Question

A 0.100g sample of weak acid HA (molar mass = 100.0 g/mol) is dissolved in 500.0 g of water. The freezing point of the resulting solution is -0.0056C. Calculate the value of K_a for this acid. Assume molarity equals the molality of the solution.

Short answer

The value of K_a is 0.50

Step-by-step solution

– Calculating the molality of the solution:

Molality = number of moles of solute/amount of solvent in g X 1000

Moles = Mass in g of solute/ molar mass

= 0.100/M₂

Here M₂ is the molar mass of acid.

Given depression in freezing point = -0.0056${}^{o}C$

Then from equation = T_f = K_fx m

K_f for water = 1.860C then we get

0.0056= (1.86 x 0.100g/M₂ x 500g) x 1000

M₂= 66.43 g

This is observed molar mass and given molar mass is 100g

– Calculating Van’t Hoff factor and degree of dissociation:

When association or dissociation occurs equation for depression at the freezing point will be

T_f = i K_f x m

Here “i” is Van’t Hoff factor and written as

i=normal molecular mass/ observed molecular mass

i=100/66.43 = 1.50

For weak acid, dissociation takes place as

HAH⁺+ A^-

Initial moles= 1 0 0

After dissociation=1-

Total number of moles after dissociation = 1 +

Then Van’t Hoff factor = i= moles after dissociation/ initial moles = 1+/1

1.50= 1+/ 1

On calculating we get = 0.50

Then percentage of dissociation will be 0.50 x 100 = 50%

Step 3- Calculating K_a

After dissociation HAH⁺ + A^-

1-

1-0.50 0.50 0.50

K_a= (H⁺) (A^-)/HA

= (0.50) (0.50)/0.50

So, K_a = 0.50