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.
van't Hoff factor (i) = 1.075 Dissociation constant (Ka) = ⁻³
Explanation
This question involves calculating the van't Hoff factor and dissociation constant using freezing point depression data. The solution follows the same methodology as Example 1.13 in the textbook for acetic acid. First, we calculate the theoretical freezing point depression assuming no dissociation, then compare it with the observed value to find the van't Hoff factor. Using the van't Hoff factor, we determine the degree of dissociation, which is then used to calculate the dissociation constant.
Solution Steps
Step 1: Calculate molar mass of CH₂FCOOH Molar mass = 2(12) + 3(1) + 19 + 2(16) = 24 + 3 + 19 + 32 = 78 g mol⁻¹
Step 2: Calculate molality of solution Moles of CH₂FCOOH = 19.5 g / 78 g mol⁻¹ = 0.25 mol Molality (m) = 0.25 mol / 0.5 kg = 0.5 mol kg⁻¹
Step 3: Calculate theoretical freezing point depression ΔTf (calculated) = Kf × m = 1.86 K kg mol⁻¹ × 0.5 mol kg⁻¹ = 0.93 K
Step 4: Calculate van't Hoff factor i = Observed ΔTf / Calculated ΔTf = 1.0 K / 0.93 K = 1.075
Step 5: Calculate degree of dissociation For CH₂FCOOH ⇌ H⁺ + CH₂FCOO⁻ i = 1 + x x = i - 1 = 1.075 - 1 = 0.075
Step 6: Calculate dissociation constant Ka = [H⁺][CH₂FCOO⁻] / [CH₂FCOOH] = (cx)² / c(1-x) = cx² / (1-x) Ka = 0.5 × (0.075)² / (1 - 0.075) = / 0.925 = ⁻³