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A chemist must prepare 700.0 mL of nitric acid solution with PH of 1.70 at 25 "He will do this in three steps:Fill a 700.0 mL volumetric flask about halfway wi...

Question

A chemist must prepare 700.0 mL of nitric acid solution with PH of 1.70 at 25 "He will do this in three steps:Fill a 700.0 mL volumetric flask about halfway with distilled water:Measure out small volume of concentrated (6.0M) stock nitric acid solution and add it to the flask:Fill the flask to the mark with distilled water:Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step_ Round your answer to 2 significant digits_mL

A chemist must prepare 700.0 mL of nitric acid solution with PH of 1.70 at 25 " He will do this in three steps: Fill a 700.0 mL volumetric flask about halfway with distilled water: Measure out small volume of concentrated (6.0M) stock nitric acid solution and add it to the flask: Fill the flask to the mark with distilled water: Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step_ Round your answer to 2 significant digits_ mL



Answers

Concentrated nitric acid is a $70.0 \%$ solution of nitric acid, $\mathrm{HNO}_{3},$ in water. The density of the solution is $1.41 \mathrm{~g} / \mathrm{mL}$ at $25^{\circ} \mathrm{C} .$ Calculate the molarity of nitric acid in this solution.

To calculate the polarity of nitric acid solution. Polarity is moles per leader. We need to know the moles of nitric acid in the leaders of the nitric acid solution. The leaders of the nitric acid solution was given to us at 20 mil leaders. Or when we convert that to leaders, it'll be 200.2 leaders, but to calculate the moles of nitric acid, it is a little bit more challenging but still not too difficult. We'll start with the volume of K. O. H that was used to titrate the 20 mL solution of nitric acid. Convert those mill leaders into leaders. Then we can convert the leaders of K O H into moles of CO h using the molar ity of K o h. 00.1 moles per leader at 0.1 Moeller concentration. Then we look at the balanced chemical reaction. This is a strong acid reacting with a strong base. The stoy geometry is one toe one So one mole K o h is required for one mole h n 03 This then gives us the moles h n 03 which we can put here in the numerator to calculate Armel Arat e in the denominator, as I mentioned, that was given to us a 20 mL, which we convert to leaders, which is 200.2 leaders. We then divide these two numbers in order to get the mole arat e of the nitric acid solution at 0.2167 Moeller.

This is Chapter 18 number 44. So we're looking for a ah neutralization problem here. So we know that our volume of base is equal to 43.33 mil leaders and our polarity of base physical to 0.1000 Mueller, we know that our volume of acid is equal to 20 milliliters and I'm Alaric have asked it isn't unknown. So here we could just say that the volume of acid time similarity of acid is equal to the volume of the base times the polarity of the base because one of the moles are equal. And when you multiply similarity by a volume, you get moles. So here we just saw for unknown you get the polarity of acid is equal to the volume of the base times the polarity of the base over the volume of the acid. This you don't get here as 43.33 milliliters times your 0.10 Oh Moeller. And then you get 20.0 milliliters! Down here. Those units cancel you get polarity. So we know that the polarity of our acid is equal to 0.2167 Moeller for significant figures because all of these values have four significant figures

Chapter 15 problems 16 tells us about a nitric acid solution that has a peach of 2.7 and asks us to determine a few different things about So we have our pH of 2.7. So a is asking us to determine the concentration of hydrogen ions. So remember that we can convert from P H two concentration of ions using the Formula 10 to the power of negative peach. So that's going to be 10 to the power of negative 2.7, which gives us a concentration of two times 10 to the negative three. Moeller now be asks us for the concentration of hydroxide ions, and there's a couple of ways to go about it. Remember that if we do one times 10 to the negative 14 divided by concentration of hydrogen ions will get the concentration of hydroxide. There's another way to solve it, though, and we can do that by determining that p o. H equals 14 minus pH, which in this case would be 14 minus 2.7, which gives us a value of 11.3. Then we could do the same processes, calculating the I H ion concentration where we have 10 to the power of negative P O H, which is 10 to the power of negative 11.3, which is equal to five tens, tend to the negative 12 more. All right now, problems See tells us that he's determined the number of moles of H I know three that were required to prepare a 5.5 leader solution. So notice that in H and 03 each bowl of a Jenna three has one mole of hydrogen that is the mole of hydrogen ions. We calculated earlier. He's the same polarity has the nitric acid here, so we know that nitric acid has a two tens tend to the negative three concentration. So how many moles have been required for 5.5? Leaders will remember that mill Arat E is the same as moles per leader. So if we dio polarity times the number of leaders that will get us two moles, which is what we're looking for, So that's two times 10 to the negative three times that 5.5 leaders, and we find that that gives us 0.11 Smalls, now D is asking us for the mass of this acid, and this is pretty straightforward. We'll just use the moulder mass here. So 0.11 moles and will use the molar mass here where one mole is equivalent to 63 grams and we'll see that. That gives us 0.693 grams of our acid. Okay, Now, the last question is asking this to determine how many milliliters of a concentrated solution will we needed to give us this many moles. So for e of the reminder, we're looking for 0.11 moles of our acid. So of R H a No. Three. And what we have available to us is a 69.5% solution that has a density of 1.42 grams per mil. So that's her concentrated solution. How many milliliters of that will get us 0.11 molds. So the way that we do this is all we have is this percent by mass and the density of from there we can get how many grams per mil of asset that we have. It's just that much per cent of the 1.4 till once we have grams per mille. We can convert that to malaria t two moles per liter. Then we can use our dilution formula to go from are concentrated acid toward delete passage. So let's go ahead and start that process. So first we need to determine the density of just the nitric acid. So our density of 1.42 grams per mil well, multiply it by this percent or 0.695 and we'll find that the density of Onley the acid in this solution here is 0.9 869 grams per milk. Now we can convert this to polarities, so malaria's moles per leader. So let's convert the leaders part first. So we have 1000 mills is equivalent to one leader and then to go from grams to moles will use the molar mass where again when wool is equivalent to 63 crams and this tells us that we have a polarity here in are concentrated solution of 15.7 Moeller. Now we can convert this to the number of milliliters that we need by using M one V one equals and to me too. So what? We're looking for here is V one, the milliliters of concentrated acid. So we can rewrite vests as V one is equivalent Teoh m two V 2/1. So here to interview to our dilute solution. And remember, we had previously calculated molar ity of our hydrogen ions to be two times 10 to the negative. Three more. And we have a 5.5 leader solution and the concentration of our very daily concentrated acid here was 15.7 more. So calculating this out, we get 0.7 leaders, which is equal to 0.7 milliliters of our constant.


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