So we had determined the pH of this buffer we'd looking at here in an earlier problem. And now we're going to take and add KOH to that buffer. Now KOH is a strong base. And we know that as a base, it can be neutralized by the acid portion of the buffer. So this is the base portion of the buffer. This is the acid portion of the buffer. So the first thing that we're going to need to do is to write the one-way reaction that neutralizes the hydroxide. So + NH4+ is going to produce NH3 + H2O. So that's the reaction, and we'll need to set up an I, C, F table and plug moles into the table. So, we're going to set off to determine the number of moles of each of these substances. Let's begin with the moles of now this will be the same as the moles of KOH. Because there's one In every KOH. So, we'll take the molarity of the KOH, which is .1 times the volume of KOH that we're adding, we're adding 3 milliliters of 0.003 liters, and that's going to give us 0.00030 moles of the hydroxide. And we can put that into our table. Okay, for the moles of ammonium in NH4+, we will look at the information provided and we see that the concentration was .748 moles per liter. And we have 100 milliliters, or .1 liter, and that will give us 0.0748 moles. And we can put that into the table. 0.07, I said I said it right, but wrote it incorrectly over here. Let's put an 8 there. And that's moles of NH4+. And the moles of ammonia, for it we will take the molarity of ammonia, which is 0.750. Times the 100 milliliters, or 0.1 liter, and that gives us 0.0750 moles. So I'll put 0.0750 there. So the smallest is going to be reacting, so I will have none of the hydroxide, it has been neutralized as a buffer will do. We will subtract 0.0003 from here, and add 0.0003 to this value. That is going to give to us, in the F line under ammonium, 0.0745 and for ammonia, NH3, 0.0753. Okay at this point we have buffer so we can determine the pH of the buffer. pH = -log of the KA of ammonium, so that's 1.0 x 10 to the -14, divided by 1.76 x 10 to the -5. That's the pKa, that whole portion here is the pKa, plus the log, and we can use moles, the moles of the base is on top, so that's 0.0753, over the moles of the acid, 0.0745. So this will give a pH equal to this first term equals to 9.246 plus the second term, which is 0.0046 for final pH to three decimal places of 9.250 Okay, so the pH is now 9.250. The pH of the buffer originally, before I added this base, was 9.247. So it rose ever so slightly to 9.250, and we would expect to see a slight elevation as we added base, but it again it is a buffer, so it's resisting change to pH.