A 0.010M solution of maleic acid, a monoprotic organic acid is 14% ionised. What is `K_(a)` for maleic acid ?

Options

A. `2.3 xx 10^(-3)`
B. `2.3 xx 10^(-4)`
C. `2.0 xx 10^(-4)`
D. `2.0 xx 10^(-6)`

Answer

B

Explanation

For monoprotic acid
`K_(a) =(c alpha^(2))/(1-alpha)`
Here `c=0.010M` and `alpha =14%=0.14`
`:. K_(a)=(0.01 xx (0.14)^(2))/(1-0.14)=(10^(-2)xx 1.96 xx 10^(-2))/(8.6 xx 10^(-1))`
`=(19.6 )/(8.6 ) xx10^(-5+1)`
`=2.28 xx 10^(-4) =2.3 xx 10^(-4)`