Question

suppose 25.0 ml of .20 M acetic acids is being titrated with 0.200 M NaOH. what is pH after 10.0 mL NaOH has been added (ka for acetic acids = 1.8 x10^-5)

Answer 1

no. of mole = molarity volume of solution in liter

no. of mole of acetic acid = 0.20 0.025 = 0.005 mole

no. of mole of NaOH = 0.20 0.010 = 0.002

acetic acid and NaOH react in equimolar proportion to form salt and water thus 0.002 mole of acetic acid react with 0.002 mole of NaOH

no. of mole of actic acid remain in solution = 0.005 - 0.002 = 0.003 mole

volume of solution = 25 ml + 10 ml = 35 ml = 0.035 liter

molarity = no. of mole / volume of solution in liter

molarity of acetic acid = 0.003 / 0.035 = 0.0857 M

acetic acid dissociates as

CH₃COOH + H₂O CH₃COO^- + H₃O⁺

K_a = [CH₃COO^- ][H₃O⁺] / [CH₃COOH]

but  [CH₃COO^- ] = [H₃O⁺] = x

K_a = [x][x] / [CH₃COOH]

Substitute the value in equation

1.8 10^-5 = [x]²/ 0.0857

[x]² = 1.8 10^-5 0.0857 = 1.5426 10^-6

[x] = 0.001242 M

Concentration of H₃O⁺ = 0.001242 M

pH = - log[H⁺]

pH = - log (0.001242)

pH = 2.90