A 40 ml solution of weak base BOH is titrated with 0.1 N HCl solution. The pH of solution is 10.04 and 9.14 after addition of 5 ml & 20 ml of acid. If `K_(b)` for weak base is `P x 10^(-5)M,` find P.

Answer

`1.82`

Explanation

Initial (mM) of BOH = a (mM)
`implies`
`{:(BOH , + , HCl , to , BCl , + , bot), ((a -0.5),, 0.5-0.5,, 0.5,,l), (mM,, 0mM,, mM,, r):}`
The slution is buffer with
`[BOH] = [ (a - 0.5)/(v)] [BCl] = [ (0.5)/(v)]`
`pH = 10.04 therefore pOH = 3.96`
`3.96 =- log K _(b) + log "" ( 0.5 //v)/( a - 0.5//v) `
`3.96 =- log K _(b) + log "" ( 0.5)/(a -0.5) ...(i)`
` implies `
`{:(BOH , + , HCl , to , BCl, + , H ), ((a -2)mM,, 0mM,, 2mM,,2r):}`
Again it is buffer solution
`[BOH] = a -2//V _(1) [BCl] = 2 //V _(1)`
`pH = 9. 14 therefore pOH = 4.86`
`pOH = 4. 86 =- log K _(b) + log "" (2)/( a -2)....(ii)`
(By) (i) & (ii) i.e., (ii)- (i) `implies`
`0. 90 = log "" (2)/(a -2) ((a - 0.5))/( 0.5) = log "" ( 2 a - 1)/( 0.5 a - 1)`
`7. 94 = ( 2a -1)/( 0.5 a - 1) implies 3. 97a - 7.94 = 2a -1`
`1 . 97 a = 6. 94 therefore a = 3.52.`
(ii) `implies 4 . 86 = - log K _(b) + log "" (2)/(1. 52) =- log `
`K _(b) + 0.12`
`-log K _(b) = 4.74 therefore log K _(b) =- 4. 74`
`K _(b) = 1 . 82 xx 10 ^(-5) M.`