ph at equivalence point when 25 ml of a 0.175M solution of acetic acid is titrated with 0.10M of NaOH at its end point

Answer 1

HAc + NaOH ==> NaAc + H2O

mols HAc to start = M x L = ?
moles NaOH must be the same; therefore, the volume of NaOH used is
mLNaOH x MNaOH = mLHAc x MHAc
mL 0.1M NaOH used = 43.75 mL but you need to confirm that.
Thus the salt formed will have a molarity of 0.004375/(43.75mL + 25.00mL) = approximately 0.07 M.
The pH at the equivalence point will be determined by the hydrolysis of the acetate ion of the salt.
............Ac^- + HOH ==> HAc + OH^-
initial...0.07.............0......0
change......-x.............x......x
equil....0.07-x.............x......x

Kb for the Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.07-x)
Solve for x = (OH^-) and convert to pH.
You need to obtain a better answer for the 0.07M, too.