enough of a monoprotic acid is dissolved in water to produce a 0.0111M solution. The pH of the resulting solution is 2.62. Calculate the Ka for the acid.
pH = 2.62 = -log (H^+); therefore,
(H^+) 0.0024
............HA ==> H^+ + A^-
I........0.0111.....0.....0
C.........-x........x......x
E.......0.0111-x....x......x
x = 0.0024
Ka = (H^+)(A^-)/(HA)
(H^+) = 0.0024 = (A^-)
(HA) = 0.0111-0.0024 = ?
Substitute and solve for Ka.
To determine the Ka for the monoprotic acid in the given solution, we can use the pH and the concentration of the acid solution.
The pH of a solution is related to the concentration of H+ ions in the solution by the equation: pH = -log[H+]. Since the solution is acidic, the concentration of H+ ions must be higher than the concentration of OH- ions.
In this case, the pH of the solution is given as 2.62, which means [H+] = 10^(-pH) = 10^(-2.62) M.
In a monoprotic acid, the concentration of the acid ([HAc]) and the concentration of H+ ions are equal. Therefore, [HAc] = [H+].
We are given that the concentration of the acid solution ([HAc]) is 0.0111 M.
Using the formula for Ka, which is the acid dissociation constant, we have:
Ka = [H+][A-]/[HAc]
Since the monoprotic acid only dissociates into H+ and A- (conjugate base), the concentration of A- is equal to the concentration of H+.
Therefore, we can rewrite the equation as:
Ka = [H+]^2/[HAc]
Substituting the known values into the equation, we have:
Ka = (0.0111)^2 / 0.0111
Simplifying the expression, we get:
Ka = 0.0111
Therefore, the Ka value for the monoprotic acid is 0.0111.