Question

Copper reacts with dilute nitric acid according to:
3 Cu(s) + 8 HNO3(aq) ? 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

If a copper penny weighing 3.045-g is dissolved in 37.23 mL of 3.750 M nitric acid and the resultant solution is diluted to 50.00 mL in a volumetric flask, what is the final concentration of NO3- in the solution?

Answer 1

This problem can be easily solved by going through the following steps

The given reaction is

3 Cu(s) + 8 HNO₃(aq) 3 Cu(NO3)₂(aq) + 2 NO(g) + 4 H₂O(l)

Step -1: Finding the moles of Cu and HNO₃

Moles of Cu = 3.045 g/63.55gmol^-1 = 0.04791 mol

Moles of HNO₃ = MxV = 3.750molL^-1x37.23 x 10^-3 L = 0.1396 mol

Step -2: Finding the limiting reagent

3 moles of Cu reacts with 8 moles of HNO₃ in the balanced chemical reaction.

moles of HNO₃ /moles of Cu = 0.1396 /0.04791 = 2.914 which is greater than the molar coefficient ratio 8/3.

Hence Cu is the limiting regent and is exhausted first leaving the excess of HNO₃ in the solution.

Hence the final concentration of NO₃^- will be due to both HNO3 and Cu(NO3)₂, beause both are completely ionised in the solution.

Step-3: Finding the moles of Cu(NO3)₂ formed and excess moles of HNO3

Moles of Cu(NO3)₂ formed = (3/3)x0.0479 mol = 0.0479 mol

Moles of HNO3 reacted with Cu = (8/3)x0.0479 = 0.1277 mol

moles of HNO3 remain unreacted in the solution = 0.1396 mol - 0.1277 mol = 0.0119mol

Step -3: Calculating the moles of NO₃^- in the final solution

moles of NO₃^- in 0.0479 mol of Cu(NO3)₂ = 2x0.0479 mol = 0.0958 mol

moles of NO₃^- in 0.0119mol of unreacted HNO3 = 1x0.0119mol = 0.0119 mol

Total moles of NO₃^- in the final solution = 0.0958 mol+0.0119 mol = 0.1077 mol

Step-4: Calculating the final concentration of NO₃-

Total volume of the solutin = 50.00mL = 50.00x10^-3 L

final concentration of NO₃- = 0.1077 mol/50.00x10^-3 L= 2.154M (answer)