Calculate the pH of the titration mixture after 20.0 mL of base have been added.

Question

A 25.0 mL sample of a  0.100 M  solution of acetic acid is titrated with a  0.125 M  solution of NaOH. Calculate the pH of the titration mixture after 20.0 mL of base have been added. The K_a for acetic acid is 1.76×10^–5.

Answer 1

First, calculate the number of moles of acetic acid initially present.

0.100 moles/L /1000 mL//L x 25.0 mL = 0.0025 moles acetic acid

Next calculate the number of moles of NaOH added..

0.125 moles/L /1000 mL/L x 20.0 mL = 0.0025 moles NaOH added.

The reaction equation is CH3COOH + NaOH ---à CH3COONa + H2O,

Since equimolar amounts of acid and base were mixed, only CH3COONa is left.

The molarity of the final solution is 0.0025 moles / (25.0 mL + 20.0 mL) x 1000 mL/L

= 0.0556 M. (CH3COONa)

CH3COONa partially reacts with water (hydrolysis) according to the equation:

CH3COONa +H2O ---à CH3COOH + OH-, so the solution is basic.

Ka = [CH3COOH] [OH-] / [CH3COONa] or Ka = [OH-]^2 / [CH3COONa] if the [CH3COOH] is small compared to the concentration of CH3COONa.

Solve for the concentration of OH-.

[OH-] = sq.rt. Ka x 0.0556 = 1.76 x 10^-5 x 0.0556 = 9.79 x 10^-7.

pOH = - log 9.79 x 10^-7 = 6.01, pH = 14 – 6.01 = 7.99.

Check all math.