Nitric acid can be prepared by bubbling dinitrogen pentoxide into water.
N2O5(g) + H2O -> 2H+(aq) + 2NO3-(aq)
a) How many moles of H+ are obtained when 1.50 L of N2O5 at 25 degrees C and 1.00 atm pressure is bubbled into water?
b) The solution obtained in (a) after reaction is complete has a volume of 437 mL. What is the molarity of the nitric acid obtained?
Here is my work for (a)- does it look ok?
n= PV/RT
(1.00 atm x 1.50 L/ .08206 L x atm/mol x K x 298.15 K)= .0613 mol N2O5 x 2= .123 mol H+
If this is correct [I don't know if I made assumptions that I shouldn't have made], how do I do (b)? I know I need to get an answer in mol/L but I'm not sure how to do that.
Your work looks ok to me, Dave. You are also correct that molarity must come out in mols/L so mols/L =
0.123 mols/1.5 L = ??
0.123 mols / 0.437 L = 0.282 M (molarity of nitric acid)
To calculate the molarity of the nitric acid obtained, you need to divide the number of moles of H+ by the volume of the solution.
Given:
Number of moles of H+ = 0.123 mol
Volume of the solution = 437 mL = 0.437 L
To find the molarity (M), use the formula:
Molarity = moles of solute / volume of solution
Molarity = 0.123 mol / 0.437 L = ??
To calculate the molarity of the nitric acid obtained, you need to divide the number of moles of H+ by the volume of the solution in liters.
So, using the given values:
Number of moles of H+ = 0.123 mol
Volume of the solution = 437 mL = 0.437 L
Molarity (M) = Moles of solute (H+) / Volume of solution (L)
Molarity = 0.123 mol / 0.437 L
Molarity = 0.281 M
Therefore, the molarity of the nitric acid obtained is 0.281 M.