Posted by Kristen on Saturday, April 21, 2012 at 8:25pm.
40 mL x 0.25M = 10.0 millimols HF.
80 mL x 0.125M = 10.0 mmols NaOH
So you are at the equivalence point and the pH will be determined by the hydrolysis of the NaF.
(NaF) = 10 mmols/120 mL - 0.0833M
..........F^- + HOH ==> HF + OH^-
init..0.0833............0.....0
change...-x.............x.....x
equil.0.0833-x...........x....x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(F^-)
Substitute the equil line from the ICE chart into the K expression and solve for x = (OH^-) then convert to pH.
See line 5.
(NaF) = 10 mmols/120 mL - 0.0833M
Replace the - with =
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