Calculate the `pH` at equilibrium point when a solution of `10^(-6) M CH_(3)COOH` is titrated with a solution of `10^(-6)M NaOH. K_(a)` for acid `2 xx 10^(-5) (pK_(a) = 4.7)` (Answer given in whole number).

Explanation

Salt of `W_(A)//S_(B) (CH_(3)COONa)` is formed.
If we calculate `pH` by using hydrolysis formula for salt of `W_(A)//S_(B)`
`pH = (1)/(2) (pK_(w) + pK_(a) + logC)]` Salt of `S_(A)`
`[["Salt"]=(10^(-6))/(2)(underset("doubled")("Volume is"))]`
`= (1)/(2) (14 +4.7 + "log" (10^(-6))/(2))`
`= (1)/(2) (18.7 - 6- 0.3) = 6.2`
But `pH` of salt of `W_(A)//S_(B)` is always `gt 7`.
So for such dilute solutions of `CH_(3)COOH` and `NaOH`.
Contribution of `overset(Theta)OH` ions from `H_(2)O` must be considered.
`{:(CH_(3)COOH +, NaOH rarr, CH_(3)COONa +H_(2)O),(10^(-6)M xxV, 10^(-6)M xxV, 10^(-6)M xxV):}`
`[CH_(3)COONa] = (10^(-6)M xx V)/(2V) = 0.5 xx 10^(-6)`
`= 5 xx 10^(-7) lt 10^(-6)M`
`[overset(Theta)OH] = ((K_(w)xxC)/(K_(a)))^(1//2) = ((10^(-14)xx5xx10^(-7))/(2xx10^(-5)))^((1)/(2))`
`=(2.5)^((1)/(2)) xx 10^(-8)M = 1.58 xx 10^(-8)M lt 10^(-6)M`
Total `[overset(Theta)OH] = [1.58 xx 10^(-8) +10^(-7)` (from `H_(2)O)]`
`= 10^(-7) (1.58 xx 10^(-1) +1) = 1.158 xx 10^(-7)`.
`pOH =- log (1.158 xx 10^(-7)) = 6.93`
`pH = 14 - 6.93 = 7.07 ~~7`.