A 40.0 mL sample of 0.150 M Ba(OH)2 is titrated with 0.400 M HNO3. Calculate the pH after the addtion of the following volumes of acid, and plot the pH versus millimeters of HNO3 added.
a. 0.00 mL
b. 10.2 mL
c. 19.9 mL
d. 30.0 mL
e. 39.8 mL
First determine where the equivalence point is (how many mL HNOe are required).
At zero mL, calculate pH from pure Ba(OH)2.
All points from zero to the equivalence point, use an ICE chart to determine how much Ba(OH)2 remains, then determine pH from that. Remember to take into account the dilution of the base.
At equivalence point, the pH is the pH of pure water since this is a strong acid/strong base titration.
All points after the equivalence point, use an ICE chart to determine how much HNO3 is in excess and determine pH from that. Post your work if you get stuck.
To calculate the pH after the addition of different volumes of acid in the titration, we need to consider the reaction that occurs between barium hydroxide (Ba(OH)2) and nitric acid (HNO3). These two compounds react to form barium nitrate (Ba(NO3)2) and water (H2O). This reaction can be represented by the following balanced equation:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O
We can use stoichiometry to determine the amount of reactants and products involved in the reaction. In this case, we have a 1:2 mole ratio between Ba(OH)2 and HNO3, which means that every mole of Ba(OH)2 reacts with 2 moles of HNO3.
Now, let's calculate the moles of Ba(OH)2 initially present in the solution:
Molarity (M) = moles/volume (L)
0.150 M x 0.040 L = 0.006 moles of Ba(OH)2
Since the reaction has a 1:2 mole ratio between Ba(OH)2 and HNO3, we need twice the moles of HNO3 to react completely. Therefore, we have:
2 x 0.006 = 0.012 moles of HNO3
a. 0.00 mL:
Since no acid has been added yet, the moles of HNO3 remain at 0.012. To find the pH, we need to calculate the concentration of H3O+ (hydronium ions) in the solution. Since the reaction between HNO3 and Ba(OH)2 is a strong acid-base reaction, we assume complete dissociation of the acid.
Concentration (M) = moles/volume (L)
0.012 moles / 0.040 L = 0.30 M
Using the definition of pH:
pH = -log[H3O+]
pH = -log(0.30) ≈ 0.52
b. 10.2 mL:
To determine the moles of HNO3 after adding 10.2 mL, we need to determine the moles of HNO3 neutralized by the Ba(OH)2 solution:
Moles of HNO3 neutralized = (Volume of acid added/Total volume of solution) x Moles of HNO3 initially present
Moles of HNO3 neutralized = (10.2 mL/50.2 mL) x 0.012 moles = 0.00243 moles
Remaining moles of HNO3 = Moles of HNO3 initially present - Moles of HNO3 neutralized
Remaining moles of HNO3 = 0.012 moles - 0.00243 moles = 0.00957 moles
To find the pH, we need to calculate the concentration of H3O+ in the solution:
Concentration (M) = moles/volume (L)
Concentration = 0.00957 moles / 0.040 L = 0.24 M
pH = -log(0.24) ≈ 0.62
c. 19.9 mL:
Following the same procedure as above, we find:
Moles of HNO3 neutralized = (19.9 mL/50.2 mL) x 0.012 moles = 0.00478 moles
Remaining moles of HNO3 = Moles of HNO3 initially present - Moles of HNO3 neutralized
Remaining moles of HNO3 = 0.012 moles - 0.00478 moles = 0.00722 moles
Concentration = 0.00722 moles / 0.040 L = 0.18 M
pH = -log(0.18) ≈ 0.74
d. 30.0 mL:
Using the same method described earlier:
Moles of HNO3 neutralized = (30.0 mL/50.2 mL) x 0.012 moles = 0.00718 moles
Remaining moles of HNO3 = 0.012 moles - 0.00718 moles = 0.00482 moles
Concentration = 0.00482 moles / 0.040 L = 0.12 M
pH = -log(0.12) ≈ 0.92
e. 39.8 mL:
Again, using the same steps:
Moles of HNO3 neutralized = (39.8 mL/50.2 mL) x 0.012 moles = 0.00957 moles
Remaining moles of HNO3 = 0.012 moles - 0.00957 moles = 0.00243 moles
Concentration = 0.00243 moles / 0.040 L = 0.061 M
pH = -log(0.061) ≈ 1.21
To plot the pH versus millimeters (mm) of HNO3 added, you can simply plot the pH values calculated for each volume of acid added (0.00 mL, 10.2 mL, 19.9 mL, 30.0 mL, 39.8 mL) on the y-axis against the corresponding volumes of acid added on the x-axis.