Question

Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL

Answer 1

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

Where 1 mole of HX reacts with 1 mole of NaOH

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

Where [H⁺] = [X⁻] because comes from the same equilibrium

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

pKa = 5.01