Consider the titration of 100.0 mL of 0.260 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.130 M NaOH. Calculate the pH of the resulting solution after each of the following volumes of NaOH has been added. (Assume that all solutions are at 25°C.)

(a) 0.0 mL

(b) 50.0 mL

(c) 100.0 mL

(d) 150.0 mL

(e) 200.0 mL

(f) 250.0 mL

I have done a through d already and got them right,
However when I get to e I start having problems, theres something I am missing.

Here is what I have tried:
Initial Concentration of Acid= (.100*.260)=.026 M
Initial Concentration of Base=
(.200*.130)=.026 M

Then:
Kb= 1.0e-14/1.3e-5= 7.69e-10
x^2/.026= 7.69e-10

x=sqrt(.026*7.69e-10)=4.47e-6
pOH= -log(4.47e-6)=5.35
pH=14-5.35=8.65

The homework system says I am incorrect so I would like some assistance in figuring out what I am doing wrong.

Thank You

All of that looks good EXCEPT for using 0.026M for concentration of the propanoate ion. That is not 0.026 because it has been diluted by the addition of the NaOH.

You started with 100 mL of 0.026M acid, when you've added 200 mL of NaOH the salt concn is now 0.026 x (100/300) = ?
Thanks for showing your work. It makes it easy to spot the problem and I don't need to work all of that stuff you already know how to do.

To solve for the pH of the resulting solution after adding 200.0 mL of NaOH, we need to consider the reaction between propanoic acid (CH3CH2COOH) and NaOH:

CH3CH2COOH + NaOH → CH3CH2COONa + H2O

(a) 0.0 mL (initial state)
The solution contains only propanoic acid. We can calculate the pH using the Ka value:

Ka = [CH3CH2COO-][H3O+] / [CH3CH2COOH]

Since the concentration of propanoic acid is 0.026 M, the initial concentration of H3O+ is equal to the initial concentration of propanoic acid:

[H3O+] = 0.026 M

Therefore, the pH is given by:

pH = -log([H3O+]) = -log(0.026) = 1.59

(b) 50.0 mL
At this point, we have added half of the volume of NaOH. The reaction has not reached the equivalence point yet, so we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

Initially, [A-] = 0 M, since no NaOH has been added. [HA] = 0.026 M.

pKa = -log(Ka) = -log(1.3 x 10^-5) = 4.89

pH = 4.89 + log(0/0.026) = 4.89 - ∞ (undefined)

(c) 100.0 mL
At this point, we have added the same volume of NaOH as the initial volume of propanoic acid. This is called the equivalence point. The reaction is complete, resulting in a solution containing the propanoate ion (CH3CH2COO-) and water (H2O). The NaOH was completely neutralized by the propanoic acid.

At the equivalence point, the solution will be neutral, so the pH is 7.

(d) 150.0 mL
We have added more than the initial volume of propanoic acid. Excess NaOH is present, resulting in a basic solution. To calculate the pH, we need to determine the concentration of the excess OH- ions. We know that the initial concentration of OH- is 0 M (before NaOH is added) and we need to subtract the concentration of H3O+ ions that form as a result of the reaction:

[H3O+] = [OH-] - [A-]

Since the initial volume of propanoic acid is 100.0 mL and we added 150.0 mL of NaOH, the remaining concentration of OH- can be calculated by the difference in volume:

[VNaOH] = 150.0 mL - 100.0 mL = 50.0 mL = 0.050 L (convert to liters)

[OH-] = 0.130 M (from NaOH) - (0.026 M / 0.100 L) * [VNaOH]

[OH-] = 0.130 M - (0.026 M / 0.100 L) * 0.050 L = 0.130 M - 0.013 M = 0.117 M

[H3O+] = 0.117 M

pOH = -log([OH-]) = -log(0.117) = 0.93
pH = 14 - pOH = 14 - 0.93 = 13.07

(e) 200.0 mL
At this point, we have added twice the volume of the initial propanoic acid. Excess NaOH is still present, resulting in a basic solution. We repeat the same steps as in (d):

[VNaOH] = 200.0 mL - 100.0 mL = 100.0 mL = 0.100 L (convert to liters)

[OH-] = 0.130 M - (0.026 M / 0.100 L) * [VNaOH]
[OH-] = 0.130 M - (0.026 M / 0.100 L) * 0.100 L = 0.130 M - 0.026 M = 0.104 M

[H3O+] = 0.104 M

pOH = -log([OH-]) = -log(0.104) = 0.98
pH = 14 - pOH = 14 - 0.98 = 13.02

(f) 250.0 mL
At this point, we have added 2.5 times the volume of the initial propanoic acid. Excess NaOH is still present, resulting in a basic solution. We repeat the same steps as in (d) and (e):

[VNaOH] = 250.0 mL - 100.0 mL = 150.0 mL = 0.150 L (convert to liters)

[OH-] = 0.130 M - (0.026 M / 0.100 L) * [VNaOH]
[OH-] = 0.130 M - (0.026 M / 0.100 L) * 0.150 L = 0.130 M - 0.039 M = 0.091 M

[H3O+] = 0.091 M

pOH = -log([OH-]) = -log(0.091) = 0.96
pH = 14 - pOH = 14 - 0.96 = 13.04

Note: In the solution provided above, we calculated the concentration of [OH-] by subtracting [A-] from [NaOH]. We did not involve Kb in the calculations because propanoic acid is a weak acid, and the contribution of hydroxide ions from NaOH is significantly larger than the hydroxide ion concentration resulting from the dissociation of the weak acid.

To calculate the pH of the resulting solution after adding 200.0 mL of 0.130 M NaOH, you need to consider the reaction between propanoic acid (a weak acid) and sodium hydroxide (a strong base).

When NaOH is added, it reacts with propanoic acid to form sodium propanoate and water:

CH3CH2COOH + NaOH → CH3CH2COONa + H2O

In this reaction, propanoic acid acts as an acid and donates a proton (H+) to NaOH, which acts as a base and accepts the proton.

To determine the amount of propanoic acid that reacts with NaOH, consider the initial concentrations and the stoichiometry of the reaction.

Initial concentration of propanoic acid = 0.260 M in 100.0 mL = 0.026 mol

Initial concentration of NaOH = 0.130 M in 200.0 mL = 0.026 mol

Since the initial moles of propanoic acid and NaOH are equal, they will react completely in a 1:1 ratio.

Therefore, the moles of propanoic acid that react with NaOH = 0.026 mol.

The remaining moles of propanoic acid after the reaction = Initial moles of propanoic acid - Moles of propanoic acid reacted with NaOH = 0.026 mol - 0.026 mol = 0 mol

This means that all of the propanoic acid has reacted with NaOH, and there is no propanoic acid left in the solution.

To find the resulting pH, you can find the concentration of hydroxide ions (OH-) in the solution using the concentration of NaOH that has been added.

Concentration of NaOH = 0.130 M in 200.0 mL = 0.026 mol

Since NaOH is a strong base, it dissociates completely in water to form Na+ and OH- ions in equal amounts.

Therefore, the concentration of OH- ions in the solution = Concentration of NaOH = 0.130 M

Now, use the concentration of OH- to find the pOH of the solution:

pOH = -log[OH-] = -log(0.130) = 0.886

Finally, calculate the pH using the relationship:

pH = 14 - pOH = 14 - 0.886 = 13.114

So, the pH of the resulting solution after adding 200.0 mL of 0.130 M NaOH is approximately 13.114.