Question

A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.40 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.

Answer 1

First, we need initial concentrations to determine final pH, for that we have two useful data. First we know that the concentration of both together, acid and base, is 0.1 M:

x + y = 0.1

Also, we know that the pKa is 4.76 and pH is 5, which we can plot in the pH for buffers expression:

ph = pKa + log [Base/Acid]

5 = 4.76 + log [x/y]

0.24 = log [x/y]

We use both equations, solve them simultaneously to get inital concentrations:

[Acetate] = 0.06347 M

[Acetic Acid] = 0.03652 M

Now, with the mL of solution, we get number of moles:

moles of acetate = 0.06347 mol/L * 0.14 L = 0.0088858 moles

moles of acid = 0.03652 mol/L * 0.14 L = 0.0051128 moles

We are adding:

0.0044 L * 0.31 mol/L of HCl = 0.001364 moles of HCl

These moles will alter equilibrium, as follows:

Moles of Acetate = 0.00889 - 0.001364 = 0.007526 moles

Moles of Acid = 0.005113 + 0.001364 = 0.006477 moles

With these moles and total volume, we get new concentrations:

[Acetate] = 0.007526 mol / 0.1444 L = 0.05212 M

[Acid] = 0.006477 mol / 0.1444 L = 0.04485 M

We now get the new pH:

pH = pKa + log [Acetate]/[Acetic Acid]

pH = 4.76 + log [0.05212 / 0.04485]

pH = 4.76 + 0.06524 = 4.82524

pH change = 4.82524 - 5 = -0.17476