Exercise 7 | Buffers and Titration of Acids and Bases

Before addition of NaOH:

The equation of the initial reaction equilibrium is:

CH₃COOH + H2O $\rightleftarrows$ CH₃COO^- + H₃O⁺

K_a = $\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right] \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$ = $\frac{{\mathrm{X}}^2}{0.100 - \mathrm{X}}$ = 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^- $\rightleftarrows$ CH₃COO^- + H₂O

n_CH3COOH,0 = [CH₃COOH] x V_CH3COOH = 0.100 x 50.0 x 10^-3 = 5.0 x 10^-3 mol

n_HO-,0 = [HO^-] x V_HO- = 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:

n_CH3COOH = n_CH3COOH,0 - n_HO-,0 = 4.25 x 10^-3 mol

n_CH3COO- = n_HO-,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