Question

Q. Depression of freezing point of 0.01 molal aqueous Acetic Acid solution is 0.0204 degree Celsius. 1 molal urea solution freezes at -1.86 degree Celsius. Assuming molality equal to molarity, pH of Acetic Acid solution.

Answer 1

freezing point depression equation:

ΔT = i*K_f*m
where
ΔT = Change in temperature in °C
i = van 't Hoff factor
K_f =molal freezing point depression constant or cryoscopic constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent
1 molal urea solution freezes at -1.86 degree celsius
1.86=K_{f *} 1
K_f=1.86
Using value of K_f
Find value of i
ΔT=K_f*m*i
ie,0.0204=1.86*i*.01
{Given m=0.01 K_f=1.86
and ΔT=0.0204}
so i=0.0204/(1.86*.01)
=1.1
CH₃CO₂H → CH₃CO₂⁻ + H^{+ {ionization releasing proton}}
(c-α) cα cα
i=1+α
α=1.1-1=0.1
[H+]= cα =0.01*0.1=0.001
pH=-log[H+]=-log 0.001= 3 Answer