Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
.02M
CH₃COOH = CH₃COO⁻ + H⁺
C xC xC
Ka = xC . xC / C = x² C
1.8 x 10⁻⁵ = x² . .02
x² = 9 x 10⁻⁴
x = 3 x 10⁻²
= .03
concentration of H⁺ = xC = .03 . .02
= 6 x 10⁻⁴ M , volume = 45 x 10⁻³ L
moles of H⁺ = 6 X 10⁻⁴ x 45 x 10⁻³
= 270 x 10⁻⁷ moles
= 2.7 x 10⁻⁵ moles
concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L
moles of Na OH = 2 X 10⁻² x 35 x 10⁻³
= 70 x 10⁻⁵ moles
=
NaOH is a strong base so it will dissociate fully .
there will be neutralisation reaction between the two .
Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles
= 67.3 x 10⁻⁵ moles of NaOH
Total volume = 45 + 35 = 80 x 10⁻³
concentration of NaOH after neutralisation.= 67.3 x 10⁻⁵ / 80 x 10⁻³ moles / L
= 8.4125 x 10⁻³ moles / L
OH⁻ = 8.4125 x 10⁻³
H⁺ = 10⁻¹⁴ / 8.4125 x 10⁻³
= 1.1887 x 10⁻¹²
pH = - log ( 1.1887 x 10⁻¹² )
= 12 - log 1.1887
= 12 - .075
= 11.925 .
