1.) Quantify the pH for a solution prepared by dissolving 0.050 moles of Acetic Acid (CH3COOH) and 0.20 moles of Sodium Acetate (CH3COONa) in water and adjusting the volume to 500 mL. The pka for Acetic Acid (CH3COOH) is 4.75.

2.) Should 0.010 moles of NaOH be added to this buffer solution, quantify the resultant pH for the solution.

How many grams of KMnO4 should be used to prepare 2.00L of a 0.500M solution? Can you show your work how to do it?

To calculate the pH of a solution, we need to determine the concentration of the hydrogen ion (H+) present in the solution. In the case of a buffer solution, we can use the Henderson-Hasselbalch equation:

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

Where:
- pH is the desired pH value
- pKa is the acid dissociation constant of the weak acid
- [A-] is the concentration of the conjugate base (in this case, CH3COO- or acetate ions)
- [HA] is the concentration of the weak acid (in this case, CH3COOH or acetic acid)

Let's calculate the pH for the solution prepared in question 1:

1.) Quantify the pH for the solution prepared by dissolving 0.050 moles of Acetic Acid (CH3COOH) and 0.20 moles of Sodium Acetate (CH3COONa) in water and adjusting the volume to 500 mL. The pKa for Acetic Acid (CH3COOH) is 4.75.

First, we need to determine the concentration of CH3COOH and CH3COO- present in the solution.

CH3COOH:
moles = 0.050 mol
volume = 500 mL = 0.5 L
concentration of CH3COOH = moles/volume = 0.050 mol / 0.5 L = 0.10 M

CH3COO-:
moles = 0.20 mol
volume = 500 mL = 0.5 L
concentration of CH3COO- = moles/volume = 0.20 mol / 0.5 L = 0.40 M

Now, we will use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log ([A-]/[HA])
pH = 4.75 + log (0.40/0.10)
pH = 4.75 + log (4)
pH ≈ 4.75 + 0.602
pH ≈ 5.35

Therefore, the pH of the solution prepared in question 1 is approximately 5.35.

2.) Should 0.010 moles of NaOH be added to this buffer solution, quantify the resultant pH for the solution.

To calculate the pH after adding NaOH, we need to consider the reaction that occurs between NaOH and the acetic acid in the buffer solution. NaOH is a strong base and will react with acetic acid (CH3COOH) to form sodium acetate (CH3COONa) and water. The balanced equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

Since NaOH reacts completely, the reaction will consume an equal amount of acetic acid in moles. Therefore, we need to subtract the moles of NaOH added (0.010 mol) from the moles of acetic acid initially present (0.050 mol) to get the remaining moles of acetic acid.

Moles of acetic acid after the reaction = Moles of acetic acid initially present - Moles of NaOH added
Moles of acetic acid after the reaction = 0.050 mol - 0.010 mol
Moles of acetic acid after the reaction = 0.040 mol

Now we can calculate the concentration of acetic acid and use the Henderson-Hasselbalch equation to find the new pH:

Concentration of acetic acid = Moles of acetic acid after the reaction / Volume of the solution
Concentration of acetic acid = 0.040 mol / 0.5 L
Concentration of acetic acid = 0.080 M

Using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])
pH = 4.75 + log (0.40/0.080)
pH = 4.75 + log (5)
pH ≈ 4.75 + 0.699
pH ≈ 5.45

Therefore, the pH of the solution after adding 0.010 moles of NaOH is approximately 5.45.

See my response above.