Based on the standard analytical practice during a titration of taking pH-volume measurements every 0.3 pH units, how many measurements should you take between the addition of 5mL and 7.5mL of titrant,

a) in the titration of a 10mL aliquot of standard 0.1500M TRIS base (Kb=1.2x10^-6) (in beaker) with 0.1000M HCl (in buret).

b) in the titration of a 10mL aliquot of 0.1500M NaOH (in buret) with standard 0.1000M HCL (in buret).

Hint: you need to calculate the pH at the 5 and 7.5mL points in each of the two titrations.

a) Well, let's start with the titration of TRIS base with HCl. Since TRIS base is a weak base and HCl is a strong acid, we need to calculate the pH at the 5mL and 7.5mL points.

At 5mL:
We have a 10mL aliquot of TRIS base (0.1500M), and 5mL of HCl (0.1000M) has been added. The moles of TRIS base will not change much, given its large concentration compared to HCl. So we approximate the concentration of TRIS base to be almost the same as the initial concentration. To calculate the pH, we can use the pOH equation:

pOH = -log(0.1500)
pOH ≈ 0.82391

pOH + pH = 14
pH ≈ 14 - 0.82391
pH ≈ 13.176

At 7.5mL:
Now, we have added 7.5mL of HCl (0.1000M). The moles of TRIS base have decreased, but we still have a considerable concentration. So, the approximation still holds. Let's calculate the pH using the same methodology:

pOH = -log(0.1500)
pOH ≈ 0.82391

pOH + pH = 14
pH ≈ 14 - 0.82391
pH ≈ 13.176

Since the pH remains the same at both points, there is no need to take any additional measurements between 5mL and 7.5mL. In this case, it's a pH stalemate!

b) Now, let's move on to the titration of NaOH with HCl. Here, NaOH is a strong base, and HCl is a strong acid.

At 5mL:
We have added 5mL of HCl (0.1000M) to a 10mL aliquot of NaOH (0.1500M). The reaction between them will result in a neutralization. Hence, the pH will be around 7.

At 7.5mL:
With 7.5mL of HCl (0.1000M) added, the pH will remain around 7 since the excess OH- ions from NaOH have already been neutralized.

Again, there is no need for any additional measurements between 5mL and 7.5mL. Seems like we've got ourselves a pH stalemate again! 😉

Remember, these calculations are approximations based on the given concentrations and simplified assumptions.

To determine the number of measurements needed between the addition of 5mL and 7.5mL of titrant, we need to calculate the pH at those points first.

a) Titration of TRIS base with HCl:
Step 1: Calculate the moles of TRIS base initially present in the 10mL aliquot.
Moles = Molarity × Volume
Moles of TRIS base = 0.1500 M × 10 mL = 1.5 mmol

Step 2: Determine the volume of HCl needed to reach the 5mL and 7.5mL points:
For each mL of HCl added, it reacts with one equivalent of TRIS base.

At 5mL:
Volume of HCl needed = 5 mL
Moles of HCl added = (0.1000 M)(5 mL) = 0.5 mmol
Total moles of TRIS base reacted = Initial moles of TRIS base - Moles of HCl added = 1.5 mmol - 0.5 mmol = 1.0 mmol

At 7.5mL:
Volume of HCl needed = 7.5 mL
Moles of HCl added = (0.1000 M)(7.5 mL) = 0.75 mmol
Total moles of TRIS base reacted = Initial moles of TRIS base - Moles of HCl added = 1.5 mmol - 0.75 mmol = 0.75 mmol

Step 3: Calculate the concentration of TRIS base and HCl at each point and use the Kb value to find the pH.
At 5mL:
Concentration of TRIS base = Moles of TRIS base / Volume of solution = 1.0 mmol / 15 mL = 0.067 M
Concentration of HCl = 0.1000 M (unchanged)
pOH = -log(Kb) + log([TRIS base])
pOH = -log(1.2x10^-6) + log(0.067) ≈ 4.92
pH = 14 - pOH ≈ 14 - 4.92 ≈ 9.08

At 7.5mL:
Concentration of TRIS base = Moles of TRIS base / Volume of solution = 0.75 mmol / 17.5 mL = 0.043 M
Concentration of HCl = 0.1000 M (unchanged)
pOH = -log(Kb) + log([TRIS base])
pOH = -log(1.2x10^-6) + log(0.043) ≈ 4.22
pH = 14 - pOH ≈ 14 - 4.22 ≈ 9.78

b) Titration of NaOH with HCl:
Step 1: Calculate the moles of NaOH initially present in the 10mL aliquot.
Moles = Molarity × Volume
Moles of NaOH = 0.1500 M × 10 mL = 1.5 mmol

Step 2: Determine the volume of HCl needed to reach the 5mL and 7.5mL points:
For each mL of HCl added, it reacts with one equivalent of NaOH.

