pH and dissoaction constand?

Question

The dissociation constant Ka for the weak acid is equal 10^-5. pH of 0,1M solution of this acid is

a, 2 b 3

b is correct. Why?

Answer 1

you have

HA<---> H+ OH-

Ka= [H+][OH-]/[HA] [H+]=[OH-]

Ka= [H+]^2/[HA] --->[H+]^ 2=Ka*[HA]
[HA] is equal to c (this is important since the acid is WEAK)

so 1/[H+]^ 2 = (1/Ka)(1/c)

you go to the log

log1/[H+]^ 2 = log(1/Ka)(1/c) = log(1/ka) -log c

remember log1/[H+]=pH log(1/Ka)=pKa

2pH= pKa-log c=5-(-1) = 6

pH =3

Answer 2

Ha<---> H+ OH-

Ka= [H+][OH-]/[HA] [H+]=[OH-]

Ka= [H+]^2/[HA] --->[H+]^ 2=Ka*[HA]


so 1/[H+]^ 2 = (1/Ka)(1/c)



log1/[H+]^ 2 = log(1/Ka)(1/c) = log(1/ka) -log c

log1/[H+]=pH log(1/Ka)=pKa

2pH= pKa-log c=5-(-1) = 6

pH =3