The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added and make a graph of pH versus Va; Va=0, 1, 5, 9, 10, 11, 15, 19, 20, and 22mL.
To calculate the pH at each volume of acid added (Va), we will use the Henderson-Hasselbalch equation:
pH = pKa + log ([base] / [acid])
For a given volume of acid added, we will first determine the moles and concentrations of each species present, and then use the equation above to determine the pH. Since we are adding HCl, we will use the pKa values, which are the negative logarithms of the base dissociation constants, Kb:
pKa = -log(Kb)
Since we have two pKb values (pKb1=4.00 and pKb2=8.00), we will also have two pKa values:
pKa1 = -log(10^(-4.00)) = 4.00
pKa2 = -log(10^(-8.00)) = 8.00
Here are the steps for each volume of acid added:
Va = 0 mL ("before titration"):
- No acid has been added yet, so only the dibasic compound B is present.
- The concentration of B is given as 0.100 M.
- Because B is a dibasic compound, it will be associated with two conjunct species: HB+ and H2B^2+.
- We will consider that B will react only once at this point (to form HB+), so we calculate the concentrations using the pKa1 value:
pH = 4.00 + log ([B] / [HB+])
0.100 M B and 0 mL of 1.00 M HCl are mixed, which means that no reaction occurs.
Therefore, the pH at Va=0 mL is:
pH = 4.00 + log (0.100 / 0) ≈ 4.00
Va = 1 mL:
- 1.00 mL of 1.00 M HCl is added, which means 0.001 moles of HCl are added.
- B will react with HCl to form HB+:
B + HCl → HB+ + Cl-
- From the stoichiometric ratio, 0.001 moles of B will react with 0.001 moles of HCl to form 0.001 moles of HB+.
- The total volume will now be 101 mL, and so our new concentrations will be as follows:
[B] = (0.100 - 0.001) moles / 101 mL ≈ 0.0009802 M
[HB+] = 0.001 moles / 101 mL ≈ 0.000009902 M
Now we can calculate the pH using the pKa1 value:
pH = 4.00 + log (0.0009802 / 0.000009902) ≈ 5.63
_va = 5 mL:
- 5.00 mL of 1.00 M HCl is added, which means 0.005 moles of HCl are added.
As Va increases, it is necessary to follow a similar procedure to find the concentration of all species (B, HB+, and H2B^2+) at every volume of HCl added.
You can follow the same procedure to calculate the pH for the other given volumes of acid added (Va = 9, 10, 11, 15, 19, 20, and 22 mL). Then, we will have the pH values at each volume.
Finally, to create a graph, you can use any graphing software or even graph paper. Plot the pH values on the y-axis and the volume of acid added (Va) on the x-axis. Your graph will give you a visual representation of the pH changes during the titration of the dibasic compound B with 1.00 M HCl.
To find the pH at different volumes of acid added, we need to calculate the concentration of B at each point based on the volume of acid added. We can use the Henderson-Hasselbalch equation to do this.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where pH is the negative logarithm of the concentration of H+ ions, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is HCl and the base is compound B. The conjugate base of B is represented by B-. The acid dissociation constants for their reactions are given as pKb1 = 4.00 and pKb2 = 8.00.
Let's calculate the pH at each volume of acid added as follows:
1. For Va = 0 mL, there is no acid added. The concentration of B remains at 0.100 M, and the concentration of B- is 0. Therefore, the pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ([B-]/[B])
pH = 4.00 + log (0/0.100)
pH = 4.00
So, at Va = 0 mL, the pH is 4.00.
2. For Va = 1 mL, 1 mL of the 1.00 M HCl is added. Since the volume of the solution increases to 101 mL, the final concentration of B can be calculated:
Final volume of B = 0.100 L + 0.001 L = 0.101 L
Final concentration of B = 0.100 M × (0.100 L / 0.101 L) = 0.09901 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([B-]/[B])
pH = 4.00 + log ([B-]/0.09901)
To find the concentration of B-, we can use the equation:
[B-] = [B] × (Ka / [H+])
Where Ka is the acid dissociation constant for the reaction B- + H+ ⇌ HB. For the first dissociation:
Ka = 10^(-pKa1) = 10^(-4.00) = 0.0001
[B-] = 0.09901 M × (0.0001 / [H+])
[H+] = 1.00 M (since we added 1 mL of 1.00 M HCl)
[B-] = 0.09901 M × (0.0001 / 1.00 M)
[B-] = 9.90 × 10^(-7) M
Now, substituting the value of [B-] into the Henderson-Hasselbalch equation:
pH = 4.00 + log (9.90 × 10^(-7) / 0.09901)
pH ≈ 4.02
So, at Va = 1 mL, the pH is approximately 4.02.
You can now follow these steps to calculate the pH at each volume of acid added according to the given values of Va.
To find the pH at different volumes of acid added, you need to consider the dissociation of the dibasic compound B and the reaction with HCl.
Dibasic compounds have two ionizable protons, so they can dissociate in two steps:
B + H2O ⇌ BH+ + OH-
BH+ + H2O ⇌ B2+ + H3O+
Given the pKb1 and pKb2 values, we can find the corresponding pKa1 and pKa2 values using the relationship:
pKb = 14 - pKa
Thus, pKa1 = 14 - pKb1 = 14 - 4.00 = 10.00
And pKa2 = 14 - pKb2 = 14 - 8.00 = 6.00
At the initial volume (Va = 0 mL), there is no acid added, and the only species present is B. We can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([A-]/[HA])
Since B is a base, [A-] (conjugate base) is 0, and [HA] (acid) is equal to the initial concentration of B.
At Va = 0 mL:
pH = pKa1 + log([B]/[BH+])
= 10.00 + log([B]/[BH+])
Now, let's calculate the pH at the given volumes of acid added:
1. Calculate [B] and [BH+] at each volume (Va):
- Consider the initial volume of B, 100.0 mL:
[B] = 0.100 M
[BH+] = 0 M (no acid added)
- At Va = 1 mL:
[B] = 0.100 M - (1 mL / 1000 mL/L) * (0.100 M / 100 mL/L) = 0.0999 M
[BH+] = (1 mL / 1000 mL/L) * (0.100 M / 100 mL/L) = 1.0 x 10^-6 M
- Repeat this process for each volume of acid added to obtain the values of [B] and [BH+].
2. Substitute these values into the Henderson-Hasselbalch equation to calculate the pH at each volume.
3. Plot the pH values against Va to create the graph.
Performing these calculations for all the given volumes (Va = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL) will give you the pH values corresponding to each volume.