Deuterium oxide 1D2O, where D is deuterium, the hydrogen-2 isotop...

Question

Deuterium oxide 1D2O, where D is deuterium, the hydrogen-2 isotope) has an ion-product constant, Kw, of 8.9 * 10-16 at 20 °C. Calculate 3D+4 and 3OD-4 for pure (neutral) D2O at this temperature.

Accepted Answer

Hi everyone for this problem it reads X 20 is an oxide of a hypothetical isotope of hydrogen. If it has an ion product constant kw of 5. times 10 to the negative 17 at 40 degrees Celsius. What are the concentrations of X plus and X minus for neutral X +20. Okay, so our goal for this problem is to solve all for the concentrations of X plus and X minus. So let's go ahead and start off by defining KW, which is our ion product constant. Okay, so when this oxide dissociates, what it's going to disassociate into is X plus and O X minus R K W R I on product constant represents the product of these two. Okay, so it's going to represent the product of these two concentrations and in this problem we were given what that product is equal to. Ok, So we have it's equal to 5.67 times 10 to the -17 for this pure oxide, the concentrations are going to equal each other. Okay, so for pure for pure X our concentration of X plus is going to equal our concentration of O X minus. And our goal here is to solve for the concentrations of both. Okay, since they're going to equal each other and so what we can do here is we can say our concentration of X plus squared is equal to the product of R K w which we were given. Okay, because essentially they're equal to each other. So if we just square the concentration of X plus, then that's going to be the same thing. So now we want to isolate our concentration of X plus because we just want the concentration of X plus and right now it's squared. So we want to get rid of this square and we'll do that by squaring both sides of our equation. So we'll take the square root of both sides. And when we take the square root we're going to get the concentration of X plus is equal to 7.529 times 10 to the negative nine. And this is going to also equal the concentration of O X minus because they're the same when it's pure Okay, so by isolating our X plus squared and getting rid of that square root, we're able to solve for the concentration of just X plus, which is going to also equal our concentration of O X minus. And this is going to be our final answer. Okay, so the concentrations for both Equals 7.529 times 10 to the -9. That's the end of this problem. I hope this was helpful

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