Asked by Savannah S.
What is the original molarity of a solution of weak acid whose Ka= 3.5 x 10^ -5 and pH is 5.20...
I'm lost, please help!
I'm lost, please help!
Answers
Answered by
DrBob222
Call the weak acid HA.
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
pH = 5.20 = -log(H^+)
Solve for (H^+), substitute into Ka expression above for H^+ as well as for A^-. For (HA) substitute X-(H^+) and solve for X. I think you can neglect the (H^+) in the X-(H^+) and simply call X - (H^+) = X
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
pH = 5.20 = -log(H^+)
Solve for (H^+), substitute into Ka expression above for H^+ as well as for A^-. For (HA) substitute X-(H^+) and solve for X. I think you can neglect the (H^+) in the X-(H^+) and simply call X - (H^+) = X