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When 100.0ml of a weak acid was titrated with 0.09381M NaOH, 27.63ml

were required to reach the equivalence point. The pH at the equivalence point was 10.99. What was the pH when only 19.47ml of NaOH had been added ?

I'll just post my work and I need someone to check and see if It's correct

HA+ NaOH => H2O + NaA
NaA => Na+ + A-
A- + H2O => HA + OH-

need Kb to get Ka

[OH-]= rad(Kb*cNaA)
cNaA= .02763L(0.09381molNaOH/L)= 2.59e-3 mol NaA/ 0.12763L= 0.0202M NaOH

pH= 10.99
pOH=14-10.99= 3.01
[OH-]=10^-pOH= 10^-3.01= 9.77e-4

9.77e-4=rad(Kb*0.0202M NaA)
Kb= 4.72e-5
Kw/Kb=Ka
Ka= 1.0e-14/4.72e-5= 2.12e-10

Need the M of HA so using the equillibrium ammounts..(not sure if this is correct)

(0.02763L)(0.09381M NaOH)=(0.1L)(M HA)
M HA= 0.0259M

Since if I need the pH when only 19.47ml of NaOH is added...it's a buffer sol so...
[H3O+]= Ka(cHA/ cNaA)

Ka= 2.12e-10
cHA= .1L(0.0259mol/L)= (2.59e-3molHA - 1.826e-3mol NaOH)/ (0.11947L)= 6.39e-3M HA

cNaA= .01947L(0.09381mol NaOH/L)= (1.826e-3 mol NaOH)/(0.11947L)= 0.01528M NaA

plugging it all in and..
[H30+]= (2.12e-10)(6.39e-3MHA/ 0.01528M NaA)= 8.86e-11

pH= -log[8.86e-11]= 10.05

Is it correct? it looks assuming I'm using common sense, fine since it is before the eq point and only 19.47ml of NaOH added so I would expect the pH to be less but by how much I'm not sure.

Thanks =)

Except for a spot or two in which we disagree on the last digit or two, you are ok. For example, you have 4 significant figures in NaOH molarity and mL, so I calculated to 4 places. The 2.59 I used 2.592 and the molarity I found 0.02031 with Ka being 2.126 x 10^-10. (OH^-) = 9.772 x 10^-4. and final pH = 10.0497 which rounds to 10.05 and that's the same as your answer. It's interesting how you did the final (H^+) calculation for that is the way I always taught it but I never see that used these days. Instead, everyone uses the Henderson-Hasselbalch equation. It's a little faster than the method you used but either is fine.

well the reason I calculated it like that was I kept getting confused with the other eqzn. I used that at the beginning but you have switch the concentrations around so rather than do it that way I just think that if I remember the other eqzn which is simpler and get the answer right then I really don't care if it takes longer.

Thanks Dr.Bob =D

You're right. The equation you used IS simpler. The H-H equation has the advantage of not requiring us to change from Ka to Kb or the other way around. We can ALWAYS use Ka.

nods*

what is the meaning of "rad" in [OH-]= rad(Kb*cNaA)

You're tight

Well, it looks like you did all the necessary calculations and arrived at the correct answer. Good job! It's always important to double-check your work when it comes to chemistry. But hey, whether you used the Henderson-Hasselbalch equation or the equilibrium equation, you got the right answer in the end. That's what counts! Keep up the good work!

Yes, the calculation appears to be correct. You correctly calculated the concentration of NaOH (cNaA) as 0.0202M and the concentration of OH- as 9.77e-4. You then used these values to calculate the Kb, Ka, and the concentration of HA (cHA) as 0.0259M. Finally, you used the Henderson-Hasselbalch equation to calculate the pH when only 19.47ml of NaOH had been added, and obtained a pH value of 10.05, which is consistent with the expected decrease in pH as more acid is added. Well done!

Yes, it's great that you were able to understand and use the simplified equation to calculate the pH at a specific point in the titration. It's important to choose a method that you are comfortable with and can use accurately. In this case, both methods give the correct answer, so it's a matter of personal preference. Good job on your calculations!