Question

Ka=7.2*10^-4

What is the pH at the equivalence point in the titration of a 28.7 ml. sample of a 0.314 M aqueous hydrofluoric acid solution with a 0.342 M aqueous barium hydroxide solution?

Answer 1

HF + Ba(OH)2----> BaF2 + H2O

0.314M. 0.342 M

0. 0.28 M 0.314 M

Finally we get mixture of two bases, so we will calculate pH from strong base Ba(OH)2 here.

Conc. of OH- = 2×0.28=0.56

pOH = -log 0.56 = 0.252

pH= 14-0.252= 13.748

Answer 2

Ba(OH)2 + 2HF Bar + 2tt20 Imole Ba(OH) 2 can neutralize 2 moleste I Mive = 2 M2 V2 > 20% 0.314 x 29.7 - 2x0.342 X V2 0.314 x 24.7 2 x 0.392 - 13.2 mL . [fo] = 28.740.314. (28.7+ 13.2 = 0.215 M i pt - 1/2 pk w + 2 pka + 1 & 2 log [ro] Las is diacidie = 12 x 14 + + + 3.2 + log (0.285) = 7+ 1.6-0.67 = 7.93 babe