Traffic in the cities releases NOx gases into the atmosphere (in the form of NO and NO2). NOx gases
contribute to rainwater acidification by dissolving and transforming into a strong acid (HNO3, nitric
acid) 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?
a) To find the molar concentration of nitric acid in the rainwater, we first need to calculate the number of moles of nitric acid produced.
Given:
Mass of NO = 0.001 g
Molar mass of NO = 30.01 g/mol
Number of moles of NO = Mass of NO / Molar mass of NO
= 0.001 g / 30.01 g/mol
≈ 3.33 x 10^-5 mol
According to the balanced equation, 4 moles of NO produce 4 moles of HNO3.
Therefore, the number of moles of HNO3 produced is also 3.33 x 10^-5 mol.
Now we need to find the volume of the rainwater in liters. Given that 0.001 g of NO dissolves into 1 liter of rainwater, the volume of rainwater is 1 liter.
Molar concentration (c) of a solute is given by the formula:
c = (Number of moles of solute) / (Volume of solvent in liters)
c = (3.33 x 10^-5 mol) / 1 L
≈ 3.33 x 10^-5 mol/L
Therefore, the molar concentration of nitric acid in the rainwater is approximately 3.33 x 10^-5 mol/L.
b) The pH of a solution can be calculated using the formula:
pH = -log[H+]
Since nitric acid (HNO3) is a strong acid, it will dissociate completely in water to produce H+ ions. The balanced equation for the dissociation of HNO3 is:
HNO3 → H+ + NO3-
Since a 1:1 molar ratio exists between HNO3 and H+ ions, the concentration of H+ ions is the same as the molar concentration of nitric acid.
Using the molar concentration obtained in part a, the concentration of H+ ions is approximately 3.33 x 10^-5 M.
pH = -log(3.33 x 10^-5)
≈ 4.48
Therefore, the pH of rainwater containing only nitric acid is approximately 4.48.
To find the molar concentration of nitric acid in the rainwater, we first need to calculate the number of moles of nitric acid formed from the given mass of NO. Then, we can determine the molar concentration.
a) Calculating moles of nitric acid:
First, we convert the given mass of NO (0.001 g) to moles. To do this, we need to know the molar mass of NO. The molar mass of nitrogen (N) is approximately 14 g/mol, and the molar mass of oxygen (O) is approximately 16 g/mol.
The molar mass of NO = (14 g/mol) + (16 g/mol) = 30 g/mol
To convert grams to moles, we use the equation:
moles = mass / molar mass
moles of NO = 0.001 g / 30 g/mol ≈ 0.00003333 mol
Since the reaction is balanced in a 1:1 ratio between NO and HNO3, the number of moles of nitric acid formed is also approximately 0.00003333 mol.
Next, we need to determine the molar concentration of nitric acid in the rainwater. Molar concentration is defined as moles of solute dissolved in 1 liter of solution (mol/L or M).
Molar concentration = moles of solute / volume of solution
Since we have 1 liter of rainwater, the molar concentration of nitric acid is:
Molar concentration = 0.00003333 mol / 1 L ≈ 0.00003333 M
Therefore, the molar concentration of nitric acid in the rainwater is approximately 0.00003333 M.
b) To find the pH of the rainwater, we need to make an assumption that none of the dissolved nitric acid dissociates or reacts further to form additional hydrogen ions (H+).
Since 1 molecule of HNO3 produces 1 molecule of H+ in the dissociation reaction, the concentration of H+ ions in the solution is the same as the molar concentration of HNO3.
Therefore, the pH of rainwater with a molar concentration of 0.00003333 M HNO3 is given by the equation:
pH = -log[H+]
pH = -log(0.00003333)
= 4.48
Thus, the pH of the rainwater, given the assumptions made, is approximately 4.48.