1.

The ionisation constant for acetic acid is 1.8xx10^(-5). At what concentration will it be dissociated to 2% ?

Answer»

1 M
0.018 M
0.18 M
0.045 M

Solution :`c=(K_(a))/(ALPHA^(2))=(1.8xx10^(-5))/((2xx10^(-2))^(2))`
`=4.5xx10^(-2)=0.045 M`


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