A 0.2 molal aqueous solution of weak acid hx is 20% ionized. the freezing point of solution is\
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dissociation reaction of weak acid HX is given by
HX ⇄ H⁺ + X⁻
at equilibrium , α dissociation is happened
now, according to van Hoff's factor , i = 1 - α + α + α = 1 + α
A/C to question ,
20% ionized , it means α = 20/100 = 0.2
Now, i = 1 + 0.2 = 1.2
Use formula
∆Tf = i × Kf × m
You didn't mention freezing constant ,Kf
well, your answer is ∆Tf = 1.2 × Kf × 0.2 = 0.24 Kf
Put ,Of value and get the freezing point of solution
HX ⇄ H⁺ + X⁻
at equilibrium , α dissociation is happened
now, according to van Hoff's factor , i = 1 - α + α + α = 1 + α
A/C to question ,
20% ionized , it means α = 20/100 = 0.2
Now, i = 1 + 0.2 = 1.2
Use formula
∆Tf = i × Kf × m
You didn't mention freezing constant ,Kf
well, your answer is ∆Tf = 1.2 × Kf × 0.2 = 0.24 Kf
Put ,Of value and get the freezing point of solution
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