When `40mL` of a `0.1 MN` weak base, `BOH` is titrated with `0.01M HC1`, the `pH` of the solution at the end point is `5.5`. What will be the `pH` if `10mL` of `0.10M NaOH` is added to the resulting solution ?

Explanation

At the end point, `pH = 5.5` (i.e. and acidic solution).
It means salt formed at the end of neutralisation undergoes hydrolysis.
For `4 mmol (40 xx 0.1)` of weak base, `4 mmol` of acid is required at the end point
`{:(BOH+,HC1hArr,BC1+,H_(2)O,),(4,4,-,-,),(-,-,4,,):}`
`rArr mmol` of salt `(BC1)` formed `= 4`
`rArr [BC1] = 4//V` where `V = V_(HC1) + V_(base)`
At the end point: `4 = 0.10 xx V_(HC1)`
`rArr V_(HC1) = 40 mL`
`rArr V = 40 + 40 = 80 mL`
`rArr [BC1] = 4//80 = 0.05 M`
Notw that `BC1` is a salt of strong acid and weak base.
`rArr [H^(o+)] = sqrt((K_(w)C)/(K_(b))) (pH = 5.5, :. [H^(o+)] = 3.2 xx 10^(-6) M)`
`rArr K_(b) = (K_(w)C)/([H^(o+)]^(2)) = ((10^(-14))xx(0.05))/((3.2 xx 10^(-6))^(2)) = 5.0 xx 10^(-5)`
When `10 mL` of `0.12M NaOh` is added:
`{:(,BC1+,NaOHrarr,BOH+,NaC1),("Initially",4mmol,10xx0.10,,),("Finally",4-10xx0.10,,,),(,4-1=3.0,-,1.0,):}`
It means a baasic buffer containing `3.0 mmol` of `BC1` and `1.0 mmol` of `BOH` is formed. Find the `pH` by using Henderson's eauation for basic buffer.
`pOH = pK_(b) + log ([BC1])/([BOH])`
`rArr pOH =- log (5 xx 10^(-5)) + log .([3.0//V])/([1.0//V])`
`rArr pOH = 4.77`
`rArr pH = 14 - pOH = 9.22`