So remember that ph is just the negative log of H plus or H. +30. Plus for example, wanted to ask us if at 50 degrees Celsius, the ionization of pure water Kw is 7.94 times 10 to the negative 14. What is the ph of a neutral solution? Well for a question like this we're dealing with pure water with no free floating ions within it. Here we're dealing with neutral solution and in a neutral solution your H plus concentration is equal to euro H minus concentration. Since we don't know either one of them, we're gonna say they're both equal to X. Remember both of these ions are related to each other by the fact that H plus or H. +30 plus times O. H minus equals kW. And here since both are X. And they're multiplying that's going to be X squared and that equals 7.94 times 10 to the negative 14. Remember your K. W. Like all other equilibrium constants is temperature dependent. Once the temperature is no longer 25 degrees Celsius, R K. W will not be 1.0 times 10 to the negative 14 it becomes a brand new number. So here we're going to take the square root Of both sides to isolate X. So that's gonna give me 2.81 7. 8 times 10 to the -7 Moller. Now that equals H plus as well as O. H minus. But since we're looking for ph we're going to focus only on the H plus. So P H equals negative log of H plus Which equals negative log of 2.8178 times 10 to the -7. So that gives me 6.55 as my concentration after rounding. So that gives us option B. Now that's when we're dealing with pure water, what would happen when we're dealing with uh figuring out the ph still in pure water, but now it has free floating ions in it. We have to take into account what does this do to both our ionic strength as well as our activity coefficient. Think of those things as you read example to you could attempt it on your own but if you get stuck, don't worry, just come back and see how I approach that same example to question.
