Question

A solution is made that is 0.015 M HCI and 0.10 M HF (Ka= 6.8 x 10^-4). What is the concentration of the undissociated weak acid in the mixture?

? M

Answer 1

           HF(aq) <----> H+(aq) + F-(aq)

initiall    0.1 M          0.015 M    -

change       -x           +x         +x

equil      0.1-x          0.015+x     x

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

(6.8*10^-4) = (x*(0.015+x))/(0.1-x)

x = 0.00354

at equilibrium,

concentration of undissociated HF = 0.1-0.00354 = 0.0965 M