Now, we'll finish out this problem by finding the pH of a potassium acetate solution. What is the pH of a 0.325 molar solution of potassium acetate? The first thing we have to do is write the reaction that represents what's going on in this mixture. This is the reaction that we looked at from the first slide, all we are worried about is the reaction of acetate with water. So, CH3COO- + H2O in equilibrium with CH3COOH + Now that we have our equilibrium reaction, we can set up our ICE table. So, we have I for initial, C for change, E for equilibrium. We have our initial concentration is given as 0.325 molar. Water we don't worry about. Our initial concentration of acetic acid is 0. Our initial concentration of hydroxide ion is also 0. Now we look at the stoichiometry of the balanced chemical equation to determine our change row. So I have -x +x and +x. And for my equilibrium row, I simply sum up the initial and change rows. So, 0.325- x, x, and x. Now, I have my equilibrium concentrations with respect to x. I can use the information that I determined on the previous problem from our Kb value. And I can set up the Kb expression which will be HA, and I'm using that as an abbreviation for my acetic acid, times Minus concentration over the CH3COO- concentration. I know that Kb is equal to 5.6 times 10 to the -10th. And that HA and Are both x, so I have x squared over my acetate concentration, which is 0.325 minus x. I'm going to make the simplifying assumption that x is much much less than 0.325. And given the small size of my Kb value, this is going to be a reasonable assumption. So, I can simplify my Kb expression. So, I end up with 5.6 times 10 to the -10th = x squared over 0.325. And when I solve for x, which is going to be my hydroxide concentration, I end up with 1.3 times 10 to the -5th. Now we can take the information that we know about the hydroxide concentration and we can find the pH of the solution. But first, we have to find the pOH of the solution. So, I know that pOH = the -log of my hydroxide concentration. So, pOH = -log of 1.3 times 10 to the -5th, and I end up with a pOH = 4.89. And because no temperature is given, we're going to assume 25 degree celsius. And when I do that I know the pH + pOH = 14, so pH = 14- pOH, or 14, 4.89, and we get a pH = 9.11. Given that this is a weak base, this acetate ion is behaving as a weak base, we expect a pH greater than seven, which is exactly what we see.
