Question
A buffer solution is prepared by mixing 0.1 M acetic acid (Ka=1.8×10−5) and 0.1 M sodium acetate. Find (a) the pH of the buffer. (b) The new pH after adding 0.01 mol of HCl to 1 L of this buffer (assume volume unchanged).
Solution — Step by Step
pH=pKa+log[HA][A−]=−log(1.8×10−5)+log0.10.1
pKa=4.74. Since [A−]=[HA], the log term is zero.
pH=4.74
The HCl reacts with the conjugate base (acetate):
CH3COO−+H+→CH3COOH
Moles of acetate decrease by 0.01, moles of acetic acid increase by 0.01.
New [A−]=0.1−0.01=0.09 M.
New [HA]=0.1+0.01=0.11 M.
pH=4.74+log0.110.09=4.74+log(0.818)
log(0.818)=−0.087. So:
pH=4.74−0.087≈4.65
Initial pH =4.74. After adding HCl, pH ≈4.65.
Why This Works
A buffer resists pH change because adding strong acid converts conjugate base to weak acid (and vice versa for strong base). The Henderson-Hasselbalch equation tracks the resulting ratio.
Compare: adding 0.01 mol HCl to pure water (1 L) would give [H+]=0.01, pH = 2. The buffer changes pH by only ∼0.1 unit. That is the buffer effect.
Alternative Method
Track moles directly: total acetate after reaction =0.09, acetic acid =0.11. Use pH=pKa+log(nA−/nHA)=4.74+log(9/11). Volumes cancel — useful when the volume is unspecified.
Common Mistake
Reversing the reaction direction. Added H+ converts A− to HA, not the other way round. Students sometimes write the wrong sign of change and end up with pH >4.74, which is impossible after acid addition.