what is the pH of a 0.026 M sodium acetate solution?

Answer 1

First you'll need to set up the equilibrium. Acetate ion is a weak base; it will react with water to form the conjugate acid of acetate and hydroxide. (Sodium is a spectator ion and can be ignored.)

Next you'll need to set up the equilibrium constant expression for the reaction and look up the Kb (equilibrium constant) for acetate. (Remember water is in a large excess and its concentration doesn't change and substances with constant concentration are left out of equilibrium constant expression.)

Now you'll need to find a way to solve for [OH-] in the equation. The 0.026M is the initial concentration of acetate, and it is assume that the initial concentration of the products are both zero. The change in concentration needed to reach equilibrium is unknown and represented by x. The reactant loses x amount and the products each gain x amount. This gives you an expression for each substance that represents the equilibrium concentrations. Plug them into the equilibrium constant expression and solve for x. You'll need to use the quadratic equation or make a simplifying assumption that x is small and the concentration of acetate at equilibrium is nearly the same as the initial 0.026M.

Now that you have the concentration of hydroxide (x = [OH-]), you can use this to find pH. Remember: pH = -log[H+] and Kw = [H+][OH-].

Answer 2

What you do is the negative log of your concetration
-log(0.026)=1.585 pH