A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 x 10−5 as Ka for acetic acid. (a) What is the pH of the solution? (b) Is the solution acidic or basic? (c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer

a. pH = 5.22

b. Acidic.

c. pH = 5.14

Explanation:

a. It is possible to find the pH of a buffer using Henderson-Hasselbalch equation (H-H equation):

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka (For acetic acid =  4.74), [A⁻] is molar concentration of conjugate base (Acetate salt) and [HA] concentration of the weak acid (Acetic acid).

Replacing:

pH = 4.74 + log₁₀ [0.600M] / [0.200M]

You use the concentration of the acetic acid and sodium acetate because you're adding equal volumes, doing the ratio of the species the same

b. As the solution has a pH lower that 7.0, it is considered as a acidic solution.

c. When you add HCl to the buffer, the reaction is:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

Where acetate ion reacts with the acid producing acetic acid.

As you have 0.200L of the buffer, 0.100L are of the acetate ion and 0.100L of the acetic acid. Initial moles of both compounds and moles of HCl added are:

CH₃COO⁻: 0.100L ₓ (0.600mol / L) = 0.0600 moles

CH₃COOH: 0.100L ₓ (0.200mol / L) = 0.0200 moles

HCl: 3.0mL = 3x10⁻³L ₓ (0.034mol / L) =  0.00010 moles HCl

The moles added of HCl are the same moles you're consuming of acetate ion and producing of acetic acid. Thus, moles after the reaction are:

CH₃COO⁻: 0.0600 moles - 0.0001 moles = 0.0509 moles

CH₃COOH: 0.0200 moles + 0.0001 moles = 0.0201 moles

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.0509moles] / [0.0201moles]

pie chart because it is in percents ( % )