At 295 K and 1.00 atm, assume that 22.0 mL of NO gas reacts with 22.0 mL of oxyg

Question

At 295 K and 1.00 atm, assume that 22.0 mL of NO gas reacts with 22.0 mL of
oxygen gas and excess water to produce gaseous nitric acid according to the
following equation:
2NO(g)+3/2O2(g)+H2O(l)----->2HNO3(g)
If all of the nitric acid produced by this reaction is collected and then dissolved
into 15.0 mL of water, what would be the pH of the resulting solution?
(Hint: before you begin, think about which reactant is the limiting reagent.)

Please help! If someone could help answer and show the steps that would be great. I'm confused!

Explanation / Answer

4NO(g)+3O2(g)+ 2H2O(l)----->4HNO3(g)

4 mol NO = 3 mol O2 = 4 mol HNO3

no of mol of NO taken = PV/RT

                      = (1*22*10^-3/(0.0821*295))

             = 0.0009 mol

no of mol of O2 taken = (1*22*10^-3/(0.0821*295))

       = 0.0009 mol

limiting reactant = NO

no of mol of HNO3 produced = 0.0009 mol

concentration of HNO3 = n/V = 0.0009/15*1000 = 0.06 M

pH = -log(H+)

    = -log(0.06)

    = 1.22

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