Adding Strong Acids or Bases to Buffer Solutions

For every mole of H3O+ added, an equivalent amount of the conjugate base (in this case, F) will also react, and the equilibrium constant for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F- is used up before reacting away all of the H3O+ , then the remaining H3O+ will affect the pH directly. In this case, the capacity of the buffer will have been exceeded - a situation one tries to avoid. However, for our example, let's say that the amount of added H3O+ is smaller than the amount of F- present, so our buffer capacity is NOT exceeded. For the purposes of this example, we'll let the added H3O+ be equal to 0.01 moles (from 0.01 moles of HCl). Now, if we add 0.01 moles of HCl to 100 mL of pure water, we would expect the pH of the resulting solution to be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0).

However, we are adding the H3O+ to a solution that has F- in it, so the H3O+ will all be consumed by reaction with F. In the process, the 0.066 moles of F- is reduced:

0.066 initial moles F- 0.010 moles reacted with H3O+ = 0.056 moles F- remaining

Also during this process, more HF is formed by the reaction:

0.10 initial moles HF + 0.010 moles from reaction of F- with H3O+ = 0.11 moles HF after reaction

Plugging these new values into Henderson-Hasselbalch gives:

pH = pKa + log (base/acid) = 3.18 + log (0.056 moles F/0.11 moles HF) = 2.89

Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid.