The acid dissociation constant, Ka, for HF is 3.5x10^-4 at 25C. The molar ionic

Question

The acid dissociation constant, Ka, for HF is 3.5x10^-4 at 25C. The molar ionic conductivities for each ion are as follows: H3O+ is 349.65, F- is 55.4. For a 0.02M solution of HF, calculate: a) The percent dissocation b) The total conductivity c) The mobility of each ion

1) The acid constant, for HF is 3.5 x 104 at 25 The molar ionic conductivities for each ion are as follows: No for Hgo is 349.65 (cm2/2 mol) and for F is 55.4 cm For a 0.02 M solution calculate: mobility each ion a) The percent dissociation b) The total conductivity c) The of

Explanation / Answer

Answer (a) The percent dissocation

Overall reaction will be

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

I                      0.02 0    0

C                         x                                         x                             x

E                      0.02 - x                             x x

Now

Ka = x2 / (0.02 -x) = 3.5 x 10-4

x = [H+] = 2.65 x 10-3 M

The percent dissocation = (2.65 x 10-3 M / 0.02 M ) = 13.23 %

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