1.

The ionisation constant of 0.2 M formic acid is 1.8xx10^(-4). Calculate its percentage ionisation.

Answer»

SOLUTION :FORMIC acid is a weak acid. Let is be represented as HA.
`underset(C(I-alpha))(HA) hArr underset(C alpha)( H^(+))+underset(C alpha)(A^-)`
ACCORDING to Ostwald's dilution law,
`K_a=C alpha^2. `
`therefore alpha=sqrt(K_a)/(C )`
`K_a=1.8xx10^(-4), C=0.2 M=2xx10^(-1) M`
` therefore alpha =sqrt((1.8xx10^(-4))/(2xx10^(-1)))=sqrt(9xx10^(-4)3xx10^(-2))`
Percentage of IONISATION `= 100 alpha `
`=10^2 xx3xx10^(-2)=3`


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