Question

# Consider titration of 45.00 mL of 0.05982 M hydrofluoric acid, HF, with a standard 0.08780 M...

Consider titration of 45.00 mL of 0.05982 M hydrofluoric acid, HF, with a standard 0.08780 M solution of potassium hydroxide, KOH. The volume of the KOH solution needed to reach the equivalence point of the titration is ___________ mL. (Fill in the blank. Report the number only, with correct number of significant figures.) The pH at the equivalence point is ___________. (Fill in the blank, using the terms 7, > 7 or < 7 only.)

(M1V1)HF = (M2V2)KOH

0.05982 * 45 = 0.08780 V2

V2 = 0.05982 * 45 / 0.0878 = 30.7 mL

the equivalence point of titration is 30.7 mL

The pH at equivalence point will be determined by

F- + H2O -----------> HF + OH-

CF- = no of moles of HF / total volume = 2.6919/ 75.7 = 0.0356 M ( as whole of HF will be converted into KF)

as the dissociation wil be small, as Kb value is very small

Kb = [HF][OH-]/[F-]

[OH-] = (Kb * cF-)1/2 =(1.5 * 10-11 * 0.0356)1/2 = 7.3 * 10-7

pOH = - log [OH-] = 6.14

pH = 14 - pOH = 7.86

thus pH is greater than 7 at equivalence point

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