Respuesta :
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
2.75 = pKa
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
1.78x10⁻³ = Ka
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
a. The value of Ka and pKa of the unknown acid should be 1.78x10⁻³ and 2.75.
b. It should be not relevant.
Calculation of the Ka and the pKa:
a. The neutralization of a weak acid, HA, with a base can determine Ka of the acid.
Since
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH,
So,
HA + XOH → H₂O + A⁻ + X⁺
In the case when you add 13 mL, the moles of HA should be half of the initial moles and, the other half, will be A⁻
Like
[HA] = [A⁻]
Now
pH = pKa + log₁₀ [A⁻] / [HA]
2.75 = pKa + log₁₀ [A⁻] / [HA]
Since
2.75 = pKa + log₁₀ 1
2.75 = pKa
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
1.78x10⁻³ = Ka
b. Since the initial concentration of the acid was not necessary. The only thing should be that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
