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 = 10.00 solution. Use the Ka of hypochlorous acid found in the chempendix.Ionization Constants

Acids Acid Formula Ionization Constant Ka Acetic CH3COOH 1.8×10–5 Arsenic H3AsO4 5.5×10–3 Benzoic C6H5COOH 6.3×10–5 Boric H3BO3 5.4×10–10 Butanoic C3H7COOH 1.5×10–5 Carbonic H2CO3 Ka1 = 4.5×10–7 Ka2 = 4.7×10–11 Chloroacetic ClCH2COOH 1.4×10–3 Chlorous HClO2 1.1×10–2 Chromic H2CrO4 Ka1 = 1.8×10–1 Ka2 = 3.2×10–7 Cyanic HCNO 3.5×10–4 Formic HCOOH 1.8×10–4 Hydrazoic HN3 2.5×10–5 Hydrocyanic HCN 6.2×10–10 Hydrofluoric HF 6.3×10–4 Hydrogen peroxide H2O2 2.4×10–12 Hydrosulfuric H2S Ka1 = 8.9×10–8 Ka2 = 1.0×10–19 Hypobromous HBrO 2.8×10–9 Hypochlorous HClO 4.0×10–8 Hypoiodous HIO 3.2×10–11 Iodic HIO3 1.7×10–1 Nitrous HNO2 5.6×10–4 Oxalic C2H2O4 Ka1 = 5.6×10–2 Ka2 = 1.5×10–4 Paraperiodic H5IO6 2.8×10–2 Pentanoic C4H9COOH 1.5×10–5 Periodic HIO4 7.3×10–2 Phenol HC6H5O 1.3×10–10 Phosphoric H3PO4 Ka1 = 6.9×10–3 Ka2 = 6.2×10–8 Ka3 = 4.8×10–13 Phosphorous H3PO3 Ka1 = 5.0×10–2 Ka2 = 2.0×10–7 Propanoic C2H5COOH 1.3×10–5 Sulfuric H2SO4 Ka1 = very large Ka2 = 1.2×10–2 Sulfurous H2SO3 Ka1 = 1.4×10–2 Ka2 = 6.3×10–8 Trichloroacetic Cl3CCOOHBases Base Formula Ionization Constant Kb Ammonia NH3 1.8×10–5 Methylamine CH3NH2 5.0×10–4 Dimethylamine (CH3)2NH 5.4×10–4 Diethylamine (C2H5)2NH 6.9×10–4 Trimethylamine (CH3)3N 6.3×10–5 Triethylamine (C2H5)3N 5.6×10–4 Ethylamine CH3CH2NH2 6.3×10–4 Ethylenediamine (CH2NH2)2 8.3×10–5 Pyridine C5H5N 1.7×10–9 Hydroxylamine NH2OH 8.7×10–9 Aniline C6H5NH2 7.4×10–10 Hydrazine H2NNH2 1.3×10–6 2.2×10–1

Answer 1

X = 5.25%

NaOCl by mas...

D = 1 g/mL

volume for a pH = 10.00 solutino

Ka = 3.5*10^-8

so..

assume a basis of V = 500 mL = 0.5L

so

mass = 500 g

5.25/100*500 = 26.25 g of NaOCl

mol = mass/MW = (26.25)/74.44 = 0.3526 mol of NAOCl

so..

[OCl-] = mol/V = 0.3526/(0.5) = 0.7052 M of OCl-

so..

OCl- + H2O <-> HOCl + OH-

Kb = [HOCl ][OH-]/[OCl-]

(10^-14)/(3.5*10^-8) = x*x /(M-X)

ntoe that

[OH-] = 10^-pOH = 10^-(14-10.00) = 0.00016982

x = 0.00016982

(10^-14)/(3.5*10^-8) =0.00016982*0.00016982/(M-0.00016982)

solve for M

M = (0.00016982*0.00016982) / ((10^-14)/(3.5*10^-8) ) + 0.00016982

M = 0.10110 mol of NaOCl per liter...

M1v1 = M2V2

0.7052 *V1=0.10110 * 500

V1 =0.10110 * 500/0.7052 = 71.68 mL

so...

intially, we need 71.68 mL of solution to dilute it to 500 mL for pH = 10.23