Question

A beaker with 185 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.50 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Answer

New pH = 4.710

change in pH = - 0.290

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

log([A^-]/[HA]) = pH -pKa

log([CH3COO^-]/[CH3COOH]) = 5.000 - 4.740

log([CH3COO^-]/[CH3COOH]) = 0.26

[CH3COO^-]/[CH3COOH] = 1.8197

moles of CH3COO^-/moles of CH3COOH = 1.8197

moles of CH3COO^- = moles of CH3COOH × 1.8197

Total moles of acetate species in the buffer = (0.100mol/1000ml)× 185ml = 0.0185mol

moles of CH3COO^- + moles of CH3COOH = 0.0185mol

(1.8197mol × moles of CH3COOH) + moles of CH3COOH = 0.0185mol

2.8197× moles of CH3COOH = 0.0185mol

moles of CH3COOH = 0.006561mol

moles of CH3COO^- = 0.0185mol - 0.006561mol = 0.01194mol

moles of HCl added = (0.400mol/1000ml)× 7.50ml = 0.0030mol

HCl reacts with conjucate base CH3COO^-

H⁺ + CH3COO^- <------> CH3COOH

0.0030moles of H⁺ reacts with 0.0030moles of CH3COO^- to give 0.0030moles of CH3COOH

after addition of HCl

total volume = 185ml + 7.50ml = 192.5ml

moles of CH3COO^- = 0.01194mol - 0.0030mol = 0.00894mol

moles of CH3COOH = 0.006561mol + 0.0030mol = 0.009561mol

[CH3COOH] = (0.009561mol/192.5ml) × 1000ml = 0.04967M

[CH3COO^-] = (0.008940mol/192.5ml)×1000ml = 0.04644M

Applying Henderson - Hasselbalch equation

pH = 4.740 + log(0.04644M / 0.04967M)

pH = 4.740 - 0.030

pH = 4.710

∆pH = 5.000 - 4.710 = -0.290