A 50 mL solution of weak base BOH is titrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respectively. Find out `K_(b)` for weak base.

Explanation

Let the normality of BOH be n.
m.e. of BOH =40n.
On the addition of 5mL of 0.1N HCl into BOH solution,
m.e. of HCl `=0.1 xx 5=0.5`
m.e. of salt formed =0.5
m.e. of BOH use d=0.5
m..e of reamining BOH =(40n=0.5)
Applying Henderson.s equation
`pOH =pK_b+log (["salt"])/(["base"])`
`(14-10.04)=pK_(b)+log ((0.5)/(40n-0.5)) ........(1)`
Similarly, on the addition of 20mL of HCl,
`(14-9.14)=pK_(b)+log ((2)/(40n-2)) ......(2)`
Subtracting (1) from (2) we get
n=0.088
Substracting n in (1), we get
`pK_(b)=4.7412 `
`-log K_(b)=4.7412`
`or log K_(b)=-4.7412 =bar5=2588`
Taking antilog `K_(b)=1.82 xx 10^(-5)`