At 300.K and 1.00 atm, assume that 25 mL of NO gas reacts with 22 mL of
oxygen gas and excess water to produce nitric acid according to the following
equation:
2NO(g)+(3/2)O2(g)+H2O(l)-->2HNO3(g)
If all of the nitric acid produced by this reaction is 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.
I would start by converting 25 mL NO gas to moles with PV = nRT. Do the same for 22 mL O2.
Using the coefficients in the balanced equation, convert moles NO and moles O2 (in two separate operations) to moles HNO3. From the hint, I expect the moles HNO3 from each will NOT be the same which means one of them is wrong. The correct one, in limiting reagent problems, is ALWAYS the smaller value and the material producing that value is the limiting reagent. Remembering that molarity = moles/L, substitute moles HNO3 you have and 0.025 for the 25 mL and that should be the molarity. Convert M to pH remembering that HNO3 is a strong acid (ionizes 100%) and the molarity = (H^+).
Thank you! That was very helpful. My final answer for the pH was 1.39. Is that correct?
yes. I came up with that answer, also.
To determine the limiting reagent in the reaction, we need to compare the number of moles of each reactant.
First, let's calculate the number of moles for each reactant:
The given volume of NO gas is 25 mL. We can convert this to moles using the ideal gas law:
PV = nRT
Here, P is the pressure (1.00 atm), V is the volume (25 mL or 0.025 L), n is the number of moles (which we want to find), R is the ideal gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature (which is not given, so we'll assume room temperature, around 298 K).
So, rearranging the formula:
n = PV / RT
n(NO) = (1.00 atm) * (0.025 L) / (0.0821 L⋅atm/mol⋅K * 298 K)
n(NO) ≈ 0.001 mol
Next, let's calculate the number of moles of oxygen gas:
The given volume of oxygen gas is 22 mL. Converting to moles:
n(O2) = (1.00 atm) * (0.022 L) / (0.0821 L⋅atm/mol⋅K * 298 K)
n(O2) ≈ 0.001 mol
Now, let's use the stoichiometry of the reaction to determine the limiting reagent. The balanced equation shows that the ratio between NO and O2 is 2:3/2, which simplifies to 4:3. Since we have the same number of moles for NO and O2, it means O2 will be the limiting reagent because we need 4 moles of NO for every 3 moles of O2.
Now, let's calculate the number of moles and volume of HNO3 formed:
From the balanced equation, we see that 2 moles of NO react to produce 2 moles of HNO3. Therefore, the number of moles of HNO3 formed will be the same as the number of moles of NO, which is 0.001 mol.
Since the nitric acid is dissolved in 25 mL of water, the resulting volume of the solution will be 25 mL + 25 mL = 50 mL or 0.050 L.
Now, we can calculate the molarity (M) of the HNO3 solution:
Molarity (M) = moles of solute / volume of solution (in liters)
M(HNO3) = 0.001 mol / 0.050 L
M(HNO3) = 0.02 M
The pH of a solution depends on the concentration of H+ ions it contains. In the case of a strong acid like HNO3, it dissociates completely in water, so the concentration of H+ ions will be the same as the concentration of HNO3.
Therefore, the pH of the resulting solution will be determined by the concentration of HNO3, which is 0.02 M.
To calculate the pH, we can use the equation:
pH = -log[H+]
pH = -log(0.02)
pH ≈ 1.70
Therefore, the pH of the resulting solution is approximately 1.70.