Calcualte the pH at the equivalence point when a solution of `0.1M` acetic is titrated with a solution of `0.1M NaOH.(K_(a)for acid = 1.9xx10^(-5))`.

Explanation

Let V mL of acid and V mL of NaOH be used. Conc.of both acid and NaOH are same.
`:. {:(,CH_(3)COOH+,NaOHrarr,CH_(3)COONa+,H_(2)O),("Conc. before reaction",(0.1xxV)/(2V),(0.1xxV)/(2V),0,0),("Conc. after reaction",0,0,(0.1xxV)/(2V),(0.1xxv)/(2V)):}`
`:. [CH_(3)COONa]=(0.1)/(2)=0.05M`
Now calculate pH by hydrolysis of `CH_(3)COONa`.
`CH_(3)COONa+H_(2)OhArrCOOH+NaOH`
`[OH^(-)]=C.h=Csqrt([(K_(H))/(C )])=sqrt([(K_(w)xxC)/(K_(a))])`
`=sqrt([(10^(-14)xx0.5)/(1.9xx10^(-5))])=5.12xx10^(-6)`
`:. pOH=5.29`
`:. pH=8.71`