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.

1. Use the Henderson-Hasselbalch equation. Substitute into the HH equation and calculate pH.

2. Set up an ICE chart using
.......CH3COOH + OH^- --> CH3COONa + H2O

and substitute the E line of the ICE chart into the HH equation and sole for the new pH.

For #1, would I have to multiply 500 mL to 0.050 moles and 0.20 moles?

Or can I just set it up like this below and solve for it?

pH=4.75 + log [0.20]/[0.050]

To determine the pH of 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, you need to consider the properties of a buffered solution.

A buffer solution is a mixture containing a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH when small amounts of acid or base are added. It maintains its pH due to the equilibrium between the weak acid and its conjugate base.

In this case, Acetic Acid (CH3COOH) is a weak acid, and Sodium Acetate (CH3COONa) is its conjugate base. The presence of both the weak acid and its conjugate base creates a buffer solution.

To calculate the pH of the buffer solution, you can use the Henderson-Hasselbalch equation:

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

Where:
pH is the desired pH of the solution
pKa is the acid dissociation constant of the weak acid
[A-] is the concentration of the conjugate base (Sodium Acetate)
[HA] is the concentration of the weak acid (Acetic Acid)

In this case, the pKa for Acetic Acid is given as 4.75. The concentration of Sodium Acetate is 0.20 moles in 500 mL, which can be converted to 0.40 moles per liter (since 500 mL = 0.5 L). The concentration of Acetic Acid is 0.050 moles in 500 mL, which can be converted to 0.10 moles per liter.

Now, substitute the values into the Henderson-Hasselbalch equation:

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 by dissolving 0.050 moles of Acetic Acid and 0.20 moles of Sodium Acetate in water and adjusting the volume to 500 mL is approximately 5.35.

Now, let's move on to the second question.

If 0.010 moles of NaOH (sodium hydroxide) is added to this buffer solution, you need to calculate the resultant pH.

Adding NaOH will react with the weak acid (Acetic Acid) in the buffer solution and convert it into its conjugate base (Sodium Acetate). This will disrupt the equilibrium and change the pH of the solution.

To calculate the resultant pH, you need to consider the reaction and use the new concentrations of the weak acid and its conjugate base.

The balanced equation for the reaction between NaOH and Acetic Acid is:

CH3COOH + NaOH → CH3COONa + H2O

For every 1 mole of NaOH, 1 mole of Acetic Acid will react to form 1 mole of Sodium Acetate.

In this case, you are adding 0.010 moles of NaOH. Since the reaction has a 1:1 stoichiometry, this means that 0.010 moles of Acetic Acid will react.

The initial concentration of Acetic Acid was 0.050 moles in 500 mL, which can be converted to 0.10 moles per liter. Subtracting the 0.010 moles that react, the new concentration of Acetic Acid becomes 0.090 moles per liter.

The concentration of Sodium Acetate remains the same since it is not consumed or formed in the reaction.

Now, you can use the Henderson-Hasselbalch equation to calculate the new pH:

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

pH = 4.75 + log(0.40/0.090)

pH = 4.75 + log(4.44)

pH = 4.75 + 0.648

pH ≈ 5.40

Therefore, the resultant pH of the solution after adding 0.010 moles of NaOH to the buffer solution is approximately 5.40.