\[ 10^{0.8} \approx 6.31 \]

Plugging into the alpha formula:

\[ \alpha_{HOCl} = \frac{1}{1+6.31} = \frac{1}{7.31} \approx 0.137 \]

Let's summarize:

• The carbonate pair in the table pins down the pH at 8.3 through the $pK_2=10.3$ equilibrium, without needing any other measurement.
• Since this pH is 0.8 units above the $HOCl$ equilibrium's $pK_a$ of 7.5, most of the free chlorine has already dissociated to $OCl^-$, leaving only a minority as $HOCl$.
• The alpha-fraction formula gives that minority share directly as about 13.7%.

So $HOCl$ makes up about $14\%$ of the total free chlorine in this water sample, rounded to the nearest integer.