You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 93 mL of 1.00 M NaOH solution?

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Calculate the pH of a buffer formed by combining 50 mL of 0.10

M
NaOH and 50 mL of
0.15 M acetic acid. The pK
a
of acetic acid is 4.75.

To determine the pH of the solution after the addition of NaOH, we need to consider the reaction that occurs between acetic acid (CH3COOH) and sodium hydroxide (NaOH):

CH3COOH + NaOH → CH3COONa + H2O

Since NaOH is a strong base, it dissociates completely to provide hydroxide ions (OH-) in the solution. The acetic acid also dissociates partially in water, giving acetate ions (CH3COO-) and hydrogen ions (H+).

Since the initial concentrations of acetic acid and sodium acetate are given, we can use the Henderson-Hasselbalch equation to calculate the pH:

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

Where pKa is the acid dissociation constant of acetic acid.

The pKa value of acetic acid is 4.75, so we can substitute the given values into the equation:

[A-] = concentration of acetate ions = 0.20 M
[HA] = concentration of acetic acid = 0.30 M

pH = 4.75 + log(0.20/0.30)

Now, we need to consider the effect of adding 93 mL of 1.00 M NaOH solution to the buffer. By using the volume and concentration information, we can calculate the change in concentration of acetic acid and acetate ions.

First, let's calculate the moles of NaOH added:

Moles of NaOH = (volume in liters) × (concentration in mol/L)
= (93 mL / 1000 mL/L) × (1.00 mol/L)
= 0.093 mol

The stoichiometry of the reaction tells us that one mole of NaOH reacts with one mole of acetic acid. So, the change in concentration of acetic acid will be equal to the moles of NaOH added.

Change in [HA] = -0.093 mol

Since the initial concentration of acetic acid is 0.30 M, the final concentration will be:

[HA]final = [HA]initial + Change in [HA]
= 0.30 M - 0.093 mol / (500 mL + 93 mL)
= 0.207 M

Now, we need to consider the change in concentration of acetate ions. Since one mole of NaOH reacts with one mole of acetic acid, one mole of acetate ions is formed for every mole of NaOH added.

Change in [A-] = 0.093 mol

The final concentration of acetate ions will be:

[A-]final = [A-]initial + Change in [A-]
= 0.20 M + 0.093 mol / (500 mL + 93 mL)
= 0.207 M

Now, we can substitute the new concentrations of acetic acid and acetate ions into the Henderson-Hasselbalch equation to calculate the new pH:

pH = 4.75 + log(0.207/0.207)
= 4.75 + log(1)
= 4.75 + 0
= 4.75

Therefore, the pH of the solution after the addition of 93 mL of 1.00 M NaOH solution will be 4.75.

I worked this problem (for you) about two days ago. Here is a link.

http://www.jiskha.com/display.cgi?id=1302555258