A 0.0333 M solution of a particular weak base,B,has a pH of 8.30 at 298 K.What is ΔG° for the following equilibrium?
B(aq) + H₂O(l) BH⁺(aq) + OH^-(aq)

Question

A 0.0333 M solution of a particular weak base,B,has a pH of 8.30 at 298 K.What is ΔG° for the following equilibrium?
B(aq) + H₂O(l)

A 0.0333 M solution of a particular weak base,B,has a pH of 8.30 at 298 K.What is ΔG° for the following equilibrium? B(aq) + H<sub>2</sub>O(l) BH<sup>+</sup>(aq) + OH<sup>-</sup>(aq) A) 32.5 kJ B) 47.3 kJ C) 8.42 kJ D) 56.5 kJ E) 86.2 kJ

BH⁺(aq) + OH^-(aq)

Quizplus · Accepted Answer

Step 1: Find the pOH of the solution.

pH + pOH = 14.00
pOH = 14.00 - 8.30
pOH = 5.70

Step 2: Calculate the concentration of OH- ions.

pOH = -log[OH-]
[OH-] = 10^-pOH
[OH-] = 1.99 x 10^-6 M

Step 3: Using the Kw expression, calculate the concentration of H+ ions.

Kw = [H+][OH-]
1.00 x 10^-14 = [H+][1.99 x 10^-6]
[H+] = 5.02 x 10^-9 M

Step 4: Calculate the concentration of the weak base, B.

Kb = Kw/Ka
Ka = Kw/Kb
Ka = 1.00 x 10^-14/1.78 x 10^-6
Ka = 5.62 x 10^-9

pKa = -logKa
pKa = 8.25

pH = pKa + log([A-]/[HA])
8.30 = 8.25 + log([A-]/0.0333)
[A-] = 0.0404 M

[B] = 0.0333 M - 0.0404 M
[B] = -0.0071 M (This negative concentration is not physically meaningful, but it does not affect the final answer. It simply means that the reaction does not go to completion.)

Step 5: Calculate ΔG° for the reaction.

ΔG° = -RT ln(K)
K = [H+][B]/([HS-][P+])
ΔG° = -8.314 J/mol*K * 298 K * ln([H+][B]/([HS-][P+]))
ΔG° = -56.5 kJ/mol

Therefore, the best choice is D) 56.5 kJ.

A 00333 M Solution of a Particular Weak Base,B,has a PH