Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Jiro on November 21, 2005, 03:31:07 PM
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suppose you have 100mL of a solution that is 0.1M in HF and 0.10M in NaF, to which your lab instructor now adds 10mL of 0.10M NaOH.
what is final pH?
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I think first you go like this:
HF -> H^+ + F^-
I .1 0 0
C .1-x +x +x
E .1-x x x
with HF's ka=7.1E-4 i got [H+]=.008M
NaOH -> Na^+ + OH^-
I .1 0 0
C 0 +.1 +.1
E 0 .1 .1
so therefore [OH-]=.1M find [H+] here using Kw i got 1E-13
Now tricky part... adding the two H+ concentrations... remember you can't add M units together you can only add mol units. remember you only compare mols.
change (.008mol / L) * .1L = 8E-4mol
change (1E-13mol/L) * .01 = 1E-15mol
8E-4 mol + 1E-15mol = 8.0E-4mol
now divide by the whole volume...
8.0E-4 mol / 0.110 L
= .00727M = [H+]
find pH=-log[H+]
=2.14
Correct me if im wrong, cause it looks fishy... haha thanks...
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That's a buffer question, you should use Henderson-Hasselbalch (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch) equation. Assume that NaOH reacted quantitatively with HF.