What is the pH of a 0.125 M solution of a weak base if ∆H° = -28.0 kJ and ∆S° = -175 J/K for the following equilibrium reaction at 25°C:

B (aq) + H2O (l) ↔ BH+ (aq) + OH- (aq)

To find the pH of a 0.125 M solution of a weak base, you need to calculate the concentration of OH- ions in the solution. The equilibrium constant (K) can be expressed in terms of the concentrations of the products (BH+ and OH-) divided by the concentration of the reactant (B):

K = [BH+] [OH-] / [B]

The values of ∆H° and ∆S° can be used to calculate the standard Gibbs free energy change (∆G°) using the equation:

∆G° = ∆H° - T∆S°

where T is the temperature in Kelvin. At equilibrium, the standard Gibbs free energy change (∆G°) is zero, so we can set ∆G° = 0 and solve for the equilibrium constant (K):

0 = -RT ln(K)

where R is the gas constant (8.314 J/(mol·K)) and ln is the natural logarithm.

We know that the concentration of B is 0.125 M since it is a 0.125 M solution of the weak base. So, we can rearrange the equation for K to solve for [OH-]:

K = [BH+] [OH-] / [B]
[OH-] = (K [B]) / [BH+]

Now, substitute the values of K and [B] into the equation, along with the given temperature (25°C = 298 K), to find [OH-]:

[OH-] = (K [B]) / [BH+]
[OH-] = exp(-∆G° / (RT)) [B] / [BH+]

Substitute the values of ∆H° and ∆S° into the equation for ∆G° and calculate it:

∆G° = ∆H° - T∆S°
∆G° = -28.0 kJ - (298 K)(-175 J/K)
∆G° = -28.0 kJ + 52.15 kJ
∆G° = 24.15 kJ

Convert ∆G° to J:

∆G° = 24.15 kJ × (1000 J/kJ)
∆G° = 24,150 J

Now substitute the values of ∆G°, R, T, [B], and [BH+] into the equation to find [OH-]:

[OH-] = exp(-∆G° / (RT)) [B] / [BH+]
[OH-] = exp(-24,150 J / ((8.314 J/(mol·K))(298 K))) (0.125 M) / (1 M)

Calculate the value inside the exponential:

-24,150 J / ((8.314 J/(mol·K))(298 K)) = -9.676

Calculate the exponential:

exp(-9.676) = 7.613 x 10^-5

Now substitute this value into the equation for [OH-]:

[OH-] = (7.613 x 10^-5) (0.125 M) / (1 M)
[OH-] = 9.516 x 10^-6 M

The concentration of OH- ions is 9.516 x 10^-6 M. To find the pOH, take the negative logarithm (base 10) of this concentration:

pOH = -log10([OH-])
pOH = -log10(9.516 x 10^-6)
pOH = 5.02

Finally, to find the pH, subtract the pOH from 14:

pH = 14 - pOH
pH = 14 - 5.02
pH = 8.98

Therefore, the pH of the 0.125 M solution of the weak base is 8.98.