At 50 °C, the ion-product constant for H2O has the value Kw = 5.48 * 10-14. (a) What is the pH of pure water at 50 °C? (b) Based on the change in Kw with temperature, predict whether ΔH is positive, negative, or zero for the autoionization reaction of water: 2 H2O1l2 Δ H3O+1aq2 + OH-1aq2

Question

Accepted Answer

hi everyone for this problem it reads if the ion product constant kw for water at 100 degrees Celsius is 51.3 times 10 to the negative 14, what will be its ph Okay, so what we want to do from this for this problem is solve for P. H. Given the ion product constant kw. So let's start off by defining what is this ion product constant. Okay, so this has to do with the auto dissociation of water. And so when we auto dissociate water, what it breaks down to is it breaks down to a hydro ni um ion and a hydroxide ion. Okay. And from this we get the kw. Okay. And what kw is it's the product of the hydro ni um ion and hydroxide ion. And in this problem, an important piece of information we're given is the value of KW We're told it's 51.3 times 10 to the -14. Okay. All right. So one thing we need to recall for the auto dissociation of water or a kW is that for pure water the concentration of hydro ni um is going to equal the concentration of hydroxide. Okay. And so we know what the kW value is and we can solve for P H. Because are because the for pure water the concentrations are equal, we can say that our concentration of hydro ni um squared is equal to r value for K. W. Okay, Since they equal each other, our concentration of hydro name equals our concentration of hydroxide. So we can say our concentration of hydro ni um squared equals 51.3 times negative four times 10 to the negative 14. The way that we can solve for P. H. Here is to recall P H. Is equal to the negative log of our hydro knee. Um ion concentration. Okay so if we can solve for our concentration of hydro knee um we can solve for P. H. Because we need the value to plug into our ph equation. So let's go ahead and isolate our hydro knee um Ion concentration by squaring both sides of this equation. If we square, if you take the square root of both sides we'll get our concentration of hydro knee um is equal to the square root of 51.3 times 10 to the negative 14. Okay so when we do that, when we take the square root we get 7. times 10 to the negative seven. So that's our concentration of hydro knee. Um so now that we have that number, we can go ahead and plug it in here. Okay so let's go ahead and do that. So we can solve for P H. P H is going to equal the negative log of our hydro knee um concentration which we just solved is 7.16 times to the negative seven. Okay, when we do that we get P H. Is equal to 6.145. And we want to make sure we're paying attention to our sig Figs here. So our KW constant was three significant figures. And so for ph three significant figures is going to equate to three decimal places. So that's why our ph is six 0.145. Okay, so this is our final answer. Ph is equal to 6.145. And we're able to do that by solving for our concentration of hydro knee um ions. Okay, so that's the end of this problem. I hope this was helpful.

At 50 °C, the ion-product constant for H2O has the value Kw = 5.48 * 10-14. (a) What is the pH of pure water at 50 °C? (b) Based on the change in Kw with temperature, predict whether ΔH is positive, negative, or zero for the autoionization reaction of water: 2 H2O1l2 Δ H3O+1aq2 + OH-1aq2

Verified Solution

Key Concepts

Ion-Product Constant of Water (Kw)

pH Scale

Temperature Dependence of Kw