A solution of mono basic acid (3 10⁻² M ) has a freezing point of depression of 0.06^ C . Calculate pKa for the acid. ( 1.6992=0.2302) [Molal depression constant 1.86^ C / m for water]

Step-by-step solution

We know that, T_f= i k _ f m Where i = Vant Hoff factor aligned & T =0.06 & ~K _ f =1.86 & ~m = molality aligned Molality may be taken as molarity because solution is dilute and solvent is water. =3 10⁻² Substituting the values in equation (i), we get, 0.06= i 1.86 3 10⁻² i = 0.06 1.86 3 10⁻² i=1.07526 Let acid is NH K_a= C C C(1- ) =C ^2 i = Total mole after dissociation initial mole = C (1- )+ C + C C =1+ = i -1 =1.07526-1 =0.07526 K _ a = C ^2=3 10⁻² (0.07526)^2 K _ a =1.6992 10⁻⁴ PK _ a =- Ka =- (1.6992 10⁻⁴ )=