Why would adding a strong acid to a buffered solution decrease the pH?

[monkeyMD]

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In TBR, Chem, section 4, page 320, passage 2, number 12, it asks, "addition of 1ml of .10M KOH to a solution made by mixing 15.0mL 0.10M H3CCo2H with 10.0mL 0.10M H3CCO2N results in..."
Answer: a soln with pH between 3.74 and 4.74
I would reason that the pH would INCREASE since you just added a strong base, which could also deprot the acid, but the answer indicates a dec in pH. Can someone explain?

 

[GreenRabbit]

In TBR, Chem, section 4, page 320, passage 2, number 12, it asks, "addition of 1ml of .10M KOH to a solution made by mixing 15.0mL 0.10M H3CCo2H with 10.0mL 0.10M H3CCO2N results in..."
Answer: a soln with pH between 3.74 and 4.74
I would reason that the pH would INCREASE since you just added a strong base, which could also deprot the acid, but the answer indicates a dec in pH. Can someone explain?

Yes, a base will always raise the pH. The key is using the Henderson-Hasselbalch equation properly. If we know the pKa of acetic acid is ~4.74, then pH = pKa + log [A-/HA] gives us a pH <4.74. We know this because [A-]/[HA] is (1/1.5) < 1, therefore the log is negative. If we assume the 1ml of .10M KOH neutralizes approx. 1ml of .10M acetic acid, then we still have ~14ml acetic acid. 1/1.4 is still <1, therefore the log is still negative, and the pH is still less than the pKa (less than 4.74), even though it is greater than before.