What is the pH of a  0.800 M NaCH3CO2 solution? Ka for acetic acid, CH3CO2H is 1.8 X 10^-5

To solve this problem, we need to understand that sodium acetate (NaCH3CO2) is the salt of a weak acid (acetic acid, CH3CO2H). When it dissolves in water, it will undergo hydrolysis to form its corresponding weak acid and a strong base (NaOH). The reaction can be written as follows:

CH3CO2- + H2O ⇌ CH3CO2H + OH-

The equilibrium expression for this reaction is:

Kb = [CH3CO2H][OH-] / [CH3CO2-]

We know that Ka * Kb = Kw, where Kw is the ion product of water, which is 1.0 x 10^-14 at 25°C. We can solve for Kb using the given Ka:

Kb = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10

Next, we set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of the species at equilibrium.

Initial: [CH3CO2-] = 0.800 M, [CH3CO2H] = 0, [OH-] = 0

Change: [CH3CO2-] = -x, [CH3CO2H] = +x, [OH-] = +x

Equilibrium: [CH3CO2-] = 0.800 - x, [CH3CO2H] = x, [OH-] = x

Substitute the equilibrium concentrations into the Kb expression:

5.56 x 10^-10 = x * x / (0.800 - x)

Assuming x << 0.800, the equation simplifies to:

5.56 x 10^-10 = x^2 / 0.800

Solving for x gives x = 2.11 x 10^-5 M, which is the concentration of OH- ions.

Finally, we can find the pH of the solution by first finding the pOH (pOH = -log[OH-]), and then subtracting the pOH from 14 (since pH + pOH = 14 at 25°C).

pOH = -log(2.11 x 10^-5) = 4.68

pH = 14 - pOH = 14 - 4.68 = 9.32

So, the pH of the 0.800 M NaCH3CO2 solution is approximately 9.32.