n the titration of 25.0 mL of 0.15 M butanoic acid, CH3CH2CH2CO2H, with 0.20 M sodium hydroxide, what will be the pH at the equivalence point? The Ka for butanoic acid is 1.5 x 10-5.

Answer 1

At the equivalence point, moles of NaOH = moles butanoic acid and all of the butanoic acid is converted to the sodium salt, sodium butanoate.

CH3CH2CH2CO2H + NaOH ==> CH3CH2CH2CO2Na + H2O

moles CH3CH2CH2CO2H = 0.025 L x 0.15 mol/L = 0.00375 moles

Volume of NaOH needed to produce 0.00375 moles: (x L)(0.20 mol/L) = 0.00375 mol and x = 0.01875 L

Total volume = 0.025 L + 0.01875 L = 0.04375 L

[CH3CH2CH2CO2Na] = 0.00375 mol / 0.04375 L = 0.0857 M

CH3CH2CH2CO2^- + H₂O ==> CH3CH2CH2CO2H + OH^-

KaKb = Kw = 1x10^-14

Kb = 1x10^-14 / 1.5x10^-5 = 6.67x10^-10

6.67x10^-10 = [CH3CH2CH2CO2H][OH^-] / CH3CH2CH2CO2^-] = (x)(x) / 0.0857

x² = 5.7x10^-11

x = 7.5x10^-6 M = [OH^-]

pOH = -log 7.5x10^-6 = 5.12

pH = 14 - pOH

pH = 8.9