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

Using the ionization equation:

[OH⁻] = 10^{-pOH} = 10^{-6.5} = 3.16 × 10^{-7} \, \text{mol L}^{-1}\

The equilibrium expression is given by:

K_{eq} = \frac{[OH⁻][HOCl]}{[OCl⁻]}\

Substituting the known values into the expression:

3.33 × 10^{-7} = \frac{(3.16 × 10^{-7})(1.3 × 10^{-4})}{[OCl⁻]}\

Rearranging to find [OCl⁻]:

[OCl⁻] = \frac{(3.16 × 10^{-7})(1.3 × 10^{-4})}{3.33 × 10^{-7}} = 1.23 × 10^{-4} \, \text{mol L}^{-1}\

The total concentration of chlorine species is:

[Cl \text{ species}] = [HOCl] + [OCl⁻] = 1.3 × 10^{-4} + 1.23 × 10^{-4} = 2.53 × 10^{-4} \, \text{mol L}^{-1}\

Using the formula for dilution:

c_1V_1 = c_2V_2\

Where:

• $c_1 = 2.0 \, \text{mol L}^{-1}$ (concentration of sodium hypochlorite)
• $c_2 = 2.53 × 10^{-4} \, \text{mol L}^{-1}$ (calculated concentration)
• $V_2 = 1.00 × 10^{4} \, \text{L}$ (volume of the pool)

We can rearrange it to find $V_1$:

V_1 = \frac{c_2V_2}{c_1}\

Substituting in the values:

V_1 = \frac{(2.53 × 10^{-4})(1.00 × 10^{4})}{2.0} = 1.3 \, \text{L}\