Question

When HClO2 is dissolved in water, it partially dissociates according to the equation

HClO2 → H+ + ClO2-. A solution is prepared that contains 5.850 g of HClO2 in 1.000 kg of water. Its freezing point is -0.3473 °C. Calculate the fraction of HClO2 that has dissociated.

Answer 1

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Step-by-step explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

HClO2 H+ + ClO2-

initial concentration: 0.0854 0 0

final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates