Question

A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.26 solution. Use the Ka of hypochlorous acid found in the chempendix. Ka=4.0×10–8
 

Answer 1

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
ω(NaClO) = 5,25% ÷ 100% = 0,0525.
d(H₂O) = 1g/mL.
m(NaClO) = 0,0525 · 1 g/ml · 1000 ml = 52,5 g in 1 L.
n(NaClO) = 52,5 g ÷ 74,44 g/mol = 0,705 mol.
c₁(NaClO) = 0,705 M.
pH = 10,26.
pH + pOH = 14.
pOH = 14 - 10,26 = 3,74.
[OH⁻] = [HClO] = 10∧(-3,74) = 1,8·10⁻⁴ M.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
2,5·10⁻⁷ = (1,8·10⁻⁴ M)² / [ClO⁻].
[ClO⁻] = [NaClO] = c₂(NaClO) = 0,1296 M.
V₂ = 500 mL.
c₁V₁ = c₂V₂.
V₁ = 0,1296 M · 500 mL ÷ 0,705 M.
V₁(NaClO) = 91,9 mL.

Answer 2

From the given dissociation constant, and pH the concentration of the ions in the solution can be determined.

The volume of the household bleach containing sodium hypochlorite,

NaOCl, that should be diluted with water, V₂ ≈ 91.88 mL.

Reasons:

The given parameter are;

Concentration of sodium hypochlorite, NaOCl = 5.25%

Assumption: Density of bleach = Density of water

Volume of solution = 500.0 mL

pH of solution = 10.26

Ka hypochlorous acid, kₐ(HClO) = 4.0 × 10⁻⁸

The given mass of NaOCl per 100 g of water = 5.25 g

Therefore, the mass of NaOCl in 1 liter of solution = 52.5 g

Concentration of the NaOCl solution is therefore;

$Molar \ concentration \ of \ NaOCl = \dfrac{52.5 \ g}{74.44 \ g/mol} \approx 0.7053 \, M$

pH of the solution = 10.26

pH + pOH  = 14

pOH = 14 - 10.26 = 3.74

[OH⁻] = $10^{-3.74}$ = 1.8197 × 10⁻⁴ M

NaClO(aq) $\longrightarrow$ Na⁺(aq) + ClO⁻(aq)

ClO⁻(aq) + H₂O(l) $\rightleftharpoons$ HClO(aq) + OH⁻(aq)

Therefore;

[HClO] = [OH⁻] = 1.8197 × 10⁻⁴ M

Kₐ × $K_b$ = $K_w$ = 10⁻¹⁴

Where;

Kₐ  = Acid dissociation constant

$K_b$  = Base dissociation constant

$K_w$ = Water dissociation constant

Therefore;

$K_b = \dfrac{10^{-14}}{4 \times 10^{-8}} = 2.5 \times 10^{-7}$

$K_b(ClO^-) = \mathbf{\dfrac{[OH^-]\cdot [HClO]}{[ClO^-]}}$

Therefore;

$2.5 \times 10^{-7} = \dfrac{(1.8197 \times 10^-4) \times (1.8197 \times 10^-4)}{[ClO^-]} = \dfrac{(1.8197 \times 10^-4)^2 }{[ClO^-]}$

$[ClO^-] = \dfrac{\left(1.8 \times 10^{-4} \right)^2}{2.5 \times 10^{-7}} = 0.1296$

The volume of the given solution required is given by the dilution formula as follows;

C₁·V₁ = C₂·V₂

$V_2 = \mathbf{\dfrac{C_1 \cdot V_1}{C_1}}$

Which gives;

$V_2 = \dfrac{0.1296 \times 500}{0.7053} \approx 91.88$

Volume of household bleach that should be diluted, V₂ ≈ 91.88 mL

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