.050 mole of weak acid (HA) is dissolved in enough water to make 1.0 L of solution. The pH of this solution is 3.50. What is the value of Ka for this weak acid?
...........HA ==> H^+ + A^-
initial..0.05M......0....0
change....-x........x....x
equil....0.05-x.....x....x
Ka = (H^+)(A^-)/(HA)
Start with pH and convert that to H^+ (which is x in the above), subsubstitute into the Ka expression and solve for Ka.
To find the value of Ka for a weak acid, you need to use the equation for the acid dissociation:
HA ⇌ H+ + A-
In this equation, HA represents the weak acid, H+ represents the hydrogen ion, and A- represents the conjugate base.
The equilibrium constant expression for this reaction is:
Ka = [H+][A-] / [HA]
Given that the pH of the solution is 3.50, we can determine the concentration of H+ ions using the equation:
pH = -log[H+]
Therefore, [H+] = 10^(-pH)
Substituting the given pH value:
[H+] = 10^(-3.50) = 3.16 x 10^(-4) M
Since the concentration of A- is equal to the concentration of H+ (according to the balanced equation), [A-] = 3.16 x 10^(-4) M.
To find the concentration of HA, we use the fact that the total amount of the acid in the solution is 0.050 mole in 1.0 L:
[HA] = moles / volume = 0.050 mol / 1.0 L = 0.050 M
Substituting these values into the equilibrium constant expression:
Ka = (3.16 x 10^(-4) M)(3.16 x 10^(-4) M) / 0.050 M
Calculating this expression gives:
Ka = 2.0 x 10^(-6)
Therefore, the value of Ka for this weak acid is 2.0 x 10^(-6).