according to the reaction below:
4 NO + 3 O2 + 2 H2O → 4 HNO3
a) 0,001 g of NO dissolves into 1 liter of rainwater and all of it transforms into nitric acid. What is the
molar concentration of nitric acid in the rainwater?
b) Supposing that no other acids are dissolved in the rainwater, what is the pH of rainwater?
To find the molar concentration of nitric acid in the rainwater (part a), we need to calculate the number of moles of nitric acid produced by dissolving 0.001 g of NO in 1 liter of rainwater.
To do this, we'll start by finding the molar mass of NO, which consists of one nitrogen atom (N) with a molar mass of approximately 14.01 g/mol and one oxygen atom (O) with a molar mass of approximately 16.00 g/mol. Adding these atomic masses together gives us a molar mass of NO of 30.01 g/mol.
Next, we'll convert the mass of NO (0.001 g) to moles. This can be done by dividing the mass by the molar mass:
Moles of NO = mass of NO / molar mass of NO
= 0.001 g / 30.01 g/mol
= 0.0000333 mol
Since the reaction stoichiometry shows that 4 moles of NO react to produce 4 moles of HNO3, the number of moles of HNO3 formed will also be 0.0000333 mol.
Finally, we divide the number of moles of HNO3 by the volume of the solution to find the molar concentration:
Molar concentration (C) = Moles of solute (n) / Volume of solution (V)
= 0.0000333 mol / 1 L
= 0.0000333 M
Therefore, the molar concentration of nitric acid in the rainwater is 0.0000333 M.
For part b, to find the pH of rainwater, we need to know the concentration of H+ ions in the solution. Since the reaction produces one mole of HNO3 for every mole of NO, the molar concentration of HNO3 will also be 0.0000333 M.
In aqueous solution, HNO3 is a strong acid that completely dissociates into H+ and NO3- ions. This means that the molar concentration of H+ ions will be the same as the molar concentration of HNO3, which is 0.0000333 M.
The pH scale is a logarithmic scale that represents the acidity or basicity of a solution. The pH of a solution can be calculated using the formula:
pH = -log[H+]
Substituting the concentration of H+ ions, we can find the pH:
pH = -log(0.0000333 M)
≈ 4.48
Therefore, the pH of the rainwater, assuming no other acids are dissolved, will be approximately 4.48.