Find pH from given concentration and Kb

Question

What is the pH of a 0.050 M triethylamine, (C₂H₅)₃N, solution?

K_b for triethylamine is 5.3 × 10^–4.

Please anything to help, these are starting to get more confusing instead of easy.

Answer 1

Let's solve this step by step:

1. First, recall the base hydrolysis reaction for triethylamine:
2. Triethylamine, (C2H5)3N, is a weak base. When it's dissolved in water, it reacts with water to produce hydroxide ions (OH-) and its conjugate acid, the triethylammonium ion ((C2H5)3NH+). This is a reversible reaction: (C2H5)3N(aq) + H2O(l) ⇌ (C2H5)3NH+(aq) + OH-(aq)
3. equilibrium - Kb expression
4. Kb = [((C2H5)3NH+][OH-]) / [(C2H5)3N] where the brackets [] represent the molar concentrations of each species at equilibrium.
5. We know the Kb value is 5.3 × 10–4
6. Initial concentration of base = 0.050 M
7. Since Kb is relatively small (5.3 × 10⁻⁴), the reaction doesn't proceed very far to the right. This means that x will be a small number compared to the initial concentration of (C2H5)3N (0.050 M). Therefore, we can make the approximation that 0.050 - x ≈ 0.050. This simplifies the calculations.
8. Set up the equilibrium expression:
9. Kb = [((C2H5)3NH+][OH-] / [(C2H5)3N]
10. Let x = concentration of OH- formed
11. 5.3 × 10–4 = x2 / (0.050 - x)
12. Assume x is small compared to 0.050, so 0.050 - x ≈ 0.050
13. Rearrange the equation:
14. x2 = (5.3 × 10–4)(0.050)
15. x = √((5.3 × 10–4)(0.050))
16. x = 1.63 × 10–2 M
17. pOH = -log(1.63 × 10–2) = 1.79
18. pH = 14 - pOH
19. PH= 14 - 1.79 = 12.21

The pH of the solution is 12.21.

Answer 2

You have to solve for pH by setting up the ICE chart for the base reaction with water.

(C2H5)3N + H2O ---> (C2H5)3NH+ OH-

Set up the Kb expression: Kb = ((C2H5)3NH+) (OH-)/(C2H5)3N

Since you know the Kb, 5.3 × 10^–4 = x.x/0.5

Solve for x, which is OH- concentration.

Solve for pOH, which is -log(OH-).

solve for pH using: pH + pOH = 14 (since you know the pOH from previous step).