What is the pH of a 0.015 M solution of sodium acetate (NaCH2CO2)? The Kb for acetate is 5.6 x 10^-10.

To find the pH of a 0.015 M solution of sodium acetate (NaCH₂CO₂), we first recognize that sodium acetate is the salt of a weak acid (acetic acid, CH₃COOH) and a strong base (sodium hydroxide, NaOH). In solution, sodium acetate dissociates into sodium ions (Na⁺) and acetate ions (CH₃COO^–).

The acetate ion can act as a weak base and can accept a proton from water, leading to the formation of acetic acid and hydroxide ions:

CH₃COO^– + H₂O → CH₃COOH + OH^–

We can find the concentration of hydroxide ions (OH^–) produced in this reaction using the value of K_b for acetate:

K_b = 5.6 x 10^-10 = \[ \frac{[CH₃COOH][OH^–]}{[CH₃COO^–]} \]

Let x be the concentration of OH^– produced. Thus, at equilibrium:

• [CH₃COOH] = x
• [OH^–] = x
• [CH₃COO^–] = 0.015 – x (approximately 0.015 since x will be small)

We can substitute these values into the K_b expression:

5.6 x 10^-10 = \[ \frac{x^2}{0.015} \]

Solving for x gives:

x² = (5.6 x 10^-10) * 0.015

x² = 8.4 x 10^-12

x = √(8.4 x 10^-12) = 9.17 x 10^-6 M

This concentration of OH^– can be used to find the pOH:

pOH = -log(9.17 x 10^-6) ≈ 5.04

Finally, since pH + pOH = 14, we can calculate the pH:

pH = 14 – pOH = 14 – 5.04 = 8.96

Therefore, the pH of a 0.015 M solution of sodium acetate is approximately 8.96.