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,