Calculating Required Volume for 1M HNO3 Solution: Nitric Acid Concentration

[thea831]

Homework Statement

Nitric acid is commercially available as a 72% ( w/ w) solution ( density, 1.42 g/cm3). How many milliliters of this reagent are needed to prepare 2.00 L of a 1.00M HNO3 solution?

Homework Equations

C%= (g HNO3/ g Solution) * 100

The Attempt at a Solution

So basically, I started with converting a mole of the solution to mL.
1 mol HNO3=44.37ml

But I feel like I need to first use the concentration to find how much reagent is in that 72%. so its .72g I assume that the reagent is contained in that solution.

so .72g of HNO3 in the solution is .507mL

but now I am stuck. How do I use that information to figure out what the mL of the reagent is needed to get 2L of 1 mol of HNO3

 

[epenguin]

Your statements and working were frankly too unclear for me to follow. I do not know what your para 2 means.

Do not "started with converting a mole of the solution to mL" which is 2 steps. Start by converting a mole to grams.

Now 1g of your commerical nitric acid, is .72g of HNO₃ .

You find the volume of the above 1g (containing your .72g of HNO₃) from the density. I do not think you have got that right.

Then you work out how much volume acid contains a mole (actually 2 moles since you need to make 2l).

It should all make sense. All the chemistry is in the first step and the rest is just arithmetic to find the voume of acid/water mix that contains a mole of HNO₃

 

[thea831]

i don't understand what you mean by 1g is .72g of commercial HNO3.

1g of HNO3 is 1/1.42 mL

so I need 2mols since there are 2L.

so here is what I am thinking.

Based on the w/w content concentration which is 72%, I use the formula listed above to solve for the mass of the reagent HNO3

.72*(2000ml*1.42g/mL)=2044g= mass of HNO3 reagent.

Now I need to know the volume of this reagent in a 2L of 1M concentration

So here is how I got my answer:

2044g HNO3 * (1 mol/ 63.012g) * (2L/2mol) = 32.4L which does not make sense

 

[thea831]

nvm i got the answer. I used MV=M2V2. Solved for M2 first then for V2

 

[Borek]

i don't understand what you mean by 1g is .72g of commercial HNO3.

At 72% w/w there is 0.72g of nitric acid in each 1g of the solution.

.72*(2000ml*1.42g/mL)=2044g= mass of HNO3 reagent.

Now you are assuming you need 2L of the solution that will have 1.42 g/mL density - that's density of the stock solution, not the final one.

Besides, density of the final solution doesn't matter, what matters is the number of moles of nitric acid in the final solution, it must be identical to number of moles of nitric acid in stock solution volume that you will dilute. That's simple mass conservation.

nvm i got the answer. I used MV=M2V2. Solved for M2 first then for V2

Using this formula blindly won't get you far. Somehow I doubt your solution, but not seeing the numbers I can be wrong.

 

[epenguin]

i don't understand what you mean by 1g is .72g of commercial HNO3.

1g of HNO3 is 1/1.42 mL

so I need 2mols since there are 2L. I did mention that

so here is what I am thinking.

Based on the w/w content concentration which is 72%, I use the formula listed above to solve for the mass of the reagent HNO3

.72*(2000ml*1.42g/mL)=2044g= mass of HNO3 reagent.

Now I need to know the volume of this reagent in a 2L of 1M concentration

So here is how I got my answer:

2044g HNO3 * (1 mol/ 63.012g) * (2L/2mol) = 32.4L which does not make sense

Good teacher once gave me sound advice - "don't try to do several steps in your head epenguin - you are not clever enough". (Let alone am I clever enough to follow what you do in your head. Not even Borek is that clever! )

You need, OK*, 2 moles of HNO₃. You need to find how many grams is that?

That is an exercise in chemistry, knowing what moles, molar masses etc. are - in general and for this substance HNO₃. (BTW you cannot do it from only the information you have written, I suppose you have, or remember, some other.)

Then when you have decided that, it is an exercise in simple physics - densites, masses, volumes, to find out in what volume of this dense water-HNO₃ mixture, that number of grams of HNO₃ is contained.


* OK to do it that way. The reason I personally would do it for 1l and then multiply the result by 2 at the end, is that I would be working out for 1l in all cases and then multiply by appropriate factor according as I wanted 2l of IM or 250 ml. or whatever of the desired solution.