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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ddarkpassengerr on March 30, 2011, 11:07:03 PM

Title: Titrations
Post by: ddarkpassengerr on March 30, 2011, 11:07:03 PM

Consider the titration of 37.8 mL of 0.245 M HF with 0.210 M NaOH. Calculate the pH at each of the following points. 44.1 mL of base are needed to reach the equivalence point.

a) Calculate the pH after the addition of 9.45 mL base.
b) Calculate the pH at halfway to the equivalence point.
c) Calculate the pH at the equivalence point.
d) Calculate the pH after the the addition of 75.6 mL of base

a) Wouldn't it be (.0378 L)(.245 M) = .009261 mol of HF
(.00945 L)(.210 M) .001985 mol of NaOH <--- limiting reagent

So, there are 0 mol of NaOH after the addition, and .00728 mol of HF left.
So [H3O+] = moles/Vtotal = .00728mol/(.0378+.00945L) = .154 M
pH = - log(.154)= .81
But it's not the right answer and I don't quite know what I'm doing wrong.... and since I don't know how to do that step, I'm not sure about the following 3 steps either. Can someone point me in the right direction?

Title: Re: Titrations
Post by: Nobby on March 31, 2011, 01:07:16 AM

HF is a weak acid, so I think you need Henderson-Hasselbalch equation.

Title: Re: Titrations
Post by: Borek on March 31, 2011, 03:43:45 AM

http://www.titrations.info/titration-curve-calculation

http://www.titrations.info/acid-base-titration-curve-calculation

Title: Re: Titrations
Post by: ddarkpassengerr on March 31, 2011, 06:36:36 PM

Okay, so using Henderson-Hasselbach (for part A)

pH = pKa + log(A-/HA)
pKa of HF = 3.15
[HA] = (.0378L)(.245M) = .0093 mol
[A-] = (.00945L)(.210M) = .0020 mol

pH= 3.15 + log(.0020/.0093) = 2.48
But that's still not the right answer.... I'm probably doing something completely wrong.

Title: Re: Titrations
Post by: Borek on April 01, 2011, 05:24:11 AM

Quote from: ddarkpassengerr on March 31, 2011, 06:36:36 PM

[HA] = (.0378L)(.245M) = .0093 mol

That was initial, what about neutralized acid?