Determine H3O and HF in 0500M HF Then determine these concen

Determine [H3O+] and [HF] in 0.500M HF. Then determine these concentrations in a solution that is 0.100M HCl and 0.500M HF.

Solution

HF+H2O -----> H3O+ + F-

Ka = 6.3*10^-4

now Ka = = 6.3 *10^-4 = [H3O+][F-]/[HF]

=> 6.3*10^-4 = x^2/0.5-x

solving you get x = 0.0174

hence [H3O+] = 0.0174 and [HF] = 0.5-0.0174 = 0.4826

now when 0.1M HCl is present, since HCl is a strong acid it dissociates completely

hence [H3O+] = 0.1M

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

=> 6.3*10^-4 = x(0.1+x)/0.5-x

solving you get x = 0.003

hence [H3O+] = 0.1+0.003 = 0.103

and [HF] = 0.5-0.003 = 0.497