Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0129 M solution. The pH of the resulting solution is 2.65 . Calculate the Ka for the acid.

Answer 1

Ka = 4.70x10⁻⁴M

Step-by-step explanation:

The general dissociation of a weak acid, HX, is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is written as:

Ka = [H⁺] [X⁻] / [HX]

Where [] represents the molar concentration in equilibrium of each specie.

The equilibrium is reached when X of HX is dissociate in X H⁺ and X X⁻, that is:

[HX] = 0.0129M - X

[H⁺] = X

[X⁻] = X

As pH = -log [H⁺]:

10^-pH = [H⁺] = X = 2.239x10⁻³M

Solving:

[HX] = 0.0129M - 2.239x10⁻³M = 0.01066M

[H⁺] = 2.239x10⁻³M

[X⁻] = 2.239x10⁻³M

Ka = [H⁺] [X⁻] / [HX]

Ka = [2.239x10⁻³M] [2.239x10⁻³M] / [0.01066M]

Ka = 4.70x10⁻⁴M