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

In summary: That way I am less likely to make a silly mistake, because I am doing a simpler problem. Also I am less likely to get tired, and then make a mistake.
  • #1
thea831
8
0

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
 
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  • #2
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
 
  • #3
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
 
  • #4
nvm i got the answer. I used MV=M2V2. Solved for M2 first then for V2
 
  • #5
thea831 said:
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.

thea831 said:
.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.

thea831 said:
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.
 
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  • #6
thea831 said:
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 :smile:

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! :biggrin:)

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.
 

1. What is the formula for calculating required volume for 1M HNO3 solution?

The formula for calculating required volume for 1M HNO3 solution is: Volume = (Desired concentration * Desired volume) / Stock concentration. This formula is also known as the dilution equation.

2. How do I know the stock concentration of my nitric acid solution?

The stock concentration of your nitric acid solution can typically be found on the label of the solution bottle or in the product specification sheet. You can also use a chemical calculator to determine the stock concentration based on the molecular weight and density of the solution.

3. What is the difference between molarity and normality?

Molarity (M) is a unit of concentration that measures the number of moles of a solute per liter of solution. Normality (N) is a unit of concentration that measures the number of equivalents of a solute per liter of solution. In the case of nitric acid (HNO3), 1M is equal to 1N because it has only one acidic hydrogen ion.

4. Can I use any unit of volume for the required volume calculation?

No, the units of volume used in the required volume calculation must be consistent. For example, if the desired concentration and stock concentration are in units of molarity (mol/L), then the desired volume must be in liters (L).

5. How do I prepare a 1M HNO3 solution from a concentrated nitric acid solution?

To prepare a 1M HNO3 solution from a concentrated nitric acid solution, you can use the dilution equation mentioned in the first question. First, calculate the required volume of the concentrated solution needed using the desired concentration and desired volume. Then, measure out the required volume of the concentrated solution and add it to an appropriate amount of water to reach the desired volume. Be sure to mix the solution thoroughly to ensure uniform concentration.

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