Question

1 (A) This relation gives the dissociation constant of acetic acid (CH3COOH)

Ka = [H3O+][CH3COO-]\[CH3COOH]

Starting from this relationship show that

pH = pKa + log [base]\[acid]

urgently need assistance

Answer 1

Start with the acid dissociation equation:

HAc -> H+ + Ac-

Ka is then

[H+][Ac-]/[HAc] {given in the question}

at the start

HAc -> H+ + Ac-

and we start from A mole/L and B mol/L for the acid and buffer

A M ___0__ B M

at the end

HAc -> H+ + Ac-
A-x____x___B+x

so Ka is
Ka=(x)(B+x)/(A-x)

Assume that x is small wrt A and wrt B so that the equation reduces to

Ka=x(B)/(A)

which we can rearrange to

x=Ka(A)/(B)

take logs of each side

log(x)=log(Ka(A)/(B))

which is

log(x)=log(Ka) + log((A)/(B))

multiply through by -1

-log(x)=-log(Ka) - log((A)/(B))

which is

-log(x)=-log(Ka) + log((B)/(A))

or

pH = pKa + log((B)/(A))

or

pH = pKa + log [base]\[acid]

Answer 2

As an added note (perhaps picky but I think needed) is that / is used as a divide sign while \ is used for other purposes.