Sodium hypochlorite (NaOCl) is the active ingredient in pool chlorine - HSC - SSCE Chemistry - Question 34 - 2022 - Paper 1

The equilibrium expression based on the provided reaction is given by:

$K_{eq} = \frac{[\text{OH}^-][\text{HClO}]}{[\text{OCl}^-]}$

Substituting the known concentrations, with [HClO] = 1.3 \times 10^{-4} mol L^{-1} and $K_{eq} = 3.33 \times 10^{-7}$:

$3.33 \times 10^{-7} = \frac{(3.16 \times 10^{-7})(1.3 \times 10^{-4})}{[\text{OCl}^-]}$

Rearranging the equation gives:

$[\text{OCl}^-] = \frac{(3.16 \times 10^{-7})(1.3 \times 10^{-4})}{3.33 \times 10^{-7}}$

After calculation:

$[\text{OCl}^-] \approx 1.23 \times 10^{-4} \text{ mol L}^{-1}$

Using the formula:

$c_1 V_1 = c_2 V_2$

Where:

• $c_1$ = concentration of sodium hypochlorite = 2.0 mol L^{-1}
• $c_2$ = concentration of OCl^- needed = 1.23 \times 10^{-4} mol L^{-1}
• $V_2$ = volume of the pool = 1.00 \times 10^{4} L

Rearranging gives us:

$V_1 = \frac{c_2 V_2}{c_1} = \frac{(1.23 \times 10^{-4} \text{ mol L}^{-1})(1.00 \times 10^{4} \text{ L})}{2.0 \text{ mol L}^{-1}}$

Performing the calculation:

$V_1 \approx 6.15 \text{ L}$