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 = 10.18 solution. Use the Ka of hypochlorous acid found in the chempendix

Answer 1

5.25% by mass means 5.25 g of NaOCl is present in 100 g of solution.

mass of NaOCl = 5.25 g

molar mass of NaOCl = 74.44 g/mol

moles of NaOCl = mass / molar mass = 5.25 / 74.44 = 0.0705 mol

mass of solution = 100 g

density of solution = 1.00 g/ml

So, volume of solution = mass/density = 100 ml = 0.100 L

concentration of NaOCl solution = moles of NaOCl / volume of solution

=> [NaOCl] = 0.0705 / 0.100 = 0.705 M

NaOCl is a salt of weak acid and strong base,

So, its pH expression is,

pH = 7 + 1/2(pKa + log C )

where, Ka is acid dissociation constant of HOCl = 4.0x10^-8

pKa = 7.4

C = concentration of NaOCl solution

=> 10.26 = 7 + 1/2(7.4 + log C )

=> C = 0.132 M

final concentration of NaOCl should be 0.132 M.

from dilution principle,

C1V1 = C2V2

=> 0.705 x V1 = 0.132 x 500

=> V1 = 93.62 mL

Therefore, 93.62 mL of household bleach must be diluted