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Question
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Solution
It is given that:
w1 = 500 g
w2 = 19.5 g
Kf = 1.86 K kg mol-1
ΔTf = 1.0°C
We know that:
M2 = `("K"_"f"xx"w"_2xx1000)/(triangle"T"_"f"xx"w"_1)`
`= (1.86 "K kg mol"^(-1) xx 19.5 "g" xx 1000 "g kg" ^(-1))/(500 "g" xx 1.0)`
= 72.54 g mol-1
Therefore, observed molar mass of CH2FCOOH, (M2)obs = 72.54 g mol-1
The calculated molar mass of CH2FCOOH is (M2)cal = 14 + 19 + 12 + 16 + 16 + 1
= 78 g mol-1
Therefore, van’t Hoff factor, i = `(("M"_2)_("cal"))/("M"_2)_("obs")`
= `78/72.54`
= 1.0753
Let α be the degree of dissociation of CH2FCOOH
| \[\ce{CH2FCOOH ⇌ CH2FCOO^- + H^+}\] | |||
| Initial Conc. | C mol L-1 | 0 | 0 |
| At equilibrium | C(1- α) | Cα | Cα |
∴ i = `("C"(1 + α))/"C"` = 1 + α Or α = i − 1 = 1.0753 − 1 = 0.0753
Now, the value of Ka is given as:
Kα = `(["CH"_2"FCOO"^-]["H"^+])/(["CH"_2"FCOOH"])`
= `("C"α."C"α)/("C"(1 - α))`
= `("C"α^2)/(1 - α)`
Taking the volume of the solution as 500 mL, we have the concentration:
C = `19.58/78 xx 1/500 xx 1000` = 0.5 M
∴ Kα = `("C"α^2)/(1 - α)`
`= (0.5xx (0.0753)^2)/(1-0.0753)`
`= (0.5xx0.00567)/0.9247`
= 0.00307 (approximately)
= 3.07 × 10-3
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