At 298.K and 1.00 atm, assume that 22 mL of NO gas reacts with 18 mL of oxygen gas and excess water to produce nitric acid according to the following equation:
2 NO(g) + 3/2 O2 (g)+ H2O(l)?? 2 HNO3(g).
If all of the nitric acid produced by this reaction is collected and then dissolved
into 25 mL of water, what would be the pH of the resulting solution? (Hint: before you begin, think about which reactant is the limiting reagent.)
Use PV = nRT to convert NO to moles. Do the same and find moles ox oxygen.
Using the coefficients in the balanced equation, convert moles NO to moles HNO3.
Same procedure, convert moles O2 to moles HNO3.
It is likely that the two numbers will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
To find pH, use the smaller value in the previous step, divide by 0.025 L (25 mL) to convert to molarity, then
pH = -log(H^+). The (H^+) will be the molarity of HNO3.
Post your work if you get stuck.
To determine the pH of the resulting solution, we need to calculate the number of moles of nitric acid formed and then determine the concentration of HNO3 in the solution.
Step 1: Identify the limiting reagent
To find the limiting reagent, we compare the number of moles of each reactant with the stoichiometric coefficients from the balanced equation.
Given:
Volume of NO gas = 22 mL
Volume of oxygen gas = 18 mL
First, we need to convert the volumes to moles using the ideal gas law equation:
PV = nRT
Since both gases are at the same temperature and pressure, we can simplify the equation and calculate the number of moles for each gas.
For NO:
P = 1.00 atm
V = 22 mL = 0.022 L
T = 298 K
R = 0.0821 L⋅atm/(mol⋅K)
n(NO) = (P * V) / (R * T)
n(NO) = (1.00 atm * 0.022 L) / (0.0821 L⋅atm/(mol⋅K) * 298 K)
n(NO) ≈ 0.00104 mol
For oxygen gas (O2):
P = 1.00 atm
V = 18 mL = 0.018 L
T = 298 K
R = 0.0821 L⋅atm/(mol⋅K)
n(O2) = (P * V) / (R * T)
n(O2) = (1.00 atm * 0.018 L) / (0.0821 L⋅atm/(mol⋅K) * 298 K)
n(O2) ≈ 0.00078 mol
Now, let's compare the number of moles of NO and O2. From the balanced equation, the stoichiometric ratio between NO and O2 is 2:3/2.
Given that n(NO) = 0.00104 mol and n(O2) = 0.00078 mol:
2 mol of NO reacts with 3/2 mol of O2.
Using the stoichiometry, we can determine the amount of O2 required to react with 0.00104 mol of NO:
(3/2 mol O2 / 2 mol NO) * 0.00104 mol NO
≈ 0.00156 mol O2
Since we only have 0.00078 mol of O2, which is less than the required amount for the reaction, O2 is the limiting reagent.
Step 2: Calculate the number of moles of HNO3 formed
From the balanced equation, 2 mol of NO reacts to produce 2 mol of HNO3.
Given that n(NO) = 0.00104 mol:
(2 mol HNO3 / 2 mol NO) * 0.00104 mol NO
= 0.00104 mol HNO3
Step 3: Calculate the concentration of HNO3 in the solution
To find the concentration of HNO3, we need to know the volume of the resulting solution.
Given:
Volume of water = 25 mL = 0.025 L
Number of moles of HNO3 = 0.00104 mol
Concentration of HNO3 = (number of moles of HNO3) / (volume of solution in liters)
Concentration = 0.00104 mol / 0.025 L
Concentration ≈ 0.0416 M
Step 4: Calculate the pH of the resulting solution
The pH of a solution is determined by the concentration of H+ ions. Since HNO3 is a strong acid, it dissociates completely in water, resulting in the same concentration of H+ ions.
Thus, the pH of the solution is determined by the concentration of HNO3.
To calculate the pH, we use the equation:
pH = -log10([H+])
Given a concentration of HNO3 = 0.0416 M:
pH = -log10(0.0416)
pH ≈ 1.38
Therefore, the pH of the resulting solution is approximately 1.38.