Formic acid (HCHO2) is a component of bee stings. It is a weak acid and dissociates according to the
equilibrium shown below:
HCHO2 <===> H+ + CHO2- Kc = 1.8 x 10-4
For a 0.5 M solution of formic acid, we wish to determine the equilibrium concentration of H+.
Complete the ICE table below. Show your calculation for the reaction quotient, Qc, and use that to decide on the signs (+ or -) for the changes in the reactant and products. Briefly explain why the signs for the changes in the reactant and products are the way they are.
Determine the equilibrium concentration of H+. (Hint: is the equilibrium constant small enough to use
the "x is small" approximation?)
[HCHO2](M) [H+](M) [CHO2-](M)
Initial 0.5 0 0
Change
Equilibrium
Initial
[HCHO2](M)= 0.5
[H+](M) = 0
[CHO2-](M)= 0
Change = ?
Equilibrium = ?
..............HCHO2 ==> H^+ + CHO2^-
initial.......0.5.......0......0
change.........-x........x......x
equil........0.5-x.......x......x
Substitute the ICE chart values into K
Ka = (H^+)(CHO2^-)/(HCHO2) and solve for (H^+) (which in the ICE chart is x).
To complete the ICE table, we need to determine the changes in concentration for each species in the equilibrium reaction.
The equilibrium constant expression for this reaction is:
Kc = [H+][CHO2-] / [HCHO2]
Let's start filling out the table:
[HCHO2](M) [H+](M) [CHO2-](M)
Initial 0.5 0 0
Change
Equilibrium
Now, let's calculate the changes in concentration and fill in the "Change" column.
Since the equilibrium constant expression is in terms of molar concentrations, we need to express the changes in terms of "x", which represents the change in the concentration of H+.
[HCHO2](M) [H+](M) [CHO2-](M)
Initial 0.5 0 0
Change -x +x +x
Here's an explanation for the signs of the changes:
- The concentration of HCHO2 will decrease because it is a reactant in the forward reaction. So the change in [HCHO2] is -x.
- The concentration of H+ will increase because it is a product in the forward reaction. So the change in [H+] is +x.
- The concentration of CHO2- will also increase because it is a product in the forward reaction. So the change in [CHO2-] is +x.
Finally, let's complete the ICE table with the equilibrium concentrations.
[HCHO2](M) [H+](M) [CHO2-](M)
Initial 0.5 0 0
Change -x +x +x
Equilibrium 0.5-x x x
Now, we can use the given equilibrium constant (Kc = 1.8 x 10^-4) to determine the equilibrium concentration of H+.
Kc = [H+][CHO2-] / [HCHO2]
1.8 x 10^-4 = x*x / (0.5 - x)
Since the equilibrium constant is small (less than 0.01), it is safe to use the "x is small" approximation. This means we can approximate 0.5 - x as 0.5, as the change in x is small compared to the initial concentration.
1.8 x 10^-4 = x*x / 0.5
Cross-multiplying:
0.9 x 10^-4 = x^2
Taking the square root of both sides:
x = √(0.9 x 10^-4)
x ≈ 0.0095
Therefore, the equilibrium concentration of H+ is approximately 0.0095 M.