Question

What is the pH of a solution made by dissolving 7.91 grams of calcium fluoride in enough water to make 690 mL of solution? The Ka for HF is 6.8x10–4.

Answer 1

Let a be the dissociation of the weak acid
HA <---> H + + A-
initial conc. c 0 0
change -ca +ca +ca
Equb. conc. c(1-a) ca ca

Dissociation constant , Ka = ca x ca / ( c(1-a)
= c a^2 / (1-a)
In the case of weak acids a is very small so 1-a is taken as 1
So Ka = ca^2
==> a = √ ( Ka / c )
Given Ka = 6.8x10^-4

c = concentration = (mass/molarmass)X(1000/volume in mL)

=(7.91 g/20(g/mol))*(1000/690 mL)

= 0.573 M

Plug the values we get a = 0.0344

[H+]=ca= 0.0197 M

pH= - log[H+]=1.70