Question

Ionisation constant of `CH_(3)COOH` is `1.7xx10^(-5)` and concentration of `H^(+)ions` is `3.4xx10^(-4)`. Then, find out initial concentration of `CH_(3)COOH` molecules.
A. `3.4xx10^(-4)`
B. `3.4xx10^(-3)`
C. `6.8xx10^(-4)`
D. `6.8xx10^(-3)`

Answer 1

Correct Answer - D
`CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)`
`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
Given that,
`[CH_(3)COO^(-)]=[H^(+)]=3.4xx10^(-4)M`
`K_(a)` for `CH_(3)COOH=1.7xx10^(-5)`
`CH_(3)COOH` is weak acid, so in it `[CH_(3)COOH]` is equal to initial concentration. Hence,
`1.7xx10^(-5)=((3.4xx10^(-4))(3.4xx10^(-4)))/([CH_(3)COOH])`
`[CH_(3)COOH]=(3.4xx10^(-4)xx3.4xx10^(-4))/(1.7xx10^(-5))`
`=6.8xx10^(-3)M`