Here at the 25 milliliters of added sodium hydroxide, we have gone beyond the equivalence point. Let's look at a titration curve. At 25 milliliters, we're here at this point. And if you read up, you see that we have passed up the equivalence point by 5 milliliters and the pH shoots up very rapidly. A few extra drops of sodium hydroxide and now we have a strong base in the solution with nothing for it to react with and pH will skyrocket. So, let's go back to our reaction that we have been writing for every point of this titration curve, H- plus H2O is my reaction. We will do an ICF table as has been the case in every step, we're not changing the amount of the acid. We've had 4.98 times 10 to the -3 moles every time. For the number of moles of the we'll take the molarity of sodium hydroxide, which is 0.249 moles per liter times the volume. And now it's 0.025 liters and this is going to give me 6.225 times 10 to the minus 3 moles of hydroxide. 225 times 10 to the minus 3, 0 don't care. Smallest one now is the HA, so we're subtracting 4.98 times 10 to the minus 3 of each of these. And we're producing 4.98 times 10 to the -3 of this. This is going to consume all of the week acid. It is going to leave 1.25 times 10 to the -3 of the hydroxide. And we will have generated 4.98 times 10 to the -3 of the the A- ion. Now at this point, I see a lot of students who want to use the Henderson Hozbog equation because they're used to seeing if there's something on both sides to use that. But the Henderson Hozbog equation is good for a buffer and this not a buffer. It does not have a weak acid in its conjugate base. What does it have? It has a strong base, a strong base and a weak base in solution. Now if you've got a mixture of both a the strong base and a weak base, you can most likely without any trouble just base your problem entirely upon the strong base. So what we need to know is, what is the concentration of After these have reacted and have been added together? We have moles of 1.25 times 10 to the -3 and we have a volume of, if we use this 25 milliliters plus the 20 milliliters because it's the total volume in there. So that 45 milliliters or 0.045 liters. This is going to give me the moles of 0.0278. I'm sorry not moles but molarity of the hydroxide. So then we can get the POH and the POH will be pretty high. It's negative log of this value and you take the negative log of that you'll have 1.556 as the POH and in the pH would certainly be 14 minus that 1.556 and that would be a pH of 12.444. So we add the extra hydroxide, the pH rises rapidly, you can base the whole calculation on simply the strong base amount. We didn't even considered this at all. It contributes too few hydroxides to even worry about.
