Find the pH of mixture of acids.

0.185 M in HCHO2 and 0.225 M in HC2H3O2

Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get the right answer help please?
Answer is pH of 2.19.

It you cannot determine the concentration of the mixture by forming a double replacement became you form products that are the same as the reacting because you are mixing two acids.

You must recognize that this is a mixture of two weak acids; i.e., formic acid and acetic acid.

I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH.

Calculate the H^+ from the strong acid, then add the H^+ from the weak acid.
Formic acid first since it is the stronger.
.........HCOOH ==> H^+ + HCOO^-
initial..0.185.....0......0
change...-x........x......x
equil..0.185-x.....x......x

Ka = (H^+)(HCOO^-)/(HCOOH)
Solve for H^+. This is what acts as the common ion (remember Le Chatelier's Principle). This causes the acetic acid to ionize less than it would if were just acetic acid solution.

........CH3COOH ==> H^+ + CH3COO^-
initial..0.225.......0......0
change....-x........x........x
equil...0.225-x......x........x

Ka acet acid = (H^+)(CH3COO^-)/(CH3COOH)
Substitute TOTAL H^+ into the Ka expression for CH3COOH. That will be about 0.00572 from HCOOH from the above calculation plus x from this ionization. Solve for x, add this (H^+) to the 0.00572 from HCOOH, then convert to pH. I obtained pH = 2.19

I got concentration of HCOOH rxn to be 5.77X10^-3. And the CH3COOH rxn to have concentration of 2.01X10^-3. I added those together to get 7.78X10-3. What do you mean add to ka expression? I keep getting 1.35 now.

To find the pH of a mixture of acids, you can apply the principles of acid dissociation and use the Henderson-Hasselbalch equation. Let's break down the steps to solve the problem:

1. Identify the dissociation equations for the two acids:
HCHO2 (formic acid) dissociates as follows: HCHO2 ⇌ H+ + CHO2-
HC2H3O2 (acetic acid) dissociates as follows: HC2H3O2 ⇌ H+ + C2H3O2-

2. Write down the initial concentrations of the acids:
[HCHO2] = 0.185 M
[HC2H3O2] = 0.225 M

3. Create an ICE (Initial, Change, Equilibrium) chart for each acid, assuming x as the amount of H+ produced from each acid:
- For HCHO2:
Initial: HCHO2: 0.185 M H+: 0 M CHO2-: 0 M
Change: HCHO2: -x M H+: +x M CHO2-: +x M
Equilibrium: HCHO2: 0.185 - x M H+: x M CHO2-: x M

- For HC2H3O2:
Initial: HC2H3O2: 0.225 M H+: 0 M C2H3O2-: 0 M
Change: HC2H3O2: -x M H+: +x M C2H3O2-: +x M
Equilibrium: HC2H3O2: 0.225 - x M H+: x M C2H3O2-: x M

4. Determine the expression for the equilibrium constant (Ka) for each acid:
Ka = ([H+][CHO2-])/[HCHO2] for HCHO2
Ka = ([H+][C2H3O2-])/[HC2H3O2] for HC2H3O2

5. Plug in the known values and simplify the equations:
[H+][CHO2-] = Ka × [HCHO2]
[H+][C2H3O2-] = Ka × [HC2H3O2]

6. Since the [H+][CHO2-] and [H+][C2H3O2-] concentrations are equal to [H+], the equations can be combined:
[H+]^2 = (Ka1 × [HCHO2] + Ka2 × [HC2H3O2]) / ([CHO2-] + [C2H3O2-])

7. Substitute the given values and solve for [H+]:
[H+]^2 = ((Ka1 × 0.185) + (Ka2 × 0.225)) / (x + x)

8. Simplify the equation and isolate [H+]:
[H+]^2 = ((Ka1 × 0.185) + (Ka2 × 0.225)) / 2x

9. Solve for x (by assuming [H+] = 10^-pH) using the quadratic formula:
10^-pH = sqrt(((Ka1 × 0.185) + (Ka2 × 0.225)) / 2x)

10. Substitute the Ka values for HCHO2 (1.8 × 10^-4) and HC2H3O2 (1.74 × 10^-5), and solve for x:
10^-pH = sqrt(((1.8 × 10^-4 × 0.185) + (1.74 × 10^-5 × 0.225)) / 2x)

11. Once you solve for x, you can calculate the concentration of H+ using the equilibrium expression:
[H+] = x

12. Finally, calculate the pH using the equation pH = -log[H+].

By following these steps, you should be able to calculate the pH of the mixture of acids and obtain the correct answer, which is pH = 2.19.

FIRST U should find whether there is reaction when they are mixed,and then what is the concentration of the mixture.next,find the concentration of H+ ion,finally,use this formular to solve your problem

pH =-log[concentration of H+ ]