A 50.0 mL aliquot of a 0.01 M solution of HCOOH was titrated with 0.10 M NaOH Predict the pH of the solution (a) at the half-equivalence point
(b) at the equivalence and (c) after an excess 10mL of the base is added. `K_(a) (HCOOH)=1.772 xx 10^(-4)`.

Explanation

(a) m.m. of HCOOH taken `=0.01 xx 50=0.5`
Because NaOH and HCOOH combine in 1:1 molar ratio m.m. of NaOH added at the half-equivalence point =0.25
in the mixture,
m.m. of HCOONa formed=0.25
m.m. of HCOOH left unreacted =0.5-0.25=0.25
Thus, we have for buffer solution of HCOOH-HCOONa
`pH =pK_a+log ""("m.m. of salt")/("m.m. of acid")`
`pH =-log (1.772 xx 10^(-4))+log""(0.25)/(0.25)=3.7515`
(b) At the equivalence point 0.5.m.m. of HCOOH combines with 0.5m.m. of NaOH to form 0.5m.m. of HCOONa
Volume of NaOH required at the equivalence point `=(0.01 xx 50)/(0.1)`
=5mL
Total volume at the equivalence point =50+5=55mL.
`[HCOONa]=(0.5)/(55) =[HCOO^(-)]`
As `HCOO^(-)` ions undergo hydrolysis
`pH =1/2 [pK_(w)+log K_(a)+log a]`
`pH =1/2 {14+3.7515+log ""(0.5)/(55)}=7.85`
(c) m.m. of the excess `NaOH=0.1 xx 10=1` total volume =55+10=65mL
molarity of NaOH `=1/(65)M`
In the presence of the strong electrolyte NaOH, `OH^(-)` ions produced due to hydrolysis of HCOONa may be neglected
`pOH=-log [OH^(-)]=-log (1/(65))=0.8129`
`pH =pK_(w)-pOH=14-0.8129=13.1871`