Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0110 M solution. The pH of the resulting solution is 2.69 . Calculate the K a for the acid.

Answer 1

Answer:

Ka for the monoprotic acid is 4.65 ×10⁻⁴ M.

Explanation:

Chemical reaction: HA(aq) ⇄ A⁻(aq) + H⁺(aq).

c(monoprotic acid) = 0.0110 M.

pH = 2.69

[A⁻] = [H⁺] = 10∧(-2.69).

[A⁻] = [H⁺] = 0.00204 M; equilibrium concentration.

[HA] = 0.0110 M - 0.00263 M.

[HA] = 0.00837 M.

Ka = [A⁻]·[H⁺] / [HA]. 

Ka = (0.00204 M)² / 0.00837 M.

Ka = 0.000465 M

= 4.65 ×10⁻⁴ M.

Answer 2

Answer:

0.0004655.

Explanation:

Ka is an acronym that stand for acid dissociation constant. This constant is used in the representation of an equilibrium constant when it concerns the dissociation of acids.

Assuming HC is an acid, as given in the question a monoprotic acid. Then, the dissociation is given below as:

HC + H2O <-----------> H3O^+ + C^-.

Therefore, the acid dissociation constant can be represented mathematically by using the formula below;

Ka = [ H30^+] [ C^-] / [HC].

(We are coming to this formula later).

So, in the question above we are given the value for the pH to be 2.69 and the formula for Calculating the value of pH is given below;

pH = - log [H^+].

When we make the concentration of Hydrogen ion the subject of the formula we have;

[H^+] = 10^-pH.

[H^+] = 10^- 2.69.

[H^+] = 0.002042 M.

Now that we know the concentration of the hydrogen ion,then we will have to go back to the equation for the acid dissociation above.

We have that; at time, t= 0 the concentration of the acid = 0.0110 M and the concentration of H3O^+ and C^- are both zero.

Then, after time,t = t the concentration of the acid = 0.0110 - x M and the concentration of H3O^+ and C^- are both x.

Therefore, going back to our formula for acid dissociation constant, we will have;

Ka = [x] [x] / [ 0.0110 - x].

Where our x has the value of 0.002042.

Ka= (0.002042)^2/ 0.008958 .

Ka= 0.000004169764/ 0.008958.

Ka= 0.0004655.

Because of the small value for ka it means that we are dealing with a weak acid.