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Step-by-Step Solution
Step 1: Calculate the number of moles of chloroacetic acid
The molar mass of chloroacetic acid (ClCH2COOH) is computed as follows:
C (2 atoms) = 2 × 12 = 24
H (3 atoms) = 3 × 1 = 3
Cl (1 atom) = 35.5
O (2 atoms) = 2 × 16 = 32
Total molar mass = 24 + 3 + 35.5 + 32 = 94.5 g/mol
Given 9.45 g of the acid:
Number of moles =
\frac{\text{mass}}{\text{molar mass}} = \frac{9.45}{94.5} = 0.1 \text{ moles.}
Step 2: Determine the molality of the solution
We dissolve 9.45 g of the acid in 500 mL (0.5 kg) of water.
Molality (m) =
\frac{\text{moles of solute}}{\text{kilograms of solvent}}
= \frac{0.1}{0.5}
= 0.2 \text{ m.}
Step 3: Use the freezing-point depression data to find the van’t Hoff factor (i)
The depression in freezing point ( \Delta T_f ) is given as 0.5°C. The freezing-point depression equation is
\Delta T_f = i \, K_f \, m,
where
- i = van’t Hoff factor,
- K_f = cryoscopic constant of water = 1.86 K·kg·mol−1,
- m = molality of the solution.
Plug in the values:
0.5 = i \times 1.86 \times 0.2.
Hence,
i = \frac{0.5}{1.86 \times 0.2} = \frac{0.5}{0.372} \approx 1.344.
Step 4: Find the degree of dissociation (α)
For a monoprotic acid (HA) dissociating into H+ and A−, the van’t Hoff factor
i \approx 1 + \alpha.
Thus,
\alpha = i - 1 = 1.344 - 1 = 0.344.
Step 5: Relate degree of dissociation to the acid dissociation constant (Ka)
Let the initial concentration of the acid in mol/L be c. We have 0.1 moles in 0.50 L, so
c = \frac{0.1 \text{ moles}}{0.50 \text{ L}} = 0.20 \text{ M}.
For the dissociation
HA ↔ H+ + A−,
the equilibrium concentrations are:
- [HA] = c(1 - \alpha) ,
- [H+] = c \alpha ,
- [A−] = c \alpha.
Hence,
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
= \frac{(c \alpha) (c \alpha)}{c (1 - \alpha)}
= \frac{c \alpha^2}{1 - \alpha}.
Now substitute c = 0.20 \,\text{M} and \alpha = 0.344:
K_a
= \frac{0.20 \times (0.344)^2}{1 - 0.344}
= \frac{0.20 \times 0.118336}{0.656}
\approx \frac{0.0236672}{0.656}
\approx 0.036.
In scientific notation,
K_a \approx 3.6 \times 10^{-2}.
Step 6: Express Ka in the form x × 10−3
We have
K_a = 3.6 \times 10^{-2} = 36 \times 10^{-3},
so the value of x (rounded to the nearest integer) is 36.
Final Answer
The dissociation constant of ClCH2COOH in the given conditions is
36 \times 10^{-3},
so the value of x is
36.
