Consider the titration of the weak acid HA with NaOH. At what fraction of Ve does pH = pKa - 1? At what fraction of Ve does pH = pKa + 1? Calculate the pH at these two points plus Vb = 0, 1/2Ve, Ve and 1.2Ve, if 100 mL of 0.100 M anilinium bromide (pKa = 4.601) is titrated with 0.100 M NaOH. {Ve is Volume at equilibrium,Vb is volume of base}

Surely you mean Ve = volume at the equivalence point.

The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
If pH = pKa, then log B/A = 0 and B/A = 1
If pH = pKa + 1, then log B/A = 1 and B/A = 10
If pH = pKa -1, then log B/A = -1 and B/A = 0.1

..........HA + OH^- ==> A^- + H2O
initial....1.....0.......0......0
add.............x..............
change.....-x....-x......x.....x
equil.....1-x.....0......x.....x

The three cases above are done this way.
When B/A = 1, then [(x)/(1-x)] = 1
Solve for x and obtain x = 1-x so you are half way to the equivalence point.

When B/A = 10, then [(x)/(1-x)]=10
Solve for x = 0.90909 and that is the fraction to Ve.

When B/A = 0.1, then......

Set up an ICE chart and substitute moles into it for base and acid and use the HH equation to solve for pH for the specific points.
Post your work if you get stuck.

Ph8.3

To determine the fractions of Ve at which pH equals pKa - 1 and pKa + 1, we need to understand the behavior of weak acids and bases during titration.

1) At pKa - 1:
At this point, half of the weak acid is converted into its conjugate base, and half is still in its acidic form. Hence, the pH is equal to pKa - 1.

2) At pKa + 1:
At this point, half of the weak acid is still in its acidic form, and half has been neutralized to form its conjugate base. Hence, the pH is equal to pKa + 1.

Now let's calculate the pH at Vb = 0, 1/2Ve, Ve, and 1.2Ve:

Given data:
- Initial volume of anilinium bromide (HA) = 100 mL
- Initial concentration of anilinium bromide (HA) = 0.100 M
- pKa of anilinium bromide = 4.601
- Concentration of NaOH = 0.100 M

Step 1: Calculate the number of moles of anilinium bromide (HA) initially present.
Moles = (Initial volume in L) x (Initial concentration in M)
Moles of HA = (0.100 L) x (0.100 mol/L) = 0.010 mol

Step 2: Determine the volume of NaOH required for complete neutralization (Ve).
Moles of HA = Moles of NaOH
0.010 mol = (Ve in L) x (0.100 mol/L)
Ve = 0.010 L (10 mL)

Step 3: Calculating pH at different volumes of base (Vb).

a) Vb = 0:
Since no base has been added, the concentration of HA remains the same.
Moles of HA = 0.010 mol
Volume of solution = 100 mL (initial volume) + 0 mL (Vb)
Concentration of HA = (0.010 mol) / (0.100 L) = 0.100 M
pH = pKa + log([A-] / [HA]) = 4.601 + log(0.000 / 0.100) = 4.601

b) Vb = 1/2Ve:
Moles of HA neutralized = (0.010 mol) / 2 = 0.005 mol
Moles of remaining HA = 0.010 mol - 0.005 mol = 0.005 mol
Volume of solution = 100 mL (initial volume) + 0.005 L (Vb)
Concentration of HA = (0.005 mol) / (0.105 L) = 0.0476 M
pH = pKa + log([A-] / [HA]) = 4.601 + log(0.005 / 0.0476) = 3.454

c) Vb = Ve:
Moles of HA neutralized = 0.010 mol
Volume of solution = 100 mL (initial volume) + 0.010 L (Ve)
Concentration of HA = (0 mol) / (0.110 L) = 0 M (Complete neutralization of HA)
pH = -log[H+] = -log(0) (pH = infinity)

d) Vb = 1.2Ve:
Moles of HA neutralized = (0.010 mol) x 1.2 = 0.012 mol
Moles of remaining HA = 0.010 mol - 0.012 mol = -0.002 mol (negative indicates excess NaOH)
Volume of solution = 100 mL (initial volume) + 0.012 L (Vb) - 0.002 L (excess volume)
Concentration of HA = (-0.002 mol) / (0.110 L) = -0.0182 M (negative indicates excess NaOH)
pH = -log[H+] = -log(0) (pH = infinity)

To summarize, the pH at different volumes of base are as follows:
- Vb = 0: pH = 4.601
- Vb = 1/2Ve: pH = 3.454
- Vb = Ve: Complete neutralization (pH = infinity)
- Vb = 1.2Ve: Excess base (pH = infinity)

To determine the fraction of Ve at which pH equals pKa - 1 and pKa + 1 in the titration of a weak acid with NaOH, we need to understand the concept of titration and the relationship between pH and pKa.

Titration is a technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration (titrant) until the reaction is complete. In this case, we are titrating a weak acid (HA) with NaOH.

The pH of a solution is a measure of its acidity or basicity. The pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka), which relates to the strength of the acid. In this case, HA is a weak acid with a known pKa value.

To calculate the pH at different points of the titration, we will use the concepts of acid-base reactions and the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given as follows:

pH = pKa + log [A-]/[HA]

Where [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid.

Now, let's calculate the pH at different points of the titration:

1. Vb = 0: At the start of the titration, before any NaOH is added, the volume of base (Vb) is zero. Therefore, the pH can be calculated using the initial concentration of the weak acid. In this case, it is 0.100 M anilinium bromide with a pKa of 4.601.

Using the Henderson-Hasselbalch equation:
pH = pKa + log [A-]/[HA]
pH = 4.601 + log [0]/[0.100]
Since log(0) is undefined, the pH at Vb = 0 is not calculable.

2. Vb = 1/2Ve: To find the fraction of Ve when pH = pKa - 1, we need to calculate the pH at this point.
pH = pKa - 1
pH = 4.601 - 1
pH = 3.601

Using the Henderson-Hasselbalch equation:
pH = 3.601 + log [A-]/[HA]

The concentration of A- and HA at this point depends on the specific volume of NaOH added and can be calculated based on the stoichiometry of the reaction.

3. Ve: To find the fraction of Ve when pH = pKa + 1, we need to calculate the pH at this point.
pH = pKa + 1
pH = 4.601 + 1
pH = 5.601

Using the Henderson-Hasselbalch equation:
pH = 5.601 + log [A-]/[HA]

The concentration of A- and HA at this point depends on the specific volume of NaOH added and can be calculated based on the stoichiometry of the reaction.

4. Vb = 1.2Ve: To find the pH at this point, we can use the Henderson-Hasselbalch equation as described earlier for Vb = 1/2Ve.

Remember to adjust the concentrations of the weak acid and its conjugate base based on the volume ratios and the stoichiometry of the reaction.

Note: The specific calculations for concentrations at various points in the titration require knowledge of the stoichiometry of the reaction, the initial concentrations, and an understanding of how the volumes of acid and base relate. Based on the given information, these further calculations are beyond the scope of this response.