Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq). Refer to table 1 and table 2.
(a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
Using ICE i found to be 2.46
The rest I do not know how to set up or solve...
(b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?
(c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?
mL
(d) Calculate the pH at the halfway point.
(e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?
mL
(f) Calculate the pH at the stoichiometric point.
Nevermind I researched it for a few hours and finallly figured it out :)
Congratulations. All titration curves are similar.
What's the answer to part f?
To solve this problem, we need to use the concept of acid-base titration and the equilibrium expression for the dissociation of the acid. Let's go step by step:
(a) To determine the initial pH of the acetic acid solution (C6H5COOH), we need to consider its dissociation in water:
C6H5COOH(aq) ⇌ C6H5COO-(aq) + H+(aq)
Given that the initial concentration of C6H5COOH is 0.20 M, we can assume that its concentration at equilibrium (after full dissociation) will also be 0.20 M. Since it is a weak acid, we can use the equilibrium expression for its dissociation:
Ka = [C6H5COO-][H+]/[C6H5COOH]
The Ka value for C6H5COOH is usually provided, or you can look it up in a reference table. Assuming a Ka value of 1.8 x 10^-5 for simplicity, we have:
1.8 x 10^-5 = [0.20-x][x]/[0.20]
Simplifying the equation, we have x^2 = 1.8 x 10^-5 * 0.20
Solving for x, we find x = 0.0196 M
Since x represents the concentration of H+, the pH can be calculated as -log([H+]), which gives us pH = -log(0.0196) = 1.71.
So, the initial pH of the 0.20 M C6H5COOH(aq) solution is 1.71.
(b) To determine the pH after the addition of 15.0 mL of 0.30 M KOH(aq), we need to consider the reaction between the acid and the base to form a salt and water:
C6H5COOH(aq) + KOH(aq) → C6H5COOK(aq) + H2O(l)
First, calculate the moles of KOH added:
moles KOH = concentration KOH * volume KOH (in L)
= 0.30 M * (15.0 mL / 1000 mL/ L)
= 0.0045 mol
Since acetic acid is a weak acid, a small portion of it will react with the KOH and form its conjugate base C6H5COO-. To determine the concentration of the acid and the conjugate base at equilibrium, we need to apply the principles of stoichiometry. The mole ratio between the acid and the base is 1:1, which means that 0.0045 mol of C6H5COOH has reacted.
Now, calculate the new concentrations at equilibrium.
C6H5COOH: (initial concentration - moles reacted) / total volume (in L)
= (0.20 M - 0.0045 mol) / (30.0 mL / 1000 mL/ L)
= 0.195 M
C6H5COO-: moles formed / total volume (in L)
= 0.0045 mol / (30.0 mL / 1000 mL/ L)
= 0.15 M
Now, we can calculate the pH using the same equilibrium expression as in part (a), with the new concentrations:
Ka = [C6H5COO-][H+]/[C6H5COOH]
1.8 x 10^-5 = (0.15-x)(x) / (0.195)
Solving for x, we find x = 0.0059 M
Therefore, the pH after the addition of 15.0 mL of 0.30 M KOH(aq) is pH = -log(0.0059) ≈ 2.23.
(c) To determine the volume of 0.30 M KOH(aq) required to reach halfway to the stoichiometric point, we need to consider the stoichiometry of the reaction.
In a 1:1 stoichiometric reaction, halfway to the stoichiometric point means that half of the initial moles of acetic acid have reacted. Since we know the initial concentration of acetic acid and the volume of the KOH solution added, we can determine the moles of acetic acid reacted.
moles of KOH = concentration of KOH * volume of KOH (in L)
= 0.30 M * (x mL / 1000 mL/ L)
= 0.30x/1000 mol
moles of C6H5COOH reacted = 0.20 M * (0.30x/1000 L)
= 0.06x/1000 mol
At halfway to the stoichiometric point, the moles of the acid reacted will be half of the initial moles. Therefore:
0.06x/1000 = 0.20/2
Solving for x, we find x = 33.333 mL
Therefore, the volume of 0.30 M KOH(aq) required to reach halfway to the stoichiometric point is 33.333 mL.
(d) To calculate the pH at the halfway point, we can use the same approach as in part (b) but with the new volumes.
First, calculate the new concentrations at the halfway point:
C6H5COOH: (initial concentration - moles reacted) / total volume (in L)
= (0.20 M - 0.06x/1000 mol) / (30.0 mL + x mL) / 1000 mL/ L
C6H5COO-: moles formed / total volume (in L)
= 0.06x/1000 mol / (30.0 mL + x mL) / 1000 mL/ L
Using the equilibrium expression, solve for x to find the concentration of H+, and then calculate the pH using -log([H+]).
(e) To determine the volume of 0.30 M KOH(aq) required to reach the stoichiometric point, we consider the stoichiometry of the reaction.
At the stoichiometric point, the moles of the acid reacted will be equal to the moles of the base added.
moles of KOH = moles of C6H5COOH
0.30 M * (x mL / 1000 mL/ L) = 0.20 M * (30.0 mL / 1000 mL/ L)
Solving for x, we find x = 20 mL
Therefore, the volume of 0.30 M KOH(aq) required to reach the stoichiometric point is 20 mL.
(f) Finally, to calculate the pH at the stoichiometric point, we can use the same approach as in part (b) and part (d) but with the new volumes.
First, calculate the new concentrations at the stoichiometric point, using the stoichiometry relation:
C6H5COOH: (initial concentration - moles reacted) / total volume (in L)
= (0.20 M - 0.20 M * (20 mL / 1000 mL/ L)) / (30.0 mL + 20 mL) / 1000 mL/ L
C6H5COO-: moles formed / total volume (in L)
= 0.20 M * (20 mL / 1000 mL/ L) / (30.0 mL + 20 mL) / 1000 mL/ L
Use the equilibrium expression to solve for x, find the concentration of H+, and then calculate the pH using -log([H+]).