Answer:
a. pH → 1.77
b. pH → 4.27
Explanation:
Malonic acid is a dyprotic acid. It releases two protons:
H₂A + H₂O → H₃O⁺ + HA⁻ Ka1
HA⁻ + H₂O → H₃O⁺ + A⁻² Ka2
Let's find the first pH:
We expose the mass balance:
Ca = [HA] + [HA⁻] + [A⁻²] = 0.2 M
We can not consider the [A⁻²] so → Ca = [HA] + [HA⁻] = 0.2M
As the acid is so concentrated, we can not consider the HA- so:
Ca = [HA] = 0.2 M
Charge balance → [H⁺] = [HA⁻] + [OH⁻]
H₂A + H₂O → H₃O⁺ + HA⁻ Ka1
Ka = H₃O⁺ . HA⁻ / H₂A
We need the HA⁻ value to put on the charge balance. We re order the Ka expression:
HA⁻ = Ka . H₂A / H₃O⁺ (notice that H₃O⁺ = H⁺)
We replace: H⁺¨ = Ka . H₂A / H⁺
(H⁺)² = Ka . Ca
H⁺ = √(Ka . Ca) We determine Ka from pKa → 10²'⁸⁴⁷ = 1.42×10⁻³
H⁺ = √(1.42×10⁻³ . 0.2) = 0.016866
- log [H⁺] = pH → - log 0.016866 = 1.77
b. NaHA is the salt from the weak acid, where the HA⁻ works as an amphoterous, this means that can be an acid or a base:
HA⁻ + H₂O ⇄ A⁻² + H₃O⁺ Ka₂
HA⁻ + H₂O ⇄ H₂A + OH⁻ Kb2
There is a formula than can predict the pH, so now we need to compare the Ka₂ and Kb₂ data.
Ka₂ = 10⁻⁵'⁶⁹⁶ = 2.01×10⁻⁶
Kb₂ = 1×10⁻¹⁴ / 1.42×10⁻³ = 7.04×10⁻¹²
So Ka₂ > Kb₂. In conclussion the pH will be acid.
[H⁺] = √(Ka1 . Ka2) → [H⁺] = √ 1.42×10⁻³ . 2.01×10⁻⁶ = 5.34×10⁻⁵
- log 5.34×10⁻⁵ = pH → 4.27
