Question
If a solution of formic acid contains 0.2 M HCHO2, 0.006 M CHO2-, and 0.006 M H+,
What is the Ka of the acid?
What is the ph of the solution at the concentrations given?
What would happen to the equilibrium position if more CHO2- were added to the solution?
What is the Ka of the acid?
What is the ph of the solution at the concentrations given?
What would happen to the equilibrium position if more CHO2- were added to the solution?
Answers
DrBob222
........HCHO2 --> H^+ + CHO2^-
I........0.2......0......0
C.......-0.006..0.006...0.006
E......0.2-0.006..0.006..0.006
Ka = (H^+)(CHO2^-)(HCHO2)
Ka = (0.006)(0.006)/(0.2-0.006)
Ka = ?
pH = -log(H^+) = -log(0.006) = ?
HCHO2 --> H^+ + CHO2^-
Le Chatelier's Principle tells us that when we do something to a system in equilibrium that the reaction will shift so as to undo what we've done to it. Therefore, if we add CHO2^- the reaction must get rid of what we've added. It can do that by shifting to the left. In doing so (H^+) will decrease and (HCHO2) will increase.
I........0.2......0......0
C.......-0.006..0.006...0.006
E......0.2-0.006..0.006..0.006
Ka = (H^+)(CHO2^-)(HCHO2)
Ka = (0.006)(0.006)/(0.2-0.006)
Ka = ?
pH = -log(H^+) = -log(0.006) = ?
HCHO2 --> H^+ + CHO2^-
Le Chatelier's Principle tells us that when we do something to a system in equilibrium that the reaction will shift so as to undo what we've done to it. Therefore, if we add CHO2^- the reaction must get rid of what we've added. It can do that by shifting to the left. In doing so (H^+) will decrease and (HCHO2) will increase.