Chemistry

Consider the titration of a 50.0 mL 0.500 M F- with 0.500 M HCl. The pKa of HF = 3.46 and the pKb of F- is 10.54.
Calculate the pH at the following volumes of HCl added.
1. Vsa= 0.00 mL of 0.500 M HCl
2. Vsa= 25.0 mL of 0.500 M HCl
3. Vsa= 49.5 mL of 0.500 M HCl
4. Vsa= 50.0 mL of 0.500 M HCl

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1. millimols F^- initially = mL x M = 50 x 0.5 = 25
At zero mL HCl added, the pH is determined by the hydrolysis of F^-
......F^- + HOH ==> HF + OH^-
I....0.5M...........0.....0
C.....-x............x.....x
E.....0.5-x.........x.....x

Kb = approx E-11 = (x)(x)/(0.5-x)
Solvr for x = (OH^-) and convert to pH.

At 25 mL HCl you have added 12.5 mmols HCl. The reaction is
......F^- + H^+ ==> HF
I.....25.....0.......0
C....-12.5..-12.5....+12.5
E.....12.5...0.......12.5

Substitute this into the Henderson-Hasselbalch equation. The pH, when you get it worked out, will be 3.46

is done the same way but of course the answer is not 3.46.

4. You will have converted all of the F^- to HF so (HF) = mmols HF/total mL. You work that as a weak acid problem.
Post your work if you get stuck.

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posted by DrBob222

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