CHEMISTRY LAB urgent assistance
posted by Petty on .
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

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
Ax____x___B+x
so Ka is
Ka=(x)(B+x)/(Ax)
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] 
As an added note (perhaps picky but I think needed) is that / is used as a divide sign while \ is used for other purposes.