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A 50 mL solution of weak base BOH is tit...

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.

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Let 40 mL of base contain x mmol of BOH.
`BOH + HCl rarr BCl + H_(2)O`
x- 0.5 0.5 when 5 mL acid is added
x-2 2.0 when 20 mL of acid is added
When pH is 10.04, pOH = 3.96 and when pH is 9.14, pOH is 4.86. Therefore,
`3.96 = p(K_(b)) + log 0.50/x-0.5`....(i)
`3.96 = pK_(b) + log 2.0/x-2`......(ii)
Subtracting Eq. (i) from Eq. (ii) gives
`0.90 = log(2/x-2 xx x-0.5/0.5) implies 28 = 4(x-0.5)/x-2`
`implies x = 3.5` , substitution in equation (i) gives
`3.96 = pK_(b) + log 0.5/3 implies K_(b) = 1.8 xx 10^(-5)`
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