A weak acid has a dissociation constant Ka = 2.5x10-2. Calculate the percentage dissociation for a 0.0750 m solution of this acid assuming a) ideal conditions and b) non-ideal conditions
dissociation of weak acid is as follows
HA + H2O ---> H30+ +A-
I 0.075 m 0 Initial concentration
C -x +x +x change in concentration
E 0.075-x x +x Equilibrium concentration
Ka = [H3O+][A-]/[HA]
we will take equilibrium concentration, and ka =2.5*10^-2,
and plug it in above equation
2.5 *10^-2= x^2/0.075-x
lets ignore x in denominator as it is too small
so the equation become
2.5*10^-2 *0.075 =x^2
x^2 = 0.1875*10^-2
x=0.43*10^-1= 0.043
percentage dissociation = [0.043/ 0.075]*100=57.3 %
THANK YOU!!
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