Consider the titration of 30.0 mL of 0.100 M NH3 with 0.150 M HCl. calculate the pH after the following volumes of titrant have been added:(a) 0mL (b) 10.0 mL (c) 19.5 mL (d) 20.0 mL (e) 20.5 mL (f) 30 mL
To calculate the pH during a titration, we need to determine the number of moles of acid (HCl) and base (NH3) at each point. We can then use the stoichiometry of the reaction between HCl and NH3 to determine the concentrations of the resulting species. Finally, we can use the equation for the ionization of water to calculate the pH.
Let's go through the calculations step by step for each of the given volumes.
(a) 0 mL:
Since no titrant has been added yet, the NH3 concentration remains at 0.100 M. NH3 is a weak base, so we can assume that none of it has reacted. Therefore, the pH can be calculated based on the initial NH3 concentration. To calculate the pH of a weak base, we need the Kb value of NH3, which is 1.8 × 10^-5.
NH3 + H2O ⇌ NH4+ + OH-
Initial NH3 concentration = 0.100 M
Initial NH4+ = 0 M
Initial OH- = 0 M
Since there is no NH4+ and OH-, we can calculate the concentration of OH- using the Kb expression:
Kb = [NH4+][OH-]/[NH3]
1.8 × 10^-5 = (x)(x)/(0.100)
x^2 = 1.8 × 10^-5 × 0.100
x = √(1.8 × 10^-5 × 0.100) = 0.013
OH- concentration = 0.013 M
Since water is a neutral solution, the concentration of H+ (which is equal to H3O+) will be the same as OH-. Therefore, the pH at this point is:
pOH = -log10(OH-) = -log10(0.013) = 1.89
pH = 14 - pOH = 14 - 1.89 = 12.11
(b) 10.0 mL:
We need to calculate the amount of NH3 remaining and the amount of NH4+ and OH- formed.
The balanced equation for the reaction is:
NH3 + HCl ⇌ NH4+ + Cl-
Initial NH3 concentration = 0.100 M
Initial NH4+ = 0 M
Initial Cl- = 0 M
NH3 is a weak base, so it undergoes partial ionization. Let's assume x moles of NH3 have reacted.
x moles of NH4+ and Cl- are formed.
0.100 - x moles of NH3 remain unreacted.
Since the reaction between NH3 and HCl is 1:1, the number of moles of NH3 reacted is equal to the number of moles of HCl added (in this case, 10.0 mL of 0.150 M HCl). We can calculate the number of moles of NH3 reacted as follows:
moles of HCl = Molarity × volume
moles of NH3 reacted = moles of HCl added = (0.150 M) × (0.0100 L) = 0.0015 moles
Since NH3 and NH4+ are present in equal amounts after the reaction, the concentration of NH4+ is also 0.0015 M.
The volume of the solution remains constant at 30 mL throughout the titration, so the NH3 concentration is now:
NH3 concentration = (0.100 M - 0.0015 mol)/(0.030 L) = 0.0985 M
Using the same procedure as in part (a), we can calculate the pOH and pH at this point.
(c) 19.5 mL:
Following the same steps as in part (b), we calculate the new concentrations:
NH3 concentration = (0.100 M - 0.0015 mol)/(0.030 L) = 0.0975 M
(d) 20.0 mL:
Following the same steps as in part (b), we calculate the new concentrations:
NH3 concentration = (0.100 M - 0.0015 mol)/(0.030 L) = 0.097 M
(e) 20.5 mL:
Following the same steps as in part (b), we calculate the new concentrations:
NH3 concentration = (0.100 M - 0.0015 mol)/(0.030 L) = 0.0965 M
(f) 30 mL:
At this point, the titration is complete. A total of 0.150 moles of HCl has been added, which is equivalent to the initial moles of NH3. Therefore, all of the NH3 has reacted, and only NH4+ is present. The concentration of NH4+ is:
NH4+ concentration = (0.150 mol)/(0.030 L) = 0.005 M
Since NH4+ is a weak acid, we can calculate the pOH and pH using the Ka expression for NH4+ and the equation for the ionization of water.
I hope this helps you in calculating the pH at each point during the titration. Let me know if you need more explanation on any step!