The concentration of fluroacetic acid `(K_(a)of acid =2.6xx10^(-3))`, which is required to get `[H^(+)]=1.50xx10^(-3)M` is:

Options

A. `0.865M`
B. `2.37xx10^(-3)M`
C. `2.37xx10^(-4)M`
D. `2.37xx10^(-2)M`

Answer

B

Explanation

`{:(,CH_(2)FCOOHhArrCH_(2),FCOO^(-)+,H^(+)),("Moles before dissociation",1,0,0),("Moles after dissociation", (1-alpha),alpha,alpha):}`
Given, `[H^(+)]=Calpha=1.5xx10^(-3)mol litre^(-1)`
`:' K_(a)=((Calpha)(Calpha))/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
`2.6xx10^(-3)= (1.5xx10^(-3)xxalpha)/((1-alpha))`
`:. alpha = 0.634`
Now, `Calpha=1.50xx10^(-3)`
`:. C= (1.5xx10^(-3))/(0.634)=2.37xx10^(-3)M`
Note: Since `K_(a)` is of the order of `10^(13)M` and thus it is not advisable to use `K_(a)=Calpha^(2)`. Because `(1-alpha)` is not equal to 1 since `alpha` is not small.