Question

Based on the values of ∆H° and ∆S° that you determined for the dissolution of borax, what would be the solubility of borax in (g/L) at 85.0°C?

Show calculations. Delta H (∆H°)= 59.279 kj/mol

Delta ∆S°= 138.6 J/mol*K

Answer 1

Given : Ho = 59.279 kJ/mol = 59279 J/mol

So = 138.6 J/mol-K

Temperature = 85.0oC = (85 + 273) K = 358 K

Go = Ho - (T * So)

Go = (59279 J/mol) - (358 K * 138.6 J/mol-K)

Go = 59279 J/mol - 49618.8 J/mol

Go = 9660.2 J/mol

ln(Ksp) = -(Go) / (R * T)

where Ksp = solubility product constant

R = constant = 8.314 J/mol-K

T = absolute temperature = 358 K

ln(Ksp) = -(9660.2 J/mol) / [(8.314 J/mol-K) * (358 K)]

ln(Ksp) = -3.2456

Ksp = e^(-3.2456)

Ksp = 0.039

The dissolution equation for borax is

Na2B4O7 (s) 2 Na+ (aq) + B4O7^2- (aq)

Each molecule of borax that dissolves given 2 ions of Na+ and one ion of B4O7^2-

Ksp = [Na+]^2[B4O7^2-]

0.039 = (2s)^2 * (s)

where s = molar solubility of borax

0.039 = 4s^3

s = (0.039 / 4)^(1/3)

s = 0.2135 M

mass solubility of borax = (molar solubility of borax) * (molar mass borax)

mass solubility of borax = (0.2135 M) * (201.22 g/mol)

mass solubility of borax = 42.97 g/L