The degree of ionization of 0.10 M lactic acid is 4.0%

The value of `K_(C)`

Options

A. `1.66 xx 10^(-5)`
B. `1.66 xx 10^(-4)`
C. `1.66 xx 10^(-3)`
D. `1.66 xx 10^(-2)`

Answer

B

Explanation

% dissociation =4%
degree of dissociation `(alpha)=(4)/(100) =0.04`
For lactic acid
`CH_(3)CH(OH)COOH Leftrightarrow CH_(3)CH(OH)COO^(-) +H^(+)`
`{:(,"Initial conc",C mol L^(-1),0,0),(,"At equ.",C(1-alpha),C alpha,C alpha):}`
`K_(c)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/((1-alpha))`
`=(0.1 xx 0.04 xx 0.04)/((1-0.04))=(1.6 xx 10^(-4))/(0.96) =1.66 xx 10^(-4)`