Unit 8: Topic 9 - Henderson Hasselbalch Equation Flashcards
Give a quantitative explanation as to why the pH of a buffer solution changes very slightly when a small amount of acid or base is added to the solution.
Recall a buffer solution has both a weak acid HA and its conjugate base A- present, or weak base and conjugate acid. For the dissociation of a weak acid, HA(aq) + H2O(l) –> A-(aq) + H3O+(aq), since the K value is very small, we can approximate Ka = [H3O+][A-]/[HA], or [H3O+] = Ka[HA]/[A]. Taking a negative logarithm gives pH = pKa + log[A-]/[HA], precisely the Henderson Hasselbalch Equation.
The only term that will affect the pH is log[A-]/[HA], since pKa is fixed. Since the logarithm function increases very slowly, addition of small amounts of either acid or base will not affect the log term significantly, so the pH will not change too much.
Note: derivation is included in case students forget the equation and may need it
Compare the change in pH when 0.1 mol of HCl is added to a solution of only 1L water versus when it is added to a solution in 1L of water with 1.0 moles of acetic acid and 1.0 moles of acetate ion.
Note: pKa = 4.76 for acetic acid.
When 0.1 moles of HCl is added to 1L of water, there is 0.1M of HCl. The pH changes from 7 to 1, a drastic change. However, when 0.1 moles of HCl is added to a buffer solution of 1M acetate ion and 1M acetic acid, the acetate ion will fully react with the hydrochloric acid. Since the reaction is A- + HCl –> HA + Cl-, there is now 0.1 moles less of acetate ion and 0.1 moles more of acetic acid. Originally, since there were equal parts of acid and conjugate base, the Henderson Hasselbalch equation gives pH = pKa = 4.76. Now, the new pH is pH = pKa + log(0.9/1.1) = 4.67, a very small change comparatively.
