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 :

Options

A. `3.4xx10^(-4)`
B. `3.4xx10^(-3)`
C. `6.8xx10^(-3)`
D. `1.7xx10^(-3)`

Answer

C

Explanation

`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)`