At 5mL:
Volume of HCl needed = 5 mL
Moles of HCl added = (0.1000 M)(5 mL) = 0.5 mmol
Total moles of NaOH reacted = Initial moles of NaOH - Moles of HCl added = 1.5 mmol - 0.5 mmol = 1.0 mmol

At 7.5mL:
Volume of HCl needed = 7.5 mL
Moles of HCl added = (0.1000 M)(7.5 mL) = 0.75 mmol
Total moles of NaOH reacted = Initial moles of NaOH - Moles of HCl added = 1.5 mmol - 0.75 mmol = 0.75 mmol

Step 3: Calculate the concentration of NaOH and HCl at each point and use the Kw value to find the pH.
At 5mL:
Concentration of NaOH = Moles of NaOH / Volume of solution = 1.0 mmol / 15 mL = 0.067 M
Concentration of HCl = 0.1000 M (unchanged)
pOH = -log(Kw) + log([OH-])
pOH = -log(1.0x10^-14) + log(0.067) ≈ 5.18
pH = 14 - pOH ≈ 14 - 5.18 ≈ 8.82

At 7.5mL:
Concentration of NaOH = Moles of NaOH / Volume of solution = 0.75 mmol / 17.5 mL = 0.043 M
Concentration of HCl = 0.1000 M (unchanged)
pOH = -log(Kw) + log([OH-])
pOH = -log(1.0x10^-14) + log(0.043) ≈ 4.19
pH = 14 - pOH ≈ 14 - 4.19 ≈ 9.81

Now, to calculate the number of measurements, we need to find the difference in pH between 5mL and 7.5mL, and divide it by the step size of 0.3 pH units.

a) Difference in pH = pH at 7.5mL - pH at 5mL = 9.78 - 9.08 = 0.7
Number of measurements = Difference in pH / Step size = 0.7 / 0.3 ≈ 2.33

b) Difference in pH = pH at 7.5mL - pH at 5mL = 9.81 - 8.82 = 0.99
Number of measurements = Difference in pH / Step size = 0.99 / 0.3 ≈ 3.30

Therefore, in both cases, you should take approximately 2 and 3 measurements, respectively, between adding 5mL and 7.5mL of titrant.

To calculate the number of measurements needed between the addition of 5mL and 7.5mL of titrant in each titration, we first need to determine the pH at these points.

For both titrations, we'll use the Henderson-Hasselbalch equation to calculate the pH at each point using the pKa value. Since Kb is given for TRIS base in the first titration, we can calculate pKa as follows:

pKa = 14 - log10(Kb)

a) Titration of TRIS Base with HCl:
Given:
- 10mL aliquot of standard 0.1500M TRIS base with Kb = 1.2x10^-6
- 0.1000M HCl in the buret

1. Calculate pKa:
pKa = 14 - log10(1.2x10^-6)
pKa ≈ 12.92

2. Calculate pH at 5mL point:
Using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
Since we're at the starting point, [A-] = 0 and [HA] = 0.1500M (concentration of TRIS base).
pH = 12.92 + log10(0.1500/1)
pH ≈ 12.92 + log10(0.1500)

3. Calculate pH at 7.5mL point:
Using the Henderson-Hasselbalch equation again:
pH = pKa + log10([A-]/[HA])
At this point, we have [A-] = 0.1500M * (7.5mL / 10mL) = 0.1125M (concentration of TRIS base used), and [HA] = 0.1500M - 0.1125M = 0.0375M.
pH = 12.92 + log10(0.1125/0.0375)

Now that we have the pH values at the 5mL and 7.5mL points, we can determine the number of measurements needed between these two points. The pH changes by 0.3 units per measurement, so we can calculate the number of measurements using the following formula:

Number of measurements = (pH at 7.5mL - pH at 5mL) / 0.3

b) Titration of NaOH with HCl:
Given:
- 10mL aliquot of 0.1500M NaOH in the buret
- 0.1000M HCl in the buret

1. Calculate pH at 5mL point:
Since we're starting with NaOH, the solution is basic and pH is calculated based on concentration of OH- ions.
pOH = -log10([OH-])
[OH-] = 0.1500M * (5mL / 10mL) = 0.0750M
pOH = -log10(0.0750)
pH = 14 - pOH

2. Calculate pH at 7.5mL point:
Using the same method as before, pH at 7.5mL can be calculated with [OH-] = 0.1500M * (7.5mL / 10mL).

Now, using the calculated pH values for the 5mL and 7.5mL points, we can determine the number of measurements needed between these two points using the same formula as in a).