1.

What will be the percentage of dissociation in 1.0 M CH_(3)CO OH at equilibrium having dissociation constantof 1.8xx10^(-5)?

Answer»


SOLUTION :As inProblem `2, K = (C alpha^(2))/(1-alpha), i.e., 1.8xx10^(-5)=(1xxalpha^(2))/(1-alpha)~~ alpha^(2) or alpha = sqrt(1.8xx10^(-5))=4.24xx10^(-3)`
% dissociation `= (4.24xx10^(-3))xx100=0.424`


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