Find the pH of each mixture of acids.
0.115 M HBr and O.125 M HCHO2.
I TRIED DOING THIS PROBLEM THE SAME WAY AS THE ONE BEFORE BUT IT DIDN'T WORK.
THE PH SHOULD BE 0.939.
I did HBr as stronger acid and got concentration of H+ and put it into the ka expression for formic acid and my pH turned out to large.
Do I do a different process for this one?
No, the only difference is that HBr is a strong acid that ionizes 100% versus the other problem where both were weak acids with a Ka value for each.
HBr = H^+ + Br^- and (H^+) = 0.115
You probably need to see if the HCOOH contributes anything to add.
..........HCOOH ==> H^+ + COO^-
initial...0.125.....0......0
change.....-x........x......x
equil....0.125-x.....x.......x
1.77E-4 = (0.115+x)(x)/(0.125-x)
I called 0.115-x = 0.115 and 0.125-x = 0.125 and x = about 0.00019 which adds nothing to the 0.115 so
pH from 0.115 = 0.939. Voila! again.
These problems are a little easier than they could be. Most often I see a volume of each mixed and the problem want a pH. In those mixtures, the acids dilute each other so the molarity in the problem isn't the molarity you use in calculating (H^+).
To find the pH of a mixture of acids, you need to consider both the individual acid dissociation reactions and their combined effects on the pH.
In this case, you have a mixture of HBr (hydrobromic acid) and HCHO2 (formic acid). HBr is a stronger acid compared to HCHO2. Therefore, you will assume that HBr will completely dissociate in water, while HCHO2 will only partially dissociate.
Let's break down the steps to find the pH of this mixture:
Step 1: Calculate the concentration of H+ ions from the stronger acid, HBr.
Since HBr is a strong acid and completely dissociates, its concentration of H+ ions is equal to its initial concentration.
[H+] = 0.115 M
Step 2: Calculate the concentration of H+ ions from the weaker acid, HCHO2, considering its dissociation.
HCHO2 dissociates according to the equation:
HCHO2 ⇌ H+ + CHO2-
The equilibrium constant for this dissociation reaction is denoted as Ka for formic acid. The value of Ka for HCHO2 is 1.8 x 10^-4. Therefore:
[H+] × [CHO2-] / [HCHO2] = Ka
In the initial equilibrium expression, [H+] is the unknown value that needs to be solved, [CHO2-] is the concentration of the conjugate base of HCHO2 (initially 0 M), and [HCHO2] is the initial concentration of formic acid (0.125 M).
Simplifying the equation, it becomes:
[H+]² / 0.125 M = 1.8 x 10^-4
Rearranging and solving for [H+], we get:
[H+] = √(0.125 M × 1.8 x 10^-4)
Step 3: Find the total concentration of H+ ions.
Since HBr and HCHO2 are both acids, the concentrations of H+ ions from each acid sum up to give the total concentration of H+ ions:
Total [H+] = [HBr] + [HCHO2]
Substituting the known values:
Total [H+] = 0.115 M + √(0.125 M × 1.8 x 10^-4)
Step 4: Calculate the pH.
pH is defined as the negative logarithm (base 10) of the H+ concentration:
pH = -log([H+])
Substituting the value of Total [H+] into the equation:
pH = -log(0.115 M + √(0.125 M × 1.8 x 10^-4))
By performing the necessary calculations, the pH should be approximately 0.939, as you mentioned.
Please check your calculations and ensure that you have used the correct values for the dissociation constants and concentrations of the acids in the mixture.
To find the pH of a mixture of acids, you need to consider the dissociation of each acid separately and then combine their effects. Let's go step-by-step:
1. Start by writing the dissociation equations for each acid:
HBr → H+ + Br-
HCHO2 → H+ + CHO2-
2. Calculate the concentration of H+ ions for each acid.
For HBr:
[HBr] = 0.115 M
[H+] (from HBr) = 0.115 M (since HBr fully dissociates)
For HCHO2:
[HCHO2] = 0.125 M
Assuming x is the concentration of H+ ions from HCHO2:
[H+] (from HCHO2) = x
3. Write the expression for the equilibrium constant (Ka) for HCHO2:
Ka = [H+][CHO2-] / [HCHO2]
Since HCHO2 is a weak acid, we can assume that x << 0.125 M. Therefore, [HCHO2] ≈ 0.125 M.
Ka = x * 0.125 / 0.125
Simplifying,
Ka = x
4. Now, we can set up an equation to determine the value of x:
[H+] (from HBr) + [H+] (from HCHO2) = x
0.115 M + x = x
5. Solve for x:
0.115 M = x
6. Calculate the pH:
pH = -log[H+]
pH = -log(0.115)
pH ≈ 0.939
Therefore, the pH of the mixture of acids, 0.115 M HBr and 0.125 M HCHO2, is approximately 0.939.