Question

Enough of a monoprotic acid is dissolved in water to produce a 1.64 M solution. The pH of the resulting solution is 2.82 . Calculate the Ka for the acid.

Answer 1

Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Where Ka is:

Ka = [H⁺] [X⁻] / [HX]

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

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

[HX] = 1.64M - 1.51x10⁻³M = 1.638M

[H⁺] = 1.51x10⁻³M

[X⁻] = 1.51x10⁻³M

And Ka is:

Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]

Ka = 1.39x10⁻⁶