The dissociation constant of acetic acid at a given temperature is `1.69xx10^(-5)`.The degree of dissociation of 0.01 M acetic acid in presence of 0.01 M HCl is equal to :

Options

A. `1.69xx10^(-7)`
B. `1.69xx10^(-5)`
C. `1.69xx10^(-3)`
D. `2.9xx10^(-2)`

Explanation

`{:(,CH_(3)COOH(aq)hArr, H^(+)(aq),+,CH_(3)COOH^(-)(aq)),(t=0,0.01,,,),(t=eq,0.01-x,x,,x):}`
`{:(,CH_(3)COOH(aq)hArr, H^(+)(aq),+,CH_(3)COOH^(-)(aq)),(t=0,0.01,,,),(t=eq,0.01-x,x,,x),(,[H^(+)]=x+0.01~~0.01M,,,):}`
`K_(a)([H^(+)][CH_(3)COO^(-)])/([CH_(3)COOH])rArr1.69xx10^(-5)=(0.01xx[CH_(3)COO^(-)])/(0.01)`
`:.[CH_(3)COO^(-)]=1.69xx10^(-5)M`
So degree of dissociation of `CH_(3)COOH=(1.69xx10^(-5))/(0.01)=1.69xx10^(-3)`