Question

8. Calculate the [H3O+] in a solution which is 0.120 M in HF and 0.015 Min NaF. Ka for HF = 6.7 x 104 a) 5.4 x 10-3 b) 8.4 x 10-5 C) 8.0 x 10-5 d) 9.0 x 10-3 e) 0.125

Answer 1

HF + H2O = H3O+ +F-

Initial concentration 0.120 M 0 0

Concentration at

equilibrium ; 0.120(1- $\alpha$ )   0.120$\alpha$ 0.120$\alpha$

As NaF is strong electrolyte it dissociates completely

  

NaF → Na+ + F-

Initial concentration 0.015M 0 0

Final concentration 0 0.015M 0.015M

Hance total F-  ion concentration =  0.120$\alpha$ + 0.015 $\simeq$ 0.015M ( as $\alpha$ is very less because HF is weak electrolyte hence it can approximately written as 0.015 M).

Ka = [H3O+][F-]/ [HF]

0.00067 = (0.120$\alpha$)(0.015)/0.120(1-$\alpha$)

As HF is weak electrolyte $\alpha$ <<<1

hence 1-$\alpha$$\simeq$ 1

[HF] $\simeq$ 0.120M

0.00067 = (0.120$\alpha$)(0.015)/ 0.120

$\alpha$ = 0.0447

[H3O+] = 0.120$\alpha$ = 0.120*0.0447 = 0.00536

[H3O+] $\simeq$

5.4 x 10-3

Hence option a is correct