Question

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 MNaOHsolution. Calculate the pH after the following volumes of base have been added.

50.0 mL

Express your answer using two decimal places.

Answer 1

moles of acetic acid = 0.150 M * 0.035 L = 0.00525 moles

Volume need for complete titration:

M1V1 = M2V2

V2 = 0.00525 / 0.150 M = 0.035 = 35 ml

moles of NaOH = 0.150 M * 0.050 L = 0.0075 moles

That means you added more NaOH than you needed:

CH3COOH + NaOH ------> CH3COONa +   H2O

        I             0.00525        0.0075                    0                 0

       C          -0.00525        -0.00525        +0.00525        +0.00525

       E               0                0.00225            0.00525          0.00525

[CH3COO-] = 0.00525/0.085 L = 0.0618 M

[NaOH] = 0.00225/0.085 L = 0.0265 M

CH3COO- + H2O -----> CH3COOH +    OH-

             I            0.0618                                  -               0.0265     

            C             -x                                       x                   x

            E          0.0618-x                               x                0.0265+x

Kb = Kw/Ka = 10-14/1.8*10-5 = 5.6 *10-10

Kb = [(0.0265+x)*x] / (0.0618 - x)

x = 1.305*10-9

[OH-] = 0.0265 + x = 0.0265 + 1.305*10-9 = 0.0265

pOH = -log (0.0265) = 1.58

pH = 14-1.58 = 12.42