Consider the titration of 100.0 mL of 0.100 M HC2H3O2 by 0.100 M KOH for the next five questions (Ka for HC2H3O2 = 1.8 x 10-5). Calculate all pH values to two decimal places.
1) Calculate the pH after 0.0 mL of KOH has been added.
2) Calculate the pH after 50.0 mL of KOH has been added.
3) Calculate the pH after 75.0 mL of KOH has been added.
4) Calculate the pH at the equivalence point.
5) Calculate the pH after 150.0 mL of KOH has been added.
To answer these questions, we need to understand the process of a titration and how the pH of the solution changes as we add the titrant.
1) To calculate the pH after adding 0.0 mL of KOH, we are still dealing with only the acidic solution of HC2H3O2. Since no KOH has been added yet, the solution remains unchanged. Since we have a weak acid (HC2H3O2), we can use the equation for the pH of a weak acid:
pH = -log[H3O+]
Here, we need to calculate the concentration of [H3O+], which can be done using the initial concentration of the acid and the dissociation constant (Ka). From the given information, we know the concentration of HC2H3O2 is 0.100 M.
HC2H3O2 + H2O ⇌ H3O+ + C2H3O2-
Using the Ka value (1.8 x 10^-5) and the initial concentration of HC2H3O2, we can calculate the concentration of [H3O+]. To do this, we use the equation:
Ka = [H3O+][C2H3O2-] / [HC2H3O2]
Since HC2H3O2 is a weak acid and can be assumed to fully dissociate in water, we can simplify this equation to:
Ka = [H3O+]^2 / [HC2H3O2]
Now, we can rearrange the equation to solve for [H3O+]:
[H3O+]^2 = Ka * [HC2H3O2]
[H3O+] = √(Ka * [HC2H3O2])
Substituting the values, we get:
[H3O+] = √(1.8 x 10^-5 * 0.100)
Calculating this value, we find:
[H3O+] ≈ 0.004
Taking the negative logarithm of this concentration, we can calculate the pH:
pH = -log(0.004)
Calculating this value, we find:
pH ≈ 2.40
Therefore, the pH after adding 0.0 mL of KOH is approximately 2.40.
2) To calculate the pH after adding 50.0 mL of KOH, we need to determine how much of the acid and base are left after the reaction. Since the stoichiometry of the reaction between HC2H3O2 and KOH is 1:1, 50.0 mL of KOH is equivalent to 50.0 mL of HC2H3O2.
This means that we have used up 50.0 mL of the original 100.0 mL HC2H3O2 solution, leaving us with 50.0 mL of HC2H3O2 remaining. We need to calculate the new concentration of HC2H3O2.
Initial concentration of HC2H3O2 = 0.100 M
Initial volume of HC2H3O2 = 100.0 mL
Volume of KOH added = 50.0 mL
Using the equation for dilution:
[C2H3O2-] = [HC2H3O2] * (V1 / V2)
Plugging in the values, we get:
[C2H3O2-] = 0.100 * (50.0 / 100.0)
Calculating this value, we find:
[C2H3O2-] = 0.050
Now, we can calculate the concentration of [H3O+] using the dissociation of HC2H3O2. Since we have a weak acid and its salt, we use the Henderson-Hasselbalch equation:
pH = pKa + log ([C2H3O2-] / [HC2H3O2])
Plugging in the values, we get:
pH = -log(1.8 x 10^-5) + log (0.050 / 0.050)
Calculating this value, we find:
pH ≈ 4.40
Therefore, the pH after adding 50.0 mL of KOH is approximately 4.40.
You can follow similar steps to calculate the pH values for the next questions.
l don't understand
I think you already know how to do these. What you must do to know how to calculate each is to recognize what you have in the solution. Here is the equation and the millimoles added. Start with 100 mL x 0.100 M = 10 millimoles HC2H3O2.
...................HC2H3O2 + KOH ==> KC2H3O2 + H2O
I.....................10................0.................0................0
add...................................x.................................
C...................-x...............-x...................+x
E....................10-x.............0.....................x
1) Calculate the pH after 0.0 mL of KOH has been added.
2) Calculate the pH after 50.0 mL of KOH has been added.
3) Calculate the pH after 75.0 mL of KOH has been added.
4) Calculate the pH at the equivalence point.
5) Calculate the pH after 150.0 mL of KOH has been added.
1) No KOH is added so the pH is determined by the ionization of the 0.1 M acid.
2) Adding 50 mL KOH addes 50 x 0.1 = 5 mmoles KOH which forms 5 mmoles KC2H3O2. You should recognize this as a buffered solution and solve accordingly.
3) 75.00 mL KOH adds 7.5 mmoles KOH. Just a rerun of (2).
4) The equivalence point. You should know that you have no HC2H3O2 and no NaOH. You have a solution of the salt so the pH will be determined by the hydrolysis of the salt. I've worked some of those problems for you. The pH of NaC2H3O2 solution are in the neighborhood of 8.2 or 8.3 so you will know if you're in the ball park or not.
5). After 150 mL NaOH that's 150 x 0.100 = 15 mmoles NaOH which is more than enough to neutralize all of the acid and have some NaOH left.
The (OH^-) = (NaOH) and pH from there
Post your work if you get stuck on any of these.