Calculate the degree of dissociation and concentration of `H_(3)O^(+)` ions in 0.01 M solution of formic acid `(K_(c) =2.1 xx 10^(-4) " at " 298 K)`

Answer

`alpha =0.144, [H_(3)O^(+)] =1.44 xx 10^(-3) M`

Explanation

The `alpha`be the degree of dissociation of formic acid `(HCOOH)` which is a weak acid. The concentration of different species at the equilibrium point is :
`{:(HCOOH(aq),hArr,HCOO^(-)(aq),+,H^(+)(aq)),(C(1-alpha),,Calpha,,Calpha):}`
Applying Law of chemical equilibrium :
`K_(a) =[[HCOO^(-) (aq)][H^(+) (aq)]]/[[HCOOH(aq)]] , 2.1 xx 10^(-4) =((0.01alpha)xx(0.01alpha))/(0.01(1-alpha))`
Since `alpha` is very small for the formic acid which is a weak acid : `(1-alpha) ~~1`
`:. " "2.1 xx 10^(-4) =((0.01alpha)(0.01alpha))/(0.01) , alpha^(2) =(2.1 xx10^(-4))/(0.01) =2.1 xx 10^(-2)`
`" or "" "alpha =(2.1 xx 10^(-2))^(1//2) =1.44 xx 10^(-1) =0.144`
`[H^(+)] =Calpha =0.01 xx 0.144 =1.44 xx 10^(-3) M`