Question

When 2.65 g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of the resulting solution is -0.257 ∘C.

Answer 1

Kf of water = 1.86c⁰/m

molality    = weight of solute*1000/grammolar mass of solute*weight of solvent in g

                 = 2.65*1000/85*250   = 0.125m

Tf = i*kf*m

   0.257   = i*1.86*0.125

i          = 0.257/1.86*0.125   = 1.05

   HA -------> H⁺ + A^-

I             1           0     0

C            -x               +x     +x

E          1-x          +x      +x

1-x + x+x   = 1.105

   1+x    = 1.105

    x      = 1.105-1

   x = 0.105

[HA]   = 0.125-0.105*0.125   = 0.1118M

[A-] = [H+] = 0.105*0.125    = 0.013125M

    ka = [H+][A-]/[HA]

          = 0.013125*0.013125/0.1118   = 0.00154

          = 1.4*1.4/