Question

In this question you are going to calculate the data that could be used to construct a titration curve (pH versus volume of titrant added) for the titration of an aqueous ammonia (NH3; weak base, pKB = 4.7) cleaning solution with a strong acid (HCl). When strong acid is added to an aqueous solution, it immediately reacts and yields H3O + ions; thus the titration is described by the reaction: NH3(aq) + H3O + (aq)  NH4 + + H2O(l) K = 2.0x109

The cleaning solution is produced by adding 0.02 moles of NH3 to 1.0 L of pure H2O, which you later titrate with 1.0 M HCl titrant solution. To simplify matters we’ll assume all activity coefficients = 1.0 for all species in this problem. Utilize the ICE (INITIAL, CHANGE, EQUILIBRIUM)Table Method (show your work) to determine the pH of the system when various amounts of the strong acid (titrant) have been added. Note: when you solve the quadratic function, make sure to select the root that is positive, and if you have two positive roots, use the root that represents a realistic change; i.e. the change isn’t greater than what was physically added to the system or greater than the concentration of reactant it reacts with.

a. Calculate pH of solution when 0 mL of titrant have been added (only the NH3 has been added)

a1. Calculate pH of solution when 10 mL of titrant have been added

Answer 1

a) When no HCl has been added to NH3, at that point, the pH of the solution is calculated by following method:

NH₃ + H₂O ----> NH₄⁺ + OH^-

concentration of NH3 = [NH3] = moles of NH3/volume of water = 0.02/1.0 = 0.02 M

pKb of NH3 = 4.7

Kb = 10^-4.7 = 2.0*10^-5

ICE table:

	[NH3]	[NH4+]	[OH-]
Initial	0.02 M	0	0
Change	-x	+x	+x
Equilibrium	0.02-x	x	x

Kb = [NH4+][OH-]/[NH3]

2.0*10^-5 = x.x/(0.02-x)

As NH3 is weak base, so 0.02-x = 0.02

2.0*10^-5 = x.x/0.02

4.0*10^-7 = x²

x = 6.3*10^-4

[OH-] = x = 6.3*10^-4

pOH = -log[OH-] = -log(6.3*10^-4) = 3.2

pH = 14-3.2 = 10.8

a1) Moles of NH3 present = 0.02 mol

When 10 ml of HCl is added, moles of HCl added = 1.0*0.010 = 0.01 mol

The reaction is:

NH3 + H3O+ ----> NH4+ + H2O, K = 2.0*10⁹

Moles of NH3 unreacted = 0.02-0.01 = 0.01 mol

Total volume = 1.0 + 0.010 = 1.010 L

concentration of NH3 = moles/volume = 0.01/1.010 = 0.0099 M

[NH4+] = moles of HCl reacted/total volume = 0.1/1.010 = 0.0099

ICE table:

	[NH3]	[NH4+]	[OH-]
Initial	0.0099 M	0.0099	0
Change	-x	+x	+x
Equilibrium	0.0099-x	0.0099+x	x

Kb = [NH4+][OH-]/[NH3]

2.0*10^-5 = x.(0.0099+x)/(0.0099-x)

2.0*10^-5 = x

x = 2.0*10^-5

[OH-] = x = 2.0*10^-5

pOH = -log[OH-] = -log(2.0*10^-5) = 4.7

pH = 14-4.7 = 9.3