# Chemistry

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HF + NaOH =NaF + H2O
What will be the pH of a solution produced when 40 mL of 0.25 M HF is titrated with 80 mL of 0.125M NaOH? Ka for HF is 3.5*10^-4

• Chemistry -

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.

• oops--Chemistry -

See line 5.
(NaF) = 10 mmols/120 mL - 0.0833M
Replace the - with =

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