Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when exactly 13 mL of base had been added was 2.75. You notice that the concentration of the unknown acid was 0.1 M.

a. What are the Ka and the pKa of your unknown acid? (Show work)
b. After finishing your calculations, you discover that the actual concentration of the unknown acid was 0.35 M. How would this affect your calculation of the pKa?

Answer 1

a. 1.78x10⁻³ = Ka

2.75 = pKa

b. It is irrelevant.

Step-by-step 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.

Answer 2

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.