A 0.100 mol/L laboratory solution of sodium propanoate, NaC2H5COO (aq), has a pH of 8.65 at 25 degrees Celsius. Calculate Kb for the propanoate ion.

1 answer

To find the value of Kb for the propanoate ion (C2H5COO-), we need to use the equation for the hydrolysis of the ion:

C2H5COO- + H2O ⇌ C2H5COOH + OH-

The equilibrium constant expression for this reaction is:

Kb = [C2H5COOH][OH-] / [C2H5COO-]

Since the initial concentration of C2H5COO- is 0.100 mol/L, we can assume that the change in concentration of C2H5COO- is negligible at equilibrium. So, we can write:

Kb = [C2H5COOH][OH-] / 0.100

We also know that the pH of the solution is 8.65. The pH is related to the concentration of hydroxide ions (OH-) by the equation:

pOH = 14 - pH

So, pOH = 14 - 8.65 = 5.35

The concentration of hydroxide ions (OH-) can then be calculated using the equation:

[OH-] = 10^(-pOH)

[OH-] = 10^(-5.35)

[OH-] = 3.55 × 10^(-6) M

Substituting the values into the equation for Kb, we have:

Kb = [C2H5COOH] × [OH-] / 0.100

Since the concentration of C2H5COOH is unknown, we need to find it using the equation for the dissociation of the weak acid:

C2H5COOH ⇌ C2H5COO- + H+

The concentration of the weak acid (C2H5COOH) at equilibrium is equal to the concentration of the propanoate ion (C2H5COO-) because the solution is initially a sodium propanoate solution.

Therefore, [C2H5COOH] = 0.100 mol/L.

Plugging in the values, we have:

Kb = (0.100 mol/L) × (3.55 × 10^(-6) M) / 0.100

Kb = 3.55 × 10^(-9)