Question

A beaker with 1.50×10² 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 4.90 mL of a 0.450 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740

Answer 1

We have two equations here:

0.1 M = [Acid] + [Base] (total molarity)

ph = pKa + log ([Base]/[Acid]) (ph)

5 = 4.74 + log ([Base]/[Acid])

We can do system of equations to get initial concentrations:

[Acid] = 0.0355 M

[Base] = 0.0645 M

Then we get the moles added to the equilibrium:

0.0049 L * 0.45 M = 0.002205 moles

These moles will be added to the acid moles in solution and substracted from the base moles, maintaining equilibrium:

Moles of acid = (0.0355M * 0.15 L) + 0.002205 = 0.00753 moles of acid

Moles of base = (0.0645M * 0.15 L) - 0.002205 = 0.00747 moles of base

New concentrations:

[Acid] = 0.00753 moles / 0.1549 L = 0.04861 M

[Base] = 0.00747 moles / 0.1549 L = 0.04822 M

New pH

pH = 4.74 + log (0.04822/0.04861) = 4.74 - 0.0035 = 4.7365

pH change = 5 - 4.7365 = 0.2635