A is a solution of trioxonitrate nitrate (v) (HNO3) of unknown concentration, B is a standard solutions of NAOH containing 4.00g/dm3 of solutions 25.0cm potion of solutions B required an average of 24.00cm of solutions A for complete neutralization
each mole of NaOH neutralizes one mole of HNO3.
I assume your volumes are in cm^3 (not cm), but I can't figure out what the question is. The molarity will tell you how many moles of NaOH you have. The volume of HNO3 will tell you the molarity needed.
Just off the cuff, since 24/25 as much HNO3 is needed, it will have to be 25/24 the concentration of the NaOH.
I think. If I have divined correctly just what you are saying ...
Two comments:
1. I think the question must be to determine the concentration of the HNO3.
The NaOH is 4 g/dm^3 (4 g/L), the molar mass of NaOH is 40; therefore, the molarity, M, of the NaOH is 4/40 mols/L or 0.1 M.
Then mols NaOH = M x L = 0.1 x 25/1000 = 0.0025 (soln B)
mols HNO3 (see rationale from oobleck) = 0.0025
Then M HNO3 (soln A) = mols/L = 0.0025/0.024 = 0.104 but watch the significant figures.
2. Where in the world did you come up with an absurd name like trioxonitrate nitrate(v). There is no such animal and this is NOT an IUPAC approved name for nitric acid. Other names like trioxonitrate(v) acid are not correct either. No matter how you slice it, the correct name for HNO3 is nitric acid. I thought the old "Stock" system was a good one and it was a good one when used to clarify naming; however, when the IUPAC adopted that for the ONLY way to name chemical compounds they started this ridiculous naming system. Na2SO4 is sodium sulfate, and not disodium tetraoxosulfate(VI). Bah Humbug.
To find the concentration of solution A, we can use the concept of stoichiometry and titration.
Here is the step-by-step process to calculate the concentration of solution A:
1. Write the balanced chemical equation for the reaction of the acid (HNO3) and base (NaOH):
HNO3 + NaOH -> NaNO3 + H2O
2. Calculate the number of moles of NaOH used in the titration:
Given that 25.0 cm³ of solution B with a concentration of 4.00 g/dm³ was used, we can convert cm³ to dm³ and then calculate the number of moles:
Number of moles of NaOH = (concentration * volume) / 1000
Number of moles of NaOH = (4.00 g/dm³ * 25.0 cm³) / 1000 = 0.100 mol
3. Use the stoichiometry of the balanced equation to determine the number of moles of HNO3:
From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HNO3. Therefore, the number of moles of HNO3 used in the titration is equal to the number of moles of NaOH, which is 0.100 mol.
4. Calculate the concentration of solution A:
Since we know the volume of solution A required for neutralization is 24.00 cm³, we can convert it to dm³ and use the equation:
Concentration of solution A = (moles of HNO3) / (volume of solution A in dm³)
Concentration of solution A = 0.100 mol / (24.00 cm³ / 1000) = 4.17 mol/dm³
Therefore, the concentration of solution A is 4.17 mol/dm³.