Question

25.0 mL of 1.00 M CH3COOH (ka=1.8x10^-5) is titrated with 1.00 M NaOH.?
Calculate the hydronium ion concentration and pH after the following volumes of NaOH have been added.

a. 0
b.14.00
c.25.90
d.26.10
e.35.00

Answer 1

a)

millimoles of acetic acid = 25 x 1 = 25

CH3COOH + H2O ---------------> CH3COO-   +   H3O+

   1.00                                                   0                      0

1.00- x                                                x                       x

Ka = x^2 / 1.00 - x

1.8 x 10^-5 = x^2 / 1.00 - x

x = 4.24 x 10^-3

[H3O+] = 4.24 x 10^-3 M

pH = 2.37

b)

mmoles of NaOH = 14

CH3COOH   +   NaOH    ----------------> CH3COO- +   H2O

25                      14                                         0               0

11                        0                                       14

pH = pKa + log [salt / acid]

   = 4.74 + log [14 / 11]

pH = 4.84

[H3O+] = 1.44 x 10^-5 M

c)

CH3COOH   +   NaOH    ----------------> CH3COO- +   H2O

25                      25.9                                        0               0

0                        0.9                                       25

[OH-] = 0.9 / 25 + 25.9 = 0.0177 M

[H3O+] = 5.66 x 10^-13 M

pH = 12.25

d)

CH3COOH   +   NaOH    ----------------> CH3COO- +   H2O

25                      26.1                                        0               0

0                        1.1                                     25

[OH-] = 0.0215

[H3O+] = 4.64 x 10^-13 M

pH = 12.33

e)

[H3O+] = 6.00 x 10^-14 M

pH= 13.22