Question

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.)

Answer 1

(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^-

^{C_F- = 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}

K_b = [HF][OH^-]/[F^-]

[OH-] = (K_b * c_F-)^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