A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

The Henderson-Hasselbalch equation is: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Where: - pH is the power of hydrogen ion activity in a solution - pKa is the negative logarithm of the acid ionization constant (Ka) - [A-] is the concentration of the conjugate base (acetate ion) in moles per liter - [HA] is the concentration of the weak acid (acetic acid) in moles per liter The acid dissociation constant, Ka, for acetic acid is \(1.8 \times 10^{-5}\). First, we find the pKa value for acetic acid: \(pKa = -\log(Ka) = -\log(1.8 \times 10^{-5}) = 4.74\) Now calculate the initial concentrations of acetate ions and acetic acid: Initial concentration of sodium acetate, \([A^-] = \frac{0.25\:mol}{2.50\:L} = 0.10\:M\) Initial concentration of acetic acid, \([HA] = \frac{0.20\:mol}{2.50\:L} = 0.080\:M\) Now using the Henderson-Hasselbalch equation to find the pH: \(pH = 4.74 + \log \frac{0.10}{0.080} = 4.74 + 0.11 = 4.85\) So, the initial pH of the buffer is 4.85.

When NaOH is added to the buffer solution, it reacts with the acetic acid: \(\mathrm{NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O}\) Since 0.05 mol of NaOH is added, it neutralizes 0.05 mol of acetic acid and produces 0.05 mol of sodium acetate: New concentration of sodium acetate, \([A^-] = \frac{0.25\:mol + 0.05\:mol}{2.50\:L} = 0.12\:M\) New concentration of acetic acid, \([HA] = \frac{0.20\:mol - 0.05\:mol}{2.50\:L} = 0.060\:M\) Using the Henderson-Hasselbalch equation to find the new pH: \(pH = 4.74 + \log \frac{0.12}{0.060} = 4.74 + 0.70 = 5.44\) So, the pH of the buffer after the addition of 0.05 mol of NaOH is 5.44.

When HCl is added to the buffer solution, it reacts with the acetate ions: \(\mathrm{HCl + CH_3COO^- \rightarrow CH_3COOH + Cl^-}\) Since 0.05 mol of HCl is added, it reacts with 0.05 mol of acetate ions and produces 0.05 mol of acetic acid: New concentration of sodium acetate, \([A^-] = \frac{0.25\:mol - 0.05\:mol}{2.50\:L} = 0.080\:M\) New concentration of acetic acid, \([HA] = \frac{0.20\:mol + 0.05\:mol}{2.50\:L} = 0.10\:M\) Using the Henderson-Hasselbalch equation to find the new pH: \(pH = 4.74 + \log \frac{0.080}{0.10} = 4.74 - 0.11 = 4.63\) So, the pH of the buffer after the addition of 0.05 mol of HCl is 4.63.