Question

A 0.10 M solution of a weak monoprotic acid has a pH of 3.40 at 25°C. What is the acid-ionization

constant, Ka, for this acid?
A) 1.6 x 10-6
B) 4.0 x 10-4
C) 3.4 x 10-5
D) 1.2 x 10-3
E) 1.8 x 10-7

Answer 1

The correct answer is A) 1.6 x 10-6

Step-by-step explanation:

A weak monoprotic acid has the following dissociation equilibrium. At the beggining (t=0), the concentration of the monoprotic acid (HA) is equal to 0.10 M and the concentration of the ions H⁺ and A⁻ is zero (no dissociation). At a time t, dissociation occur and there is x concentration of H⁺ and A⁻ which is given by the dissociation constant Ka.

HA(aq) ⇄ H⁺(aq) + A⁻(aq)

t=0 0.10 M 0 0

t -x +x +x

eq 0.10 M-x x x

Ka=
`(x^(2) )/(0.10 - x)`

As the pH is 3.40, we can calculate the concentration of both H⁺ and A⁻, as follows:

pH= - log (conc H⁺)= -log x

⇒ x =
`10^(-3.40)`= 3.98 x 10⁻⁴

Now we introduce x in the previous equation to calculate Ka:

Ka=
`((3.98 x 10^(-4) )^(2) )/((0.10 - (3.98 x 10^(-4)) )`

Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