In a 0.05M solution of a weak monoprotic acid,[H+]=.0018.What is its Ka?
............HA ==> H^+ + A^-
initial....0.05....0......0
change..-0.0018...0.0018..0.0018
equil..0.05-0.0018..0.0018..0.0018
Ka = (H^+)(A^-)/(HA)
Substitute and solve for Ka.
To find the Ka (acid dissociation constant) of a weak monoprotic acid, we need to use the concentration of the acid and the concentration of the hydrogen ions ([H+]).
The equation for the dissociation of the weak monoprotic acid can be written as follows:
HA ⇌ H+ + A-
Assuming that the initial concentration of the weak acid (HA) is 0.05 M, and the concentration of the hydrogen ions ([H+]) is 0.0018 M, we can set up an equilibrium expression using the given values:
Ka = [H+][A-]/[HA]
Now we need to solve for [A-]. Since the weak acid dissociates in a 1:1 ratio, the concentration of [A-] is the same as the concentration of [H+], which is 0.0018 M.
Substituting the values, we get:
Ka = (0.0018 M)(0.0018 M)/ 0.05 M
Simplifying the equation:
Ka = (0.00324 M^2)/0.05 M
Finally, calculating the value of Ka:
Ka = 0.0648 M/ M
Therefore, the Ka of the weak monoprotic acid is 0.0648 M/ M.