1.

A 0.025M solution of a monobasic acid had a freezing point of -0.06^(@)C. Calculate K_(a) for the acid . K_(f)(H_(2)O)=1.86

Answer»

Solution :`DeltaT_(f)` (observed)=`0.06^(@)C`, (f.p. of `H_(2)O=0^(@)C`)
`DeltaT_(f)` (calculated) `=0.025xx1.86=0.0465^(@)C`
(for DILUTE aqueous solution, molarity = molality)
`:.i=("observed depression in f.p.")/("calculated depression in f.p.")=(0.06)/(0.0465)=1.29`
For the acid HA we have
`{:(0.025,,,,0,,,,0,,"initial concentration"),(HA,,=,,H^(+),,+,,A^(-),,),(0.025(1-X),,,,0.025x,,,,0.025x,,"final concentration"):}`
(x is the degree of dissociation of HA)
`i=(1+x)/(1)=1.29`, `x=0.29`
`:.K_(a)=((0.25x)(0.025x))/(0.025(1-x))=0.025(x^(2))/(1-x)`
`=(0.025xx0.29^(2))/(1-0.29)=2.96xx10^(-3)`


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