Question

What is the pH of a solution if I mix together 45 mL of 5.67 mol/L Acetic acid (CH​ COOH) and 83 mL of 2.35 mol/L NaOH?

Ka of acetic acid = 1.8 x 10

Answer 1

Consider reaction, CH₃COOH _(aq) + NaOH _(aq)  CH₃COONa _(aq) + H₂O (l)

From stoichiometry of reaction, 1 mol CH₃COOH 1 mol NaOH 1 mol CH₃COONa

We have, mmol = concentration volume

mmol of CH₃COOH = concentration x volume = 5.67 45 = 255.15 mmol

mmol of NaOH = 2.35 83 = 195.05 mmol

Excess mmol of CH₃COOH = initial mmol of CH₃COOH - mmol of NaOH  

Excess mmol of CH₃COOH = 255.15 - 195.05 = 60.1 mmol

mmol of CH₃COONa produced = mmol of NaOH added to solution =195.05 mmol

Volume of solution = 45 + 83 = 128 ml

[CH₃COOH]= No of moles / Volume of solution in L

[CH₃COOH] = [ 60.1 /1000] / [128/1000]

[CH₃COOH ] =0.4695 M

[CH₃COONa]= [195.05 /1000] / [128 /1000]

[CH₃COONa]= 1.524 M

Solution contain weak acid (acetic acid) and its salt (sodium acetate) .Hence, solution acts as a buffer solution and it's pH is calculated by using Henderson's equation

pH = pKa + log [ salt] / [ Acid]

pH = - log Ka + log [CH₃COONa] / [CH₃COOH] ( pKa = - log Ka )

pH = -log (1.8 x 10 ^-05) + log 1.524 / 0.4695

pH = 4.74 + 0.5113

pH = 5.25