Question

What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.56 and has a freezing point of -2.0 ∘C?

Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)

Answer 1

Given that pH = 456

T_f = -2⁰C

Let us consider,

Tf = i*kf*m

Tf = T0-Tf

= 0-(-2)

= 2

Tf = i*Kf*m

i = vanthoff factor of CH3cOONa= 2

Kf = frreezing point constant of water = 1.86 C/m

molality(m) = ?

0-(-2) = 2*1.86*m

m = molality = 0.537

molality(m) = M*1000/((D*1000)-M*mWT of solute)

   0.537 = (M*1000)/((1.01*1000)-(M*122.12))

M = molarity of solution = 0.51 M

We know, pka of benzoicacid = 4.2

Now , pH = pka + log(salt/acid)

4.56 = 4.2 + log x

0.36 = log x

x = 2.29

total buffer concentration = 0.51 M

[C6H5COOH] +[C6H5COONa] = 0.51 M

[C6H5COOH]/[C6H5COONa] = 2.29

[C6H5COOH] = 2.29 [C6H5COONa]

Therefore ,

concentration of C6H5COONa = X = 0.155 M

concentration of C6H5COOH = 0.51-X

= 0.51 - 0.155

= 0.355 M