1.

ionisation constant of CH_(3)COOH is 1.7xx10^(-5)and concentration of H^(+) is 3.4xx10^(-4). Then the initial concentration of CH_(3)COOH molecules is :

Answer»

`3.4xx10^(-4)`
`3.4xx10^(-3)`
`6.8xx10^(-3)`
`1.7xx10^(-3)`

Solution :`CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)`
If `alpha` is the INITIAL CONCENTRATION of `CH_(3)COOH`, then at equilibrium.
`(alpha-3.4xx10^(-4)) ""3.4xx10^(-4) ""3.4xx10^(-4)`
`K=((3.4xx10^(-4))xx(3.4xx10^(-4)))/((alpha-3.4xx10^(-4)))=1.7xx10^(-5)`
`(alpha-3.4xx10^(-4))=((3.4xx10^(-4))^(2))/(1.7xx10^(-5))`
`=6.8xx10^(-3)`
`THEREFORE alpha = 6.8xx10^(-3)+3.4xx10^(-4)~~ 6.8xx10^(-3)`


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