# Exercise 7 | Buffers and Titration of Acids and Bases

Calculate the pH of the solution obtained by titrating 50.0 mL of 0.100 M acetic acid with 0.150 M NaOH

- after 5.00 mL of NaOH

Data: pKa (CH3COOH/CH3COO-) = 4.74

The equation of the initial reaction equilibrium is:

CH3COOH + H2O $⇄$ CH3COO- + H3O+

t = t0:

[CH3COOH] = 0.100 M

[CH3COO-] = [H3O+] = 0 M

t = equilibrium:

[CH3COOH] = 0.100 - X M

[CH3COO-] = [H3O+] = X M

Ka =  =  = 10-4.74

X2 + (1.82 x 10-5) X – 1.82 x 10-6 = 0

X = 1.34 x 10-3

pH = -log [H+] = 2.87

After addition of 5.00 mL of NaOH:

The titration equation is:

CH3COOH + HO- $⇄$ CH3COO- + H2O

nCH3COOH,0 = [CH3COOH] x VCH3COOH = 0.100 x 50.0 x 10-3 = 5.0 x 10-3 mol

nHO-,0 = [HO-] x VHO- = 0.150 x 5.0 x 10-3 = 7.5 x 10-4 mol

HO- is the limiting reagent.

After 5.00 mL of NaOH:

nCH3COOH = nCH3COOH,0 - nHO-,0 = 4.25 x 10-3 mol

nCH3COO- = nHO-,0 = 7.5 x 10-4 mol

Henderson-Hasselbalch equation can be used:

pH = pKa + log $\frac{{\left[\mathrm{base}\right]}_{0}}{{\left[\mathrm{acid}\right]}_{0}}$

pH = 4.74 + log $\frac{{\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]}_{0}}{{\left[{\mathrm{CH}}_{3}\mathrm{COOH}\right]}_{0}}$

pH = 3.99