Question

a solution containing 100.0 mL of 0.135 M CH3COOH (ka=1.8E-5) is being titrated with 0.54 M NaOH. calculate the ph:

at the equivalence point and 5 mL past the equivalence point.

how much is 5 mL past? i do not know how to do it?

Answer 1

I've posted an answer on your previous post on this similar question. Anyway, I'll just repost it here:

For the equivalence point, I've already posted steps on how to work on it on the other chemistry problem you've posted. So here, I'll just guide you on the "5 mL past equivalence point".

To determine the volume of the point 5 mL past the equivalence point, first, we get the volume of 0.54 M NaOH that was added at equivalence point:
0.135 M * 100 mL = 13.5 mmol CH3COOH
At the balanced reaction, their mole ratio is 1:1, thus,
13.5 mmol CH3COOH * (1 mmol NaOH / 1 mmol CH3COOH) = 13.5 mmol NaOH
M = n/V
V = n/M
V,NaOH = 13.5 mmmol / 0.54 M
V,NaOH = 25 mL
At equivalence point, the total volume is 125 mL.
5 mL past the equivalence point means 125 + 5 = 130 mL. That 5 mL is the 0.54 M NaOH.
At this point, there is excess NaOH (or OH- ions). Getting the moles of OH-,
M = n/V
n = MV
n = 0.54 M * 5 mL
n = 2.7 mmol OH- (excess)
Solve for the concentration:
M = n/V
M = 2.7 mmol / 130 mL
M = 0.020769 M OH-
pOH = -log(OH-)
pOH = -log(0.020769)
pOH = 1.683
pH = 14 - pOH
pH = 12.317

hope this helps~ `u`

Answer 2

thanks