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A 0.252g sample of unknown organic base ...

A `0.252g` sample of unknown organic base is dissolved in water and titrated with a `0.14 M HCl` solution .After then addition of `20mL` of acid ,a `pH` of `10.7` is recorded. The equivalence point is reached when a total of `40mL` of `HCl` is added . If the base and acid combine in `1:1` molar ratio, then 'a' g is the molar mass of the organic base of 'b' is the ioniation constant of base .Report your answer as `(a)/(1000b)` .

Text Solution

Verified by Experts

At eq.point
No. of millieq. Of Base=No of millieq of `HCl`
`(0.252xx1000)/("Molar mass")=0.14xx40`
:. "Molar mass of base" `=45`
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