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If the freezing point of 0.1 M HA(aq) so...

If the freezing point of `0.1 M HA(aq)` solution is `-0.2046^(@)C` then `pH` of solution is
`(` If `K_(f)` water `=1.86mol^(-1)kg^(-1))`

Text Solution

Verified by Experts

The correct Answer is:
2
`DeltaT_(f)=iK_(f)m`
`i=(DeltaT_(f))/(K_(f)m)=(0.2046)/(1.86xx0.1)=(204.6)/(186)`............(i)
`{:(,,HA,,hArr,,H^(+),,+,,A^(-)),("Initial",,0.1,,,,,,,,),("At eqm.",,0.1-x,,,,x,,,,x):}`
`i=(0.1-x+2x)/(0.1)=(0.1+x)/(0.1)=1+10x`.........(ii)
Eqn. (i) and (ii) , `x=10^(-2)`
`:.pH=-log[H^(+)]`
`=-log10^(-2)=2`
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