formic acid is a weak acid with a ka value of 1.8x10^-4 at 25 celsius a solution with a concentration of .424M formic acid is prepared in a laboratory.

1) Write the ionization equation for an aqeous solution of formic acid
2)determine the ph of the solution prepared above.
i think i got part 1 but i don't know how to do part 2 :l

Set up an ICE chart.

.........HCOOH + H2O --> H3O^+ + HCOO^-
I.......0.424.............0........0
C........-x...............x........x
E.....0.424-x.............x........x

Substitute the E line into the Ka expression for HCOOH and solve for x = (H3O^+), then convert that to pH.

which one is acid and which one is base?

To determine the pH of the solution prepared with formic acid (HCOOH), you need to use the ionization equation and apply the concept of weak acid dissociation. Let's break down the steps for part 2:

1) Write the ionization equation for an aqueous solution of formic acid:
The ionization equation for formic acid (HCOOH) in water is as follows:
HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq)
This equation represents the dissociation of formic acid into a hydrogen ion (H⁺) and a formate ion (HCOO⁻) in aqueous solution.

2) Determine the initial concentrations:
The concentration of formic acid (HCOOH) in the solution is given as 0.424 M. Since formic acid is a weak acid, we can assume that the dissociation is less than 5%, so we can neglect the concentration of H⁺ and consider the initial concentration of formic acid as the initial concentration of HCOOH.

3) Write the expression for the acid dissociation constant (Ka):
Ka is given as 1.8x10^-4. The Ka expression for the ionization of formic acid can be written as:
Ka = [H⁺][HCOO⁻] / [HCOOH]

4) Use the ICE table to set up the equilibrium expression:
Let "x" represent the concentration of H⁺ and HCOO⁻ ions formed. Since the initial concentration of formic acid (HCOOH) is 0.424 M and it is a weak acid, we can assume that "x" represents the change in concentration for both H⁺ and HCOO⁻ ions.

Initial: HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq)
Concentration: 0.424 M x x

At equilibrium, the concentration of HCOOH will be reduced by "x" and the concentrations of H⁺ and HCOO⁻ ions will be "x".

5) Substitute the equilibrium concentrations in the Ka expression:
Ka = [H⁺][HCOO⁻] / [HCOOH]
1.8x10^-4 = x*x / (0.424-x)

6) Solve the quadratic equation:
Rearrange the equation to:
1.8x10^-4 = x² / (0.424-x)

Solve for x by quadratic equation methods or make the simplifying assumption that x is much less than 0.424, so "x" can be neglected compared to 0.424.

Assuming x << 0.424:
1.8x10^-4 = x² / 0.424

Multiply both sides by 0.424:
0.0000768 = x²

Take the square root of both sides:
x = 0.00876 M

Notice that the assumption x << 0.424 is valid because 0.00876 M is significantly smaller than 0.424 M.

7) Calculate the pH:
To find the pH, use the relationship between pH and H⁺ ion concentration:
pH = -log[H⁺]

Since H⁺ concentration is equal to x in this case, we can substitute the value of x:
pH = -log(0.00876)
pH ≈ 2.06

Therefore, the pH of the solution prepared with formic acid (0.424 M) is approximately 2.06.

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