What is the boiling point of a 0.10 M solution of NaHSO4 if
the ...

Question

What is the boiling point of a 0.10 M solution of NaHSO4 if 
the solution has a density of 1.002 g>mL?

Accepted Answer

Hey everyone, we're told for a 0.150 molar solution of sodium ascorbic a what is the boiling point? If it has a density of 1.5 g per mil leader and were provided R K A one and r k A two. So let's go ahead and write out what's happening here. So we have our sodium s corbet and this is going to disassociate into our sodium ions plus our hydrogen s corbet ions. And as we've learned, salts of weak acids are basic in nature. So we have to consider the reverse ionization of the monosodium salt, which is sodium hydrogen escort bait first. Let's go ahead and determine our total ion concentration. Taking that hydrogen S. Corbet, we're going to react this with water and our products here are going to be ascorbic acid plus our hydroxide ions creating our ice chart. Initially we had 0.150 moller of our hydrogen S. Corbet ion. And we had zero of our products formed initially we can go ahead and disregard our water since it is a liquid and our change is going to be a minus X on our reactant side and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium, we have 0.150 minus X on our react inside and an X. And an X on our product side. Since we are looking at our hydroxide ions, we need to find our kB of our reaction and we can do so by taking our kw and dividing it by R K. A. Now plugging in those values, We get 1.0 times 10 to the -14 which is RK. W. And dividing it by R K. A. Of eight point oh times 10 to the negative five. This gets us to a K. B. Of 1.25 times 10 to the negative 10, solving for X. We can take that KB and make it equal to our products over our reactant. In this case will be X times X all over 0. minus X. Now we can go ahead and check if our X. And our denominator is negligible. And we can do so by taking 0. And dividing it by RKB. 1.25 times 10 to the negative 10. If we get a value that is greater than 500 then we can safely disregard our X. And our denominator since it is negligible And in this case it is greater than 500. So we can disregard that X. In our denominator, solving for X we have 1.25 times 10 to the negative is equal to X squared over 0.150, multiplying both sides by 0. and subsequently taking the square root on both sides. This will get us to an ex of 4.3301 times to the negative six moller. And this is going to be the concentration of our hydroxide ions and also our ascorbic acid. To calculate the concentration of our hydrogen s. Corbet at equilibrium. All we need to do is take 0.150 and subtract 4.3301 times 10 to the negative six. This gets us to a concentration of 0.150 Moller. Now let's go ahead and calculate our total ion concentration, adding all these values up. We add 0.150 moller from our sodium ions plus 0.150 moller from our hydrogen s corbet ions. And we're also going to add 4.3301 times 10 to the - moller from our hydroxide ions. And lastly We're going to add 4.3301 times 10 to the -6 molar from our ascorbic acid. When we add this all up, we get a total ion concentration of 0.300 moller. Now let's go ahead and determine the mass of our solvent. We can go ahead and assume That we have 100 ml of solution. So taking our density of 1.05 g over mill leaders, We can go ahead and multiply that by 100 ml. This gets us to 105 g. Next we want to take the polarity of our sodium s. Corbett and we want to convert this into grams of sodium escort bait. So we know that one leader contains tens of third male leaders and we had 100 ml. Next looking at sodium S Corvettes Mueller mass, we know that we have 198.12 g of sodium ask, orbit her one mole of sodium ask or bait, calculating this out and canceling out all of our units. We end up with 2. g of sodium escort bait. Now taking the mass of our solvent, which was 105 g. We're going to subtract the value we just calculated for which was 2.9718 g of sodium escort bait. And this will get us to a total of 102. g. And let's go ahead and convert this into kilograms. So we know that we have 10 to the third grams per one kg. So this will get us 0.10-0- kg of water. Now let's go ahead and determine our boiling point. First, we're going to determine our morality and we know that this is going to be mole of ions divided by kilograms of water. So, plugging in these values, we have 0.300 moller which was our total ion concentration. And we're going to multiply this. Bye 0.1000 L. Next we're going to divide this by the total kg. Which we calculated for as 0.10-0-8 two kg of water. Now, when we calculate this out, we end up with 0.294 to determine our boiling point, we can use the formula KB times M. Plugging in these values. KB stands for a proportionality constant which is 0.52°C over em And we can go ahead and multiply that by 0.294, this gets us to A value of 0.15°C. Now to find our boiling point, We're going to take 100 and add 0. And this will get us to 100.15°C, which is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions.

What is the boiling point of a 0.10 M solution of NaHSO4 if the solution has a density of 1.002 g>mL?

Watch next