At 300.K and 1.00 atm, assume that 25 mL of NO gas reacts with 22 mL of

Question

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:
O (g) H O(l) 2 HNO (g)
2
2NO(g) 3 + 2 + 2 ⎯⎯→ 3 .
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.

Answer 1

You need to provide an equation which is readable. Most of us on this board write subscripts in regular form; i.e.,

H₂O is written as H2O. We write an exponent, such as 3² = 9 as 3^2 = 9.

Answer 2

To find the pH of the resulting solution, we need to determine the concentration of the nitric acid (HNO3) produced in the reaction.

To find the limiting reagent, we compare the number of moles of each reactant to the stoichiometry of the balanced equation. The balanced equation shows that it takes 2 moles of NO gas (2NO) to produce 2 moles of nitric acid (2HNO3). Similarly, it takes 1 mole of oxygen gas (O2) to produce 2 moles of nitric acid.

First, let's convert the volumes of gases to moles using the ideal gas law (PV = nRT), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

For NO gas:
n(NO) = (V(NO) / Vm(NO)) * (P / (R * T))

Since both NO and O2 were measured at the same temperature and pressure, the molar volumes cancel out. Therefore, we only need to calculate the moles of NO gas.

So, n(NO) = (25 mL / 1000 mL/L) * (1 L / 22.4 L/mol) * (1.00 atm / (0.0821 L.atm/mol.K * 300 K))

Calculating this expression, we find that n(NO) ≈ 0.045 mol

For O2 gas:
n(O2) = (22 mL / 1000 mL/L) * (1 L / 22.4 L/mol) * (1.00 atm / (0.0821 L.atm/mol.K * 300 K))

Calculating this expression, we find that n(O2) ≈ 0.038 mol

Based on these calculations, we can see that NO is the limiting reagent since it produces fewer moles of nitric acid.

Now, let's find the concentration of the nitric acid produced by the reaction. Since we dissolve all the nitric acid in 25 mL of water, we can assume the final volume of the solution is 25 mL.

Concentration of HNO3 = (moles HNO3) / (volume of solution)

Since the moles of HNO3 produced is equal to the moles of NO used, we can determine the concentration:

Concentration of HNO3 = (0.045 mol) / (25 mL)

Converting the volume to liters, we get:

Concentration of HNO3 = (0.045 mol) / (0.025 L) = 1.8 M

Finally, to find the pH of the resulting solution, we use the fact that nitric acid is a strong acid, meaning it will dissociate completely in water. This means that the concentration of H+ ions in the solution will be equal to the concentration of HNO3.

pH = -log[H+]

So, pH = -log[1.8] ≈ -0.255

Therefore, the pH of the resulting solution would be approximately 0.255.