Question

At 295 K and 1.00 atm, assume that 20. mL of NO gad reacts with 20. mL of oxygen gas and excess water to produce gaseous 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 collected and then dissolved into 15. mL of water, what would be the pH of the resulting solution?

Answer 1

1. Calculate the no. of moles of O2 and NO using PV=nRT

For O2 = n= PV/RT = 1atm*20mL/0.0821*295 = 0.826mole

For NO, n= PV/RT = 1atm*20mL/0.0821*295 = 0.826mole

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

2mole of NO = 1.5mole of O2 that means 0.826 mole of NO require 1.5*0.826/2 = 0.619mole of O2

Hence NO is the limiting reagent i.e., 0.826-0.619=0.207mole lesser than required

In 0.0826 moles of NO reacts with 0.619 mole of O2 to form 2*0.619mole of HNO3=1.238mol of HNO3 will be formed

Molarity of HNO3 = 1.238/15mL= 0.08M=[H+]

pH = -log[H+] =-log0.08= 1.09