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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.
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Solution 1
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_2 xx 1000)/(Delta 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_2FCOO^-][H^+])/([CH_2FCOOH])`
= `(C alpha * C α)/(C (1 - alpha))`
= `(C alpha^2)/(1 - alpha)`
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 alpha^2)/(1 - alpha)`
= `(0.5 xx (0.0753)^2)/(1 - 0.0753)`
= `(0.5 xx 0.00567)/0.9247`
= 0.00307 (approximately)
= 3.07 × 10−3
Solution 2
Calculated molar mass of CH2FCOOH (M'cal)
= (2 × 12) + (3 × 1) + (1 × 19) + (2 × 16)
= 78 g mol−1
In the present case, w = 19.5 g, W = 500 g, ΔTf = l.0°C,
Kf = 1.86 K kg mol−1
∴ Observed molecular mass
`M'_"obs" = (1000 xx K_f xx w)/(W xx Delta T_f)`
= `(1000 xx 1.86 xx 19.5)/(500 xx 1.0)`
= 72.54 g mol−1
Hence, van’t Hoff factor (i) = `(M'_(cal))/(M'_(obs))`
= `78/72.54`
= 1.0753
Suppose the degree of dissociation of fluoroacetic acid is α.
| \[\ce{CH2FCOOH ⇌ CH2FCOO^- + H^+}\] | |||
| Initial Conc. | C mol L−1 0 0 | ||
| At equilibrium | C(1− α) Cα Cα | ||
∴ Initial number of moles ∝ C
Number of moles at equilibrium ∝ [C(1− α) + Cα + Cα] ∝ [C(1 + α)]
∴ `i = "Number of particles in solution"/"Normal number of particles"`
= `(C(1 + alpha))/C`
= 1 + α
or α = i − 1
= 1.0753 − 1
= 0.0753
Dissociation constant (Ka) of fluoroacetic acid is given by
`K_a = ([CH_2FCOO^-][H^+])/([CH_2FCOOH])`
= `(C alpha * C α)/(C (1 - alpha))` ...(∵ α is very small, 1 − α ≈ 1)
= Cα2
Concentration (C) of the acid = `19.5/78 xx 1000/500`
= 0.5 m
∴ Ka = Cα2
= 0.5 × (0.0753)2
= 0.5 m
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