So this question states determine the concentration of hydro Nia my on so basically find H plus or H +30 plus they both mean the same thing for a neutral solution at 25 degrees Celsius and 50 degrees Celsius. Okay, Now what we should realize here is what does neutral mean? Neutral means that your compound is not acidic or basic, and what it really means is that h 30 plus equals O H minus. They're both the same thing and because they're equal to each other, we're gonna set them both to an unknown variable. We're gonna set them two X and then what we're gonna do here is we know at 25 degrees Celsius, k w will be 1.0 times 10 to the negative 14. So a 25 degrees Celsius, That's what our K W. Is. And we just said that h 30 plus in which minus are equal when the solution is neutral and we're gonna set them both equal to X. So then you're gonna say x times X really means x squared and you only wanna find X, not X squared. So take the square root so in this case, X equals 1.0 times 10 to the negative seven. That number right there represents my concentration of H 30 plus. And technically, because it's neutral, it also represents the concentration of O H minus. Now, if you wanted to do it at 50 degrees Celsius would have to realize that since the temperature changes kw changes to this new number here. So we're gonna say 5.476 times 10 to the negative 14. Same exact process equals X squared. We just want X. So we're gonna take the square root of both sides. When we do that, it's going to give us a final answer of X equals 2. times 10 to the negative seven? No. Now So remember, this is important to know because K w equals H plus and times O H minus. This equation is important because of the temperature temperature is not given. Assume that K w equals 1.0 times 10 to the negative 14. If you know h plus concentration, then you know O h minus concentration. Because K w will be our constant and vice versa. If you know o h minus. You confined h Plus, this is the relationship we need to understand and these air the beginning steps for us to figure out the p H and p o h of different types of acquis solutions.