Answer:
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]
<em>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).</em>
Replacing:
pH = 4.74 + log₁₀ [0.600M] / [0.200M]
<em>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</em>
<em />
<h3>pH = 5.22</h3><h3 />
b. As the solution has a pH lower that 7.0, it is considered as a <em>acidic solution.</em>
<em></em>
c. When you add HCl to the buffer, the reaction is:
CH₃COO⁻ + HCl → CH₃COOH + Cl⁻
<em>Where acetate ion reacts with the acid producing acetic acid.</em>
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]
<h3>pH = 5.14</h3>
<em />
