Acid Base question

Question

 An acetate buffer was prepared by dissolving 0.1 mol of acetic acid and 0.11 mol of sodium acetate in 1L of water. Suppose 25 mL of 0.1 M HCl is added to 100 mL of acetate buffer. What are pH values of the buffer before and after the addition of acid? What would be the pH if same volume of the 0.1 M HCl solution was added to 125 mL of water. 

Answer 1

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

pKa acetic acid = 4.75

Before addition of acid:

pH = 4.75 + log [0.11]/[0.10] = 4.75 + log 1.1 = 4.75 + 0.041

pH = 4.79

After addition of 25 ml 0.1 M HCl t0 100 ml of buffer:

mols H⁺ = 0.025 L x 0.1 mol/L = 0.0025 mols H⁺ added

mols CH₃COO^- in 100 ml = 0.1 L x 0.11 mol/L = 0.011 mol

mols CH₃COOH in 100 ml = 0.1 L x 0.1 mol/L = 0.01 mol

H⁺ + CH₃COO^- ==> CH₃COOH

0.0025.......0.011...............0.01.........Initial

-0.0025.....-0.0025.........+0.0025.....Change

0.............0.0085.............0.0125.......Equilibrium

pH = pKa + log [salt]/[acid]

[salt] = 0.0085 mol / 0.125 L = 0.068 M

[acid] = 0.0125 mol / 0.125 L = 0.1 M

pH = pKa + log [salt]/[acid] = 4.75 + log (0.068/0.1) = 4.75 - 0.17

pH = 4.58