BR CBT 1 PS question 37..please help

[SaCkO]

What is the pH of a solution of 25 mL 0.10 M H3CCO2H(aq) (pKa = 4.74) after 10 mL of 0.10 M KOH(aq) has been added to it?



A. pH ≤ 3.74
B. 4.74 > pH > 3.74
C. 5.74 > pH > 4.74
D. pH ≥ 5.74

So The answer is B if you use the henderson hasselbalch equation the pH will decrease because there is still more acid than base at that point so log A-/HA will be a negative number thus the pH will be less
BUT
lets put the numbers on the side, iam having trouble understanding how when iam adding a strong base to the solution the pH is not going up and it is in fact going down making the solution more acidic. wasn't the solution before adding the base at its Pka value since its a weak acid? and why did the pH of the solution go down when we are adding a base to it?

I would really appreciate if someone explained to me where iam thinking wrong about this

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[SaCkO]

I think I figured what i was doing wrong.
The pH at the beginning is less than the pKa , I don't know why i was thinking that the pH would be equal to the pKa. Only when the Acid is half titrated and the conjugate base equals the Acid the pH would be the pKa.
Hopefully I didn't look dum thinking about this

 

[Class1P]

how do you solve this by hendersan Hasselbach?

 

[411309]

I think I figured what i was doing wrong.
The pH at the beginning is less than the pKa , I don't know why i was thinking that the pH would be equal to the pKa. Only when the Acid is half titrated and the conjugate base equals the Acid the pH would be the pKa.
Hopefully I didn't look dum thinking about this

yeah pretty much. Think of it this way..We have 2.5 M of the weak acid 1 M of the strong base. So the ph will actually be less then the pka. We would need 1.25 M for the ph to equal the pka, and anything more than will have a ph over the pka.

 

[MedPR]

Ah this got me at first too. I read pH = 4.74 instead of pKa = 4.74.