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 the specified volumes of acid added, we need to determine the species present in the solution and their concentrations at each step of the titration.
The compound B is a dibasic compound with two ionizable protons. Let's denote the original volume of the solution as V0 = 100.0 mL and the initial concentration of B as c0 = 0.100 M.
1. Initial volume (Va = 0 mL):
At this stage, no acid has been added, so the solution consists only of compound B. To find the concentration of B at this point, we can use the formula c = c0 * V0 / (V0 + Va):
c = 0.100 M * 100.0 mL / (100.0 mL + 0 mL) = 0.100 M.
Since B is a base, we can calculate its pOH initially:
pOH = pKb1 + log10(c)
pOH = 4.00 + log10(0.100)
pOH ≈ 4.00 + (-1) = 3.00
Finally, we can calculate the pH using the equation pH + pOH = 14:
pH + 3.00 = 14
pH ≈ 14 - 3.00 = 11.00
So, at Va = 0 mL, the pH of the solution is approximately 11.00.
Now, let's calculate the pH at the specified volumes of acid added (Va = 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL) using the same steps:
2. Va = 1 mL:
Concentration of B after adding 1 mL of HCl:
c = c0 * V0 / (V0 + Va)
c = 0.100 M * 100.0 mL / (100.0 mL + 1 mL) = 0.099 M
pOH = pKb2 + log10(c)
pOH = 8.00 + log10(0.099)
pOH ≈ 8.00 + (-1.00) = 7.00
pH ≈ 14 - 7.00 = 7.00
3. Va = 5 mL:
c = c0 * V0 / (V0 + Va)
c = 0.100 M * 100.0 mL / (100.0 mL + 5 mL) = 0.095 M
pOH = pKb2 + log10(c)
pOH = 8.00 + log10(0.095)
pOH ≈ 8.00 + (-1.022) = 6.978
pH ≈ 14 - 6.978 = 7.022
4. Va = 9 mL:
c = c0 * V0 / (V0 + Va)
c = 0.100 M * 100.0 mL / (100.0 mL + 9 mL) = 0.091 M
pOH = pKb2 + log10(c)
pOH = 8.00 + log10(0.091)
pOH ≈ 8.00 + (-1.041) = 6.959
pH ≈ 14 - 6.959 = 7.041
5. Va = 10 mL:
c = c0 * V0 / (V0 + Va)
c = 0.100 M * 100.0 mL / (100.0 mL + 10 mL) = 0.090 M
pOH = pKb2 + log10(c)
pOH = 8.00 + log10(0.090)
pOH ≈ 8.00 + (-1.047) = 6.953
pH ≈ 14 - 6.953 = 7.047
6. Va = 11 mL:
c = c0 * V0 / (V0 + Va)
c = 0.100 M * 100.0 mL / (100.0 mL + 11 mL) = 0.089 M
pOH = pKb2 + log10(c)
pOH = 8.00 + log10(0.089)
pOH ≈ 8.00 + (-1.053) = 6.947
pH ≈ 14 - 6.947 = 7.053
(Continued in the next message due to character limit)
To find the pH at different volumes of acid added, we need to consider the titration of the dibasic compound B with 1.00 M HCl. The compound B is a base, and when it reacts with HCl, it undergoes acid-base reactions.
Here's how to solve this question step by step:
1. Identify the species involved and their concentrations at each stage of the titration.
- Initially, the only species present is the dibasic compound B, which has a concentration of 0.100 M.
- As HCl is added, it reacts with the base B to form the respective acid and chloride salt. The concentration of B decreases while the concentration of the acid formed increases.
2. Determine the moles of base B initially present:
- Moles of B = initial concentration of B × volume of B (in L)
- Moles of B = 0.100 M × 0.100 L = 0.010 mol
3. Calculate the moles of acid added at each volume Va:
- Moles of acid added = HCl concentration × volume of acid added (in L)
Now, let's calculate the pH at each volume Va:
- At Va = 0 mL (initial condition):
- Moles of acid added = 1.00 M × 0 L = 0 mol
- Moles of base remaining = 0.010 mol
- Since B is a weak base, it will react partially with water to form OH- ions. The concentration of OH- can be calculated using the pKb1 value:
- Kb1 = [B][OH-] / [B(OH)]
- pKb1 = -log(Kb1)
- pKb1 = -log([B][OH-] / [B(OH)])
- Convert the pKb1 value to Kb1 value: Kb1 = 10^(-pKb1)
- [OH-] = sqrt(Kb1 × [B]) = sqrt(10^(-pKb1) × [B])
- [OH-] = sqrt(10^(-4.00) × 0.010 M) = 3.162 × 10^(-3) M
- pOH = -log([OH-]) = -(-log(3.162 × 10^(-3))) = 2.50
- pH = 14 - pOH = 14 - 2.50 = 11.50
- At other volumes Va (1, 5, 9, 10, 11, 15, 19, 20, and 22 mL):
- Calculate the moles of acid added.
- Calculate the moles of base remaining.
- Determine the concentration of OH- using the pKb2 value since the pOH is unknown:
- Kb2 = [B-][OH-] / [BH]
- pKb2 = -log(Kb2)
- pKb2 = -log([B-][OH-] / [BH])
- Convert the pKb2 value to Kb2 value: Kb2 = 10^(-pKb2)
- [OH-] = sqrt(Kb2 × [BH]) = sqrt(10^(-pKb2) × [BH])
- pOH = -log([OH-])
- pH = 14 - pOH
By following this process, you can calculate the pH for each volume Va and plot a graph of pH versus Va.
You have 10 problems wrapped up here in one post and I'm not about to spend the next hour doing them in step by step fashion for you; however, I can help you through it. First, how much do you know how to do? Which part confuses you.
1. The first thing you must do is to determine where the two equivalence points are (in mL acid added).
2. Va = 0. That's the pure B. Make an ICE chart and calculate H^+ and pH.
3. Va = ? at the first eq point. H^+ at that point is √k1k2.
3. All volumes between 0 and first eq pt is pH = pK2 + log base/acid
4. Va = ? at second eq pt. Make and ICE chart for this and proceed to H^+ and pH.
5. All volumes between 1st and 2nd eq pt is equation in 3; i.e., you have a buffer and the Henderson-Hasselbalch equation applies. pH = pK1 + log base/acid
6. Va = ? from 2nd eq pt onward is just excess HCl.