1. 25.0mL of 0.20M Propanic acid (HC3H5O2, Ka = 1.3×10-5) is titrated using 0.10M NaOH. Calculate the following pH. Show your work.

a. When 0.0 mL NaOH is added.

b. When 25.0 mL NaOH is added

c. When 50.0 mL NaOH is added

d. When 75.0mL NaOH is added.

Let's call the acid HP.

a.
.......HP ==> H^+ + P^-
I.....0.2M....0.....0
C......-x.....x.....x
E....0.2-x....x.....x

Plug the E line into the Ka expression and solve for x = (H^+), then convert to pH.

b is done with the Henderson-Hasselbalch equation.

c is the equivalence point and the rxn is HP + NaOH ==> NaP + H2O
The pH is determined by the hydrolysis of the salt NaP. The concentration of the salt is mols NaP/L. mols NaP = M x L = 0.2 x 0.025 = 0.005. L = 50 mL from the beginning + 50 mL added NaOH = 75 mL = 0.075 L so M NaP = 0.005/0.075 - 0.0667. The hydrolysis works this way.
.........P^- + HOH ==> HP + OH^-
I.....0.0667...........0.....0
C......-x..............x.....x
E....0.0667-x..........x.....x

Kb for P^- = (Kw/Ka for HP)
= (x)(x)/(0.0667-x). Solve for x = (OH^-) and convert to pH.

D. You have excess OH^-. Calculate that and convert to pH.

Post your work if you have questions and PLEASE tell us exactly what you don't understand.

To calculate the pH at each step, we need to consider the reaction between propanic acid (HC3H5O2) and sodium hydroxide (NaOH). Propanic acid is a weak acid, and sodium hydroxide is a strong base.

The balanced chemical equation for the reaction is:

HC3H5O2 + OH- -> C3H5O2- + H2O

We can assume that the volume of the solution remains constant throughout the titration.

a. When 0.0 mL NaOH is added:
At this stage, there is only propanic acid present in the solution.

To calculate the concentration of propanic acid ([HC3H5O2]), we use the equation:
[HC3H5O2] = (moles of HC3H5O2) / (volume of solution)

Given that the initial volume of propanic acid solution is 25.0 mL and the concentration is 0.20 M, we can calculate the moles of HC3H5O2 as follows:
moles of HC3H5O2 = (volume of solution in L) * (concentration of HC3H5O2)
= (25.0 mL / 1000 mL/L) * (0.20 mol/L)
= 0.005 mol

Since no NaOH is added, the reaction does not occur, so all of the propanic acid will be present.

The pH can be calculated using the formula:
pH = -log10 [H+]
Since the concentration of H+ is equal to the concentration of HC3H5O2, we can use the equation:
[H+] = [HC3H5O2]

Substituting the values, we get:
pH = -log10 (0.20)
= 0.70

So, when 0.0 mL NaOH is added, the pH is 0.70.

b. When 25.0 mL NaOH is added:
At this stage, 0.005 moles of NaOH have been added.

The balanced chemical equation shows that 1 mole of NaOH reacts with 1 mole of HC3H5O2, so the moles of HC3H5O2 remaining will be:
moles of HC3H5O2 remaining = moles of HC3H5O2 initial - moles of NaOH added
= 0.005 mol - 0.005 mol
= 0.000 mol

Since there are no moles of HC3H5O2 remaining, the pH will depend on the concentration of OH-. We can calculate the concentration of OH- as follows:
[OH-] = (moles of NaOH added) / (volume of solution)
= (0.005 mol) / (25.0 mL / 1000 mL/L)
= 0.20 M

Using the pOH formula:
pOH = -log10 [OH-]
= -log10 (0.20)
= 0.70

Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
= 14 - 0.70
= 13.30

So, when 25.0 mL NaOH is added, the pH is 13.30.

c. When 50.0 mL NaOH is added:
At this stage, 0.010 moles of NaOH have been added.

The moles of HC3H5O2 remaining will be:
moles of HC3H5O2 remaining = moles of HC3H5O2 initial - moles of NaOH added
= 0.005 mol - 0.010 mol
= -0.005 mol

Since the moles of HC3H5O2 remaining is negative, it means that all the HC3H5O2 has reacted. Thus, the pH will be determined by the excess OH- ions formed.

The concentration of OH- can be calculated as follows:
[OH-] = (moles of NaOH added) / (volume of solution)
= (0.010 mol) / (50.0 mL / 1000 mL/L)
= 0.20 M

Using the pOH formula:
pOH = -log10 [OH-]
= -log10 (0.20)
= 0.70

Again, pH + pOH = 14, so we can calculate the pH:
pH = 14 - pOH
= 14 - 0.70
= 13.30

So, when 50.0 mL NaOH is added, the pH is 13.30.

d. When 75.0 mL NaOH is added:
At this stage, 0.015 moles of NaOH have been added.

The moles of HC3H5O2 remaining will be:
moles of HC3H5O2 remaining = moles of HC3H5O2 initial - moles of NaOH added
= 0.005 mol - 0.015 mol
= -0.010 mol

Again, the negative moles of HC3H5O2 remaining indicate that all the HC3H5O2 has reacted. Therefore, the pH is determined by the excess OH- ions formed.

