At 298 K and 1.00 atm, assume that 22 mL of NO gas reacts with 16 mL of oxygen g
ID: 853722 • Letter: A
Question
At 298 K and 1.00 atm, assume that 22 mL of NO gas reacts with 16 mL of oxygen gas and excess water to produce gaseous nitric acid according to the following equation:
If all of the nitric acid produced by this reaction is collected and then dissolved into 25mL of water, what would be the pH of the resulting solution?
(Hint: before you begin, think about which reactant is the limiting reagent.)
At 298 K and 1.00 atm, assume that 22 mL of NO gas reacts with 16 mL of oxygen gas and excess water to produce gaseous nitric acid according to the following equation: 2 NO (g) + 3/2 O2 (g) + H2O (I) -- > 2HNO3 (g) If all of the nitric acid produced by this reaction is collected and then dissolved into 25mL of water, what would be the pH of the resulting solution? (Hint: before you begin, think about which reactant is the limiting reagent.)Explanation / Answer
You first must think about the limiting reactant because the amount of gas that is produced is based on that. Assuming all NO gets converted to HNO3, then moles NO = HNO3 since the stoichiometric ratio is 2:2. Figure out moles of NO by plugging into P=nRT/V or PV = nRT.
You plug in the the pressure (P), the gas constant, R, which is .0821, you plug is total volume in Liters which is 22mL/1000 = 0.022 and you solve for "n" which is the moles. n = .0009 moles. Now, to find the pH, you need the concentration or molarity (M) of the H3O+. So you take that # of moles of acid, .0009, divide it by .025Liters of water to get your molarity: .0009/.025 = .036 molarity.
Since HNO3 is one of the six strong acids, it completely dissociates when put into water. Therefore, the concentration of the original acid is equivalent the concentration of your H3O+ molecules. To find the pH: -log([H3O+]) --> -log(.036) =
pH = 1.44
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