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.
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Solution
freezing point depression equation:
ΔT = i*Kf*m where ΔT = Change in temperature in °C i = van 't Hoff factor Kf=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=Kf * 1 Kf=1.86 Using value of Kf Find value of i ΔT=Kf*m*i ie,0.0204=1.86*i*.01 {Given m=0.01 Kf=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