A 0.100 mol quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA?
HA ==>H^+ + A^-
Ka = (H^+)(A^-)/((HA)
You find (H^+) from the pH. The (A^-) = (H^+) and (HA) = 0.1-(H^+)
Solve for Ka.
To find the value of Ka for the acid HA, we need to first understand the equilibrium reaction between the acid and water. In this case, the acid HA will dissociate into its conjugate base A- and a hydrogen ion H+ when it reacts with water.
The balanced equation for this reaction is:
HA + H2O ⇌ A- + H3O+
Now, let's define the initial and equilibrium concentrations of the acid and its conjugate base.
Given:
Initial concentration of HA = 0.100 mol
Volume of the solution = 1.00 L
Since HA dissociates into A- and H+ in a 1:1 ratio, the initial concentration of both A- and H+ are zero.
At equilibrium, let's assume that:
Concentration of HA = (0.100 - x) mol
Concentration of A- = x mol
Concentration of H+ = x mol (since they are formed in a 1:1 ratio)
The value of x represents the change in concentration that occurs due to the dissociation of HA.
Now, let's use the given pH value to calculate the concentration of H3O+ ions in the solution.
pH = -log[H3O+]
3.75 = -log[H3O+]
Taking the antilog of both sides:
[H3O+] = 10^(-3.75)
Now, we can use the fact that the concentration of H3O+ is equal to the concentration of H+ to solve for x.
[H+] = x = 10^(-3.75) mol
The equilibrium expression for the acid dissociation constant Ka is:
Ka = [A-][H+]/[HA]
Plugging in the values we found:
Ka = (x)(x)/(0.100 - x)
Substituting x = 10^(-3.75):
Ka = (10^(-3.75))(10^(-3.75))/(0.100 - 10^(-3.75))
Calculating this expression will give you the value of Ka for the acid HA.