The concentration of OH- can be calculated as follows:
[OH-] = (moles of NaOH added) / (volume of solution)
= (0.015 mol) / (75.0 mL / 1000 mL/L)
= 0.20 M

Using the pOH formula:
pOH = -log10 [OH-]
= -log10 (0.20)
= 0.70

Once again, pH + pOH = 14, so we can calculate the pH:
pH = 14 - pOH
= 14 - 0.70
= 13.30

Therefore, when 75.0 mL NaOH is added, the pH is 13.30.

To calculate the pH at each step of the titration, we need to determine the moles of acid and base at each point and use the balanced equation to calculate the concentrations of the acid and conjugate base.

First, let's determine the moles of acid present initially and at each step of the titration:

a. When 0.0 mL NaOH is added:
In this case, no NaOH has been added yet, so the acid is still in its original concentration. The moles of propanic acid (HC3H5O2) can be calculated as:
moles of acid = volume of acid solution (L) x concentration of acid (M)
moles of acid = 0.025 L x 0.20 M = 0.005 mol

Now, let's determine the concentration of propanic acid (HC3H5O2) and its conjugate base (C3H5O2-) at each step using an ICE table (Initial, Change, Equilibrium):

a. When 0.0 mL NaOH is added:

HC3H5O2 + H2O ⇌ H3O+ + C3H5O2-

Initial: 0.20 M 0 M 0 M 0 M
Change: - - + +

At equilibrium: 0.20 - x M x M x M x M

Since the initial concentration of the acid is 0.20 M and no NaOH has been added yet, the concentration of propanic acid remains at 0.20 M. The concentration of the conjugate base (C3H5O2-) formed from the acid in the reaction is zero.

To calculate the pH, we need the concentration of H3O+. At equilibrium, the concentration of H3O+ is equal to the concentration of the acid because the acid is a monoprotic weak acid.

pH = -log[H3O+]
pH = -log(0.20) = 0.699

Therefore, the pH when 0.0 mL NaOH is added is approximately 0.699.

b. When 25.0 mL NaOH is added:
To calculate the moles of NaOH added, we need to use the following equation:

moles of NaOH = volume of NaOH solution (L) x concentration of NaOH (M)
moles of NaOH = 0.025 L x 0.10 M = 0.0025 mol

Now, let's determine the number of moles of HC3H5O2 and C3H5O2-:

moles of HC3H5O2 remaining = moles of HC3H5O2 initially - moles of NaOH added
moles of HC3H5O2 remaining = 0.005 mol - 0.0025 mol = 0.0025 mol

moles of C3H5O2- formed = moles of NaOH added
moles of C3H5O2- formed = 0.0025 mol

To calculate the concentration of propanic acid (HC3H5O2) and its conjugate base (C3H5O2-) at this step, we use the following equation:

initial volume of acid solution - volume of NaOH solution added = volume of acid solution remaining
0.025 L - 0.025 L = 0 L

Since the volume of the acid solution remaining is 0 L, the concentration of the propanic acid (HC3H5O2) and its conjugate base (C3H5O2-) are both zero.

Therefore, the pH when 25.0 mL NaOH is added is undefined because there is no acid present.

c. When 50.0 mL NaOH is added:
To calculate the moles of NaOH added:
moles of NaOH = volume of NaOH solution (L) x concentration of NaOH (M)
moles of NaOH = 0.050 L x 0.10 M = 0.005 mol

Now, let's determine the number of moles of HC3H5O2 and C3H5O2-:

moles of HC3H5O2 remaining = moles of HC3H5O2 initially - moles of NaOH added
moles of HC3H5O2 remaining = 0.005 mol - 0.005 mol = 0 mol

moles of C3H5O2- formed = moles of NaOH added
moles of C3H5O2- formed = 0.005 mol

To calculate the concentration of propanic acid (HC3H5O2) and its conjugate base (C3H5O2-) at this step, we use the following equation:

initial volume of acid solution - volume of NaOH solution added = volume of acid solution remaining
0.025 L - 0.050 L = -0.025 L

Since the volume of the acid solution remaining is negative, we cannot determine the concentration of the acid and the conjugate base.

Therefore, the pH when 50.0 mL NaOH is added is undefined because there is no acid present.

d. When 75.0 mL NaOH is added:
To calculate the moles of NaOH added:
moles of NaOH = volume of NaOH solution (L) x concentration of NaOH (M)
moles of NaOH = 0.075 L x 0.10 M = 0.0075 mol

Now, let's determine the number of moles of HC3H5O2 and C3H5O2-:

moles of HC3H5O2 remaining = moles of HC3H5O2 initially - moles of NaOH added
moles of HC3H5O2 remaining = 0.005 mol - 0.0075 mol = -0.0025 mol

moles of C3H5O2- formed = moles of NaOH added
moles of C3H5O2- formed = 0.0075 mol

To calculate the concentration of propanic acid (HC3H5O2) and its conjugate base (C3H5O2-) at this step, we use the following equation:

initial volume of acid solution - volume of NaOH solution added = volume of acid solution remaining
0.025 L - 0.075 L = -0.050 L

Since the volume of the acid solution remaining is negative, we cannot determine the concentration of the acid and the conjugate base.

Therefore, the pH when 75.0 mL NaOH is added is undefined because there is no acid present.