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

Answer:

The correct answer is A) 1.6 x 10-6

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= $\frac{x^{2} }{0.10 - x}$

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= $\frac{(3.98 x 10^{-4} )^{2} }{(0.10 - (3.98 x 10^{-4}) }$

Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