You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.36. You determine that the concentration of the unknown acid was 0.3535 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid? (Leave the unit box empty, as pKa has no unit.)

4.27

Explanation:

Let's consider the dissociation of a generic monoprotic acid.

HA(aq) → H⁺(aq) + A⁻(aq)

The pH is 2.36. The concentration of H⁺ is:

pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -2.36 = 4.37 × 10⁻³ M

We know that the concentration of the acid Ca = 0.3535 M. We can find the acid dissociation constant using the following expression.

[H⁺] = √(Ca × Ka)

Ka = [H⁺]²/Ca

Ka = (4.37 × 10⁻³)²/0.3535

Ka = 5.40 × 10⁻⁵

The pKa is:

pKa = -log Ka = -log 5.40 × 10⁻⁵ = 4.27

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons

b) nuclear ive learned that in a pedtrodic table