A 1.00 L solution of a 0.100 M buffer with pH = 6.45 has been prepared from an acid with a Ka of 3.55x10–7. What is the pH after 17.0 mL of 1.0 M HCl has been added?

Answer 1

Let the buffer be HA + A^-.

HA is the weak acid and A^- is the conjugate base.

First, we will find the concentration of HA and A^- before any HCl is added. To do this, we will use the Henderson Hasselbalch equaion:

pH = pKa + log [A^-]/[HA] and pKa = -log Ka = -log 3.55x10^-7 = 6.45

Since pH = pKa then [A^-]/[HA] = 1 (note log 1 = 0)

[A-] = 0.05 M

[HA] = 0.05M

Now we can see what happens when 17.0 mls of 1.0 M HCl is added:

moles HCl added = 17.0 ml x 1 L/1000 mls x 1.0 mol/L = 0.0170 mols HCl added

Assume that the 17 mls doesn't significantly change the volume so we still have 1 L (0.17% error)

The H+ reacts with A- to form HA

New [A-] = 0.05 - 0.017 = 0.033 M

New [HA] = 0.05 M + 0.017 = 0.067 M

pH = pKa = log [A-]/[HA] = 6.45 + log (0.033/0.067)

pH = 6.45 + (-0.31)

pH = 6.14