At 0 °C, the density of liquid water is 0.9998 g/mL and the
valu...

Question

At 0 °C, the density of liquid water is 0.9998 g/mL and the 
value of Kw is 1.14 * 10-15. What fraction of the molecules 
in liquid water are dissociated at 0 °C? What is the percent dissociation 
at 0 °C? What is the pH of a neutral solution at 0 °C?

Accepted Answer

Hello. Everyone in this video, we're given the density here as well as the kw of liquid water at five degrees Celsius and at five degrees Celsius, we will determine three things in this problem. One is the fraction of liquid water molecules that associate to is percent association and three is ph of a neutral solution. So the chemical reaction that's going on here is that we have two moles of H. Two of Which of course is a liquid. And it's an equilibrium with H. 30. Plus as well as O. H. Minus. So for our kw equation that is equal to the concentration of our two products which is H. 30 plus, multiplied by the concentration of O. H minus. Alright so we're gonna go ahead and first deal with the ph of a neutral solution. So let's go ahead and determine the concentration of H. 20. 1st. Let's go ahead and actually write this. So for number one I'll do this in the green color. So the concentration of H 20. This can be found well with dementia analysis. So we can go ahead and use the density that's given to us of our liquid water. So that's 0. g per every one millimeter. So we can go ahead and multiply this by the molar mass of water. So that is for every one mole of liquid H20. He gets 18.02 g. And we're gonna convert our male leaders into leaders to give us the concentration units. So for every 1000 ml we have one leader. So you see that female leaders will cancel out and the grounds will cancel out leaving us with the units of moles on top and leaders on the bottom. So once you put everything into the calculator we see that the concentration for H20. Is going to be equal to .48-8 Moller. Alright, so for a neutral, let's actually put this in writing as well. So four a neutral solution Which is what we said we have here the concentration of H. 30 plus is going to be equal to the concentration of O. H. Minus. So then our equation then for K. W. That we have here in red, that's going to be that K. W. Is equal to the concentration of H. 30. Plus to the power of two. Since it's just basically multiplying each other since we do have them equal to each other. All right. No. So Kw Then which vary said is equal to 1.2 times 10 to the -15. And it's given to us here in the problem already. So how we can find the concentration then is if we take the square root so then the concentration of H. 30 plus is equal to the concentration of O. H minus which is equal to the square root of R. K. W. So 1.2 times 10 to the negative 15. We go ahead and put that into the calculator. We see that that's concentration is equal to 3. times 10 to the negative eight molars. You know, we go ahead and use this value to calculate for R. P. H. So again, I'm just gonna settle down from our space. If you recall the ph equation, is that ph Is equal to the negative log of the concentration of H. 30. Plus. And it happens that we do have this concentration value that's over here. So I'm gonna go ahead and simply plug it into my equation. So P. H. Is equal to the negative log of 3.461 times 10 to the negative eight. So once I put that into my calculator, I see that my P. H. Is equal to 7.64. All right, So, we have an answer for part three here, We just thought my ph is equal to 7.64. Alright, We'll go ahead and now work on the first part and that's the fraction of liquid water molecules that associate. So we'll go ahead and do this in blue. So go ahead and maybe do this off to the right side here. Alright, so the fraction equation is equal to the concentration of H. 20. That's associated. So to put D. I. S over the concentration of H. 20. That we have initially. Alright. And just to make sure we know that the concentration of H. 20. That's associated. As you go to the concentration of H30. Plus at equilibrium. Alright. So then the fraction is equal to we have 3.4641 times to the negative eight moller over 55 .48 - eight Moller. So once you put that into a calculator receipt that my fraction and that's again for number one or part one, is that my fraction Is equal to 6.243, 6 times 10 to the -10. So this is my second answer here. Alright. And lastly is to go ahead and calculate for the percentage association, I'll go ahead and do this in this purple color. This on the bottom here, so again we calculate for the percent dissociation. So there's a formula for that. So we follow the formula we get that on top we have this 3.461 times 10 to the negative eight moller Over 55.4. 8 - eight moller times 100% of course. So simplifying our fraction here we get 6. times 10 to the negative 10 times is 100%. And if you combine those two on the right side of our equation sign we get that the percent association is equal to 6.24 times 10 to the negative 8%. So this is going to be my final answer for this problem

At 0 °C, the density of liquid water is 0.9998 g/mL and the value of Kw is 1.14 * 10-15. What fraction of the molecules in liquid water are dissociated at 0 °C? What is the percent dissociation at 0 °C? What is the pH of a neutral solution at 0 °C?

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