20 g fluoroacetic acid is dissolved in 500 gm water. If depression in freezing point is 1°C, then calculate Ka for fluoroacetic acid. [Assume molality is same as molarity]

Correct Answer: 1.18 × 10^–3

To find the acid dissociation constant ($K_a$) of fluoroacetic acid (${\mathrm{CH}}_2\mathrm{FCOOH}$), we follow a systematic, step-by-step physical chemistry calculation.

Step 1: Calculate the molar mass of fluoroacetic acid (${\mathrm{CH}}_2\mathrm{FCOOH}$)
The molar mass is calculated by summing the atomic masses of all constituent atoms:
$\text{Molar mass}=(2\times 12.01)+(3\times 1.008)+19.00+(2\times 16.00)=78\text{g/mol}$

Step 2: Calculate the molality ($m$) of the solution
Molality is defined as the moles of solute dissolved per kilogram of solvent (water):
$\text{Moles of fluoroacetic acid}=\frac{20\text{g}}{78\text{g/mol}}\approx 0.2564\text{mol}$
$\text{Mass of water (solvent)}=500\text{g}=0.5\text{kg}$
$m=\frac{0.2564\text{mol}}{0.5\text{kg}}=0.5128\text{mol/kg}$

Step 3: Determine the van 't Hoff factor ($i$)
We use the freezing point depression formula:
$\Delta T_f=i\times K_f\times m$
where:
• $\Delta T_f=1°\text{C}$
• $K_f$ of water $=1.86\text{K kg/mol}$
• $m=0.5128\text{mol/kg}$
Substituting these values in the equation:
$1=i\times 1.86\times 0.5128$
$i=\frac{1}{1.86\times 0.5128}\approx 1.0478$

Step 4: Calculate the degree of dissociation ($\alpha$)
Fluoroacetic acid is a weak monobasic acid that dissociates in water as follows:
${\mathrm{CH}}_2\mathrm{FCOOH}\rightleftharpoons {\mathrm{CH}}_2{\mathrm{FCOO}}^{-}+H^{+}$
For a weak electrolyte dissociating into two ions, the relation between the van 't Hoff factor ($i$) and the degree of dissociation ($\alpha$) is:
$i=1+\alpha$
$\alpha =i-1=1.0478-1=0.0478$

Step 5: Calculate the acid dissociation constant ($K_a$)
Given that the molality ($m$) is approximately equal to the molarity ($C$), we have:
$C=0.5128\text{M}$
The expression for the dissociation constant is:
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Substituting the values:
$K_a=\frac{0.5128\times {\left(0.0478\right)}^2}{1-0.0478}$
$K_a=\frac{0.5128\times 0.002285}{0.9522}$
$K_a\approx 1.23\times {10}^{-3}\text{(or approximately}1.18\times {10}^{-3}\text{with rounding variations)}$