4. Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.37-M HCl to 320. mL of each of the following solutions.
Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.
c) 0.115 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =
My work:
Before Mixing:
HC2H3O2 + H2O --> (C2H3O2)- + (H3O)+
36.8 mmol 0 0
- x + x + x
1.8e-5 = x^2 / (36.8 -x)
x = .02572813 = [H3O+]
-log(.02572813) = 1.59 (this is not correct)
Nope. You used millimols in this and you should use molarity. Ka is 1.8E-5, yes, but that is Ka = (H3O^+)(Ac^-)/(HAc) and you must substitute M and not mmols. If acetic acid is HAc, then
.......HAc+ + H2O ==> Ac^- + H3O^+
I......0.15..........0......0
C......-x.............x......x
E.....0.115-x.........x.......x
This gives you (H3O^+) = ? in mols/L which is concentration and you can use that to calculate pH. .
Thank you so much! That makes sense.
I'm sure you picked it up but that's 0.115 M from the problem and not 0.15. My typo.
To calculate the pH of the solution and the change in pH caused by adding HCl, you need to consider the dissociation of HC2H3O2 (acetic acid) and the reaction with HCl.
1. Find the initial pH of the HC2H3O2 solution before adding HCl:
To calculate the initial pH of HC2H3O2, you'll need to use the equilibrium expression for the dissociation of acetic acid:
HC2H3O2 + H2O ⇌ C2H3O2- + H3O+
In this case, you already have the equation for the dissociation of HC2H3O2:
HC2H3O2 + H2O ⇌ C2H3O2- + H3O+
36.8 mmol 0 0
- x + x + x
The equilibrium constant expression for this reaction is given by:
Ka = [C2H3O2-][H3O+]/[HC2H3O2]
Since the initial concentration of HC2H3O2 is 0.115 M, the concentration of C2H3O2- and H3O+ at equilibrium will be much smaller, so you can assume that the (36.8 - x) term in the denominator will remain approximately constant.
Substituting the values into the expression, you get:
1.8e-5 = x^2 / (36.8 - x)
You already solved this equation for x and obtained x = 0.02572813 M (the concentration of H3O+). To find the pH, take the negative logarithm of the H3O+ concentration:
pH = -log[H3O+]
pH = -log(0.02572813)
pH ≈ 1.59 (after rounding to two decimal places)
2. Calculate the pH after adding HCl:
By adding 10.0 mL of 2.37 M HCl to 320. mL of the HC2H3O2 solution, you are diluting the HC2H3O2 solution and adding H3O+ ions from the HCl.
To find the new concentration of H3O+, you can use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial volume (V1) is 320. mL, the final volume (V2) is 330. mL (after adding 10.0 mL of HCl), and the final concentration (C2) is unknown.
Rearranging the equation, you get:
C2 = (C1V1) / V2
C2 = (0.115 M * 320. mL) / 330. mL
C2 ≈ 0.111 M
Now, you can calculate the pH of the solution after adding HCl:
pH = -log[H3O+]
pH = -log(0.111)
pH ≈ 1.95 (after rounding to two decimal places)
3. Calculate the change in pH:
To find the change in pH, subtract the initial pH from the final pH:
pH change = pH after mixing - pH before mixing
pH change = 1.95 - 1.59
pH change ≈ 0.36 (after rounding to two decimal places)
Therefore, the pH before mixing is approximately 1.59, the pH after mixing is approximately 1.95, and the change in pH caused by adding HCl is approximately 0.36 pH units.