A 10.14 g 10.14 g sample of a weak acid (HA) is dissolved in 511.0 g 511.0 g of water. The freezing point of the solution is − 0.368 °C −0.368 °C . The molar mass of the weak acid is 93.0 g/mol. 93.0 g/mol. Calculate the acid dissociation constant, K a Ka , of the weak acid.

-1.75

Explanation:

Given

Mass of acid = 10.14g

Mass of water = 511g

Molar mass of the acid = 93.0g/mol

Freezing point = -0.368°C

 To get the molality of the solution

Molality = (Number of moles of solute ×1000) / Mass of solvent (g)

Molality  = (Mass of solute × 1000) / Molar mass of solute × Mass of solvent (g)

Molality = (10.14g × 1000) /(93.0g/mol × 511g)

                 = 0.21m

Vant Hoff factor (Ka) is used to calculate the extent of dissociation and it is given by the equation T = i×m×Kf

Where T is change in freezing point of the solution

              i is Vant Hoff factor

              m is the molality of the solute

              Kf is the cryoscopic constant.  For water Kf is taken as 1

Therefore

 -0.368°C = i × 0.21m × 1

i= -0.368/0.21

= -1.75

answer

the ions in solute supperate from eachouthere

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explanation: