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A 40.0 mL solution of a weak base, BOH i...

A `40.0 mL` solution of a weak base, `BOH` is titarted with `0.1N HCI` solution. The `pH` of the solution is found to be `10.04` and `9.14` after the addition of `5.0 mL` and `20.0 mL` of the acid, respectively. Find out the dissociation constant of the base.

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`{:("Case-I",BOH,+,HCl,to,BCl,+,H_(2)O),("Millimole before reaction",a,,0.1 xx5 = 0.5,,0,,0),("millimole after reaction ",(a-0.5),,0,,0.5,,0.5):}`
` :. pOH = - log K_(b) + log. ([BCl])/([BOH])" "` …..(i)
` :. pH = 10.04`
` :. 3.96 = - logK_(b) + log. (0.5)/((a-0.5))" "` ....(ii)
` :. pOH = 3.96 `
Case - II : `BOH + HCl to BCl +H_(2)O `
` {:("Millimole before reaction ",a,0.1xx20=2,,),("Millimole after reaction ",(a-2),0,2,2):}`
` :. pOH = - log K_(b) + log . ([BCl])/([BOH])" "` ...(iii)
` :. pH = 9.14 :. pOH = 4.86 `
` :. 4.86 = - log K_(b) + log . 2/((a-2))" "` ....(iv)
Solving Eqs.(ii) and (iv) , `K_(h) = 1.81 xx10^(-5)`
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