Answer to Question #184718 in Chemistry for leshan

Question

A 0.40 M solution of C3H5O3- has a pH of 8.728. What is the Kb of this ion?

Answer 1

Solution:

For any aqueous solution at 25^∘C:

pH + pOH = 14

pOH = 14 - pH = 14 - 8.728 = 5.272

We can convert between pOH and [OH⁻] using the following equation:

pOH = −log[OH⁻]

[OH⁻] = 10^−pOH = 10^−5.272 = 5.346×10⁻⁶ M

[OH⁻] = 5.346×10⁻⁶ M

Consider the equilibrium,

C₃H₅O₃⁻ + H₂O ⇌ HC₃H₅O₃ + OH⁻

where,

From this, we can construct an ICE(Initial, Change, Equilibrium) Table:

Hence,

K_b = [HC₃H₅O₃] × [OH⁻] / [C₃H₅O₃⁻]

K_b = (5.346×10⁻⁶ M) × (5.346×10⁻⁶ M) / (0.399995 M) = 7.145×10⁻¹¹ M

K_b = 7.145×10⁻¹¹

Answer: K_b = 7.145×10⁻¹¹