Question

Nitric acid can be prepared by bubbling dinitrogen pentoxide into water. N2O5(g) 1 H2O 9: 2H1(aq) 1 2NO3 2(aq) (a) How many moles of H1 are obtained when 1.50 L of N2O5 at 258C and 1.00 atm pressure is bubbled into water? (b) The solution obtained in (a) after reaction is complete has a volume of 437

Short answer

Question: Determine the number of moles of H+ obtained and its concentration in the solution when 1.50 L of N2O5 at 25°C and 1.00 atm pressure is bubbled into water. The final volume of the solution is 4.37 L. Answer: 0.1224 mol of H+ are obtained and the concentration of H+ ions in the solution is approximately 0.0280 M.

Step-by-step solution

Calculate the number of moles of N2O5

First, we will use the Ideal Gas Law equation ( PV=nRT ) to determine the number of moles of N2O5. For this equation, P is the pressure (1.00 atm), V is the volume (1.50 L), R is the ideal gas constant (0.0821 L atm K¯¹ mol¯¹), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K). n = PV / RT = (1.00 atm * 1.50 L) / (0.0821 L atm K¯¹ mol¯¹ * 298.15 K) n ≈ 0.0612 mol of N2O5

Determine the number of moles of H+ produced

Using the stoichiometry of the reaction, we can calculate the number of moles of H+ formed when 0.0612 mol of N2O5 reacts with water. The balanced equation is: N2O5(g) + H2O(l) -> 2H+(aq) + 2NO3−(aq) From this equation, we can see that 1 mole of N2O5 produces 2 moles of H+. Thus, 0.0612 mol N2O5 * (2 mol H+ / 1 mol N2O5) ≈ 0.1224 mol of H+

Calculate the concentration of H+ in the solution

The problem states that the final volume of the solution is 4.37 L. We can now use the formula to calculate the concentration of H+ ions in the solution. Concentration of H+ = moles of H+ / volume of solution = 0.1224 mol / 4.37 L ≈ 0.0280 M (moles per liter) The final answers for the exercise are: (a) 0.1224 mol of H+ are obtained when 1.50 L of N2O5 at 25°C and 1.00 atm pressure is bubbled into water. (b) The concentration of H+ ions in the solution after the reaction is complete is approximately 0.0280 M.