A problem about calculating for the pH?

Question

The Ka (acid dissociation constant) for benzoic acid is 6.5 x 10^-5. Calculate the pH of a 0.12 M benzoic acid solution?

Thanks for your help!

Answer 1

Much easier way, I answered one similar to this given by you aswell so check that first.

ka= [B-][H+]/[BH],         [BH] is constant  as 0.12M as next to nothing of it dissociates shown by the low ka value (below 10^-2 generally).

[B-]=[H+] so sub in [H+]

ka= [H+]^2/[BH]

rearange and sub in [BH] and ka



6.5*10^-5(ka) * 0.12M ([BH])=[H+]^2

7.8*10^-6=[H+]^2

[H+]= square root of 7.8*10^-6 = 2.79*10^-3

ph=-Log[H+]= -Log(2.79*10^-3)= 2.55

Answer 2

BH<--->B- + H+

[B-]=[H+]

so[H+]^2=Ka*[BH]

[H+]= square root  Ka*[BH]

1/[H+]=square root 1/Ka*1/[H+]

pH = (Ka-log c)/2=( 4.18+0.92)/2=2.55