Calculating pH given Ka

pKa of acetic acid=4.76

a)when no base is added,

let the dissocitation be x.so,

Ka=x^2/(c-x)

or 1.8*10^-5=x^2/(0.2-x)

or x=1.89*10^-3

so pH=-log(x)

=2.72



b) when halfway to the equivalence point,

pH=pKa+log(salt/acid)

=4.76+log(salt/acid)

since [salt]=[acid],

so pH=pKa

so pH=4.76



c)let the volume of NaOH needed be x ml.so,

0.2*50=0.222*x

or x=45.05 ml

so,

pH=4.76+log((0.222*40.05)/(0.2*50-0.222*40.05))

or pH=5.66



d)at the equivalence point,

pH=7+0.5pKa+0.5log(C)

where C is the concentration of the salt formed.

so,

pH=7+0.5*4.76+0.5*log(0.2*50/95.05)

or pH=8.89



e)at 5 ml beyond the equivalnce point,

[OH-]=(50.05*0.222-50*0.2)/(50+45.05)

=0.011

so pH=14-pOH

=12.04