Find the pH during the titration of 20.00 mL of 0.2380 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 10-5), with 0.2380 M NaOH solution after the following additions of titrant.

(a) 0 mL
Ive got the answer of 2.72
(b) 10.00 mL
Ive got the answer of 4.81
I'm lost from here...Please help!!
(c) 15.00 mL

(d) 19.00 mL

(e) 19.95 mL

(f) 20.00 mL

(g) 20.05 mL

(h) 25.00 mL

Your answer to a is correct.
Your answer to b is correct.
c,d,e are all done essentially the same way. I will go through c.
For simplicity in typing, I will call butanoic acid a simple HB. Then
butanoic acid + KOH = potassium salt of butanoic acid + H2O OR in simple terms:
HB + KOH ==> KB + H2O
mols HB at beginning of titration are L x M = 0.020 x 0.2380 = ??
mols KOH added = (for c) 0.015 x 0.2380 = ??
Subtract from HB to see how much HB is left. Add to KB to see how much B^- is there. Plug into the Henderson-Hasselbalch equation and solve for pH.
d and e are done essentially the same way.
For must recognize that 20.00 mL KOH is the EXACT equivalence point; therefore, the pH is determine SOLELY by the amount of salt (KB) present and that is determined by the hydrolysis of the salt. Using the Bronsted-Lowry theory of acids/bases, write the hydrolysis of the anion, B^-.
B^- + HOH ==> HB + OH^-
Post back if you don't know what to do with that.

For the others AFTER the equivalence point, these are just excess KOH so you need to calculate how much excess KOH is present, the total volume of the solution, and the molarity of the KOH present at those points. Then use pH + pOH = pKw
I shall be happy to check your work if you post it.

A solution of ammonia and water contains 3.10×1025 water molecules and 5.20×1024 ammonia molecules. How many total hydrogen atoms are in this solution?

To find the pH during the titration of butanoic acid with NaOH after the given additions of titrant, follow these steps:

(a) For 0 mL NaOH:
You mentioned the answer is 2.72, which is correct. This can be calculated by finding the pH of the initial solution of butanoic acid using the given Ka value.

(b) For 10.00 mL NaOH:
You mentioned the answer is 4.81, which is correct. This can be calculated by adding the moles of NaOH to the moles of butanoic acid initially present, then calculating the concentration of the resulting salt (KB), and finally using the Henderson-Hasselbalch equation to find the pH.

(c), (d), and (e):
For these additions, you need to calculate the moles of KOH added, subtract them from the initial moles of butanoic acid, and add them to the moles of the resulting salt (KB). Then, use the Henderson-Hasselbalch equation to find the pH.

For example, for (c) 15.00 mL NaOH:
Moles of butanoic acid initially present = 0.020 L x 0.2380 M = 0.00476 moles.
Moles of KOH added = 0.015 L x 0.2380 M = 0.00357 moles.

Subtracting the moles of KOH added from the initial moles of butanoic acid gives:

0.00476 moles - 0.00357 moles = 0.00119 moles of butanoic acid remaining.

Adding the moles of KOH to the moles of the resulting salt gives:

0.00357 moles + 0.00357 moles = 0.00536 moles of the resulting salt (KB).

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])

Plug in the values:

pKa = -log(Ka) = -log(1.54 x 10^-5) ≈ 4.81

[A-] = moles of resulting salt/volume of solution = 0.00536 moles/0.035 L ≈ 0.153 M

[HA] = moles of butanoic acid remaining/volume of solution = 0.00119 moles/0.035 L ≈ 0.034 M

pH = 4.81 + log(0.153/0.034) ≈ 5.48

So, the pH for (c) 15.00 mL NaOH is approximately 5.48.

You can follow the same process for (d) 19.00 mL NaOH and (e) 19.95 mL NaOH to find their respective pH values.

(f) For 20.00 mL NaOH:
At this point, you have reached the equivalence point, and the pH is solely determined by the hydrolysis of the salt (KB) formed. Using the Bronsted-Lowry theory of acids/bases, you can write the hydrolysis reaction of the anion B^- to find the pH.

The hydrolysis reaction is: B^- + H2O → HB + OH^-

To calculate the pH, you need to consider the concentration of the hydroxide ion (OH^-) formed from the hydrolysis reaction. Using the concentration of the resulting salt (KB) and the volume of the solution, find the concentration of OH^- and then use pH + pOH = 14 to determine the pH.

For (f) 20.00 mL NaOH, the concentration of KB is 0.2380 M (as mentioned initially). You can use this concentration, along with the volume of the solution (which is 0.020 L) to calculate the concentration of OH^-.

OH^- = [KB]/volume of solution = 0.2380 M/0.020 L = 11.90 M

Using pH + pOH = 14:
pH + (-log[OH^-]) = 14
pH + (-log(11.90)) = 14

Solving for pH:
pH ≈ 14 - (-log(11.90)) ≈ 14 + 1.08 ≈ 15.08

So, the pH for (f) 20.00 mL NaOH is approximately 15.08.

(g) 20.05 mL NaOH and (h) 25.00 mL NaOH:
After the equivalence point, you have excess KOH. To find the pH at these points, you need to calculate the moles of excess KOH present, determine the total volume of the solution, calculate the molarity of the KOH, and then use the equation pH + pOH = pKw to find the pH.

For example, for (h) 25.00 mL NaOH:
Moles of excess KOH added = (volume of NaOH added - volume at equivalence point) x Molarity of NaOH
Moles of excess KOH = (25.00 mL - 20.00 mL) x 0.2380 M = 1.19 x 10^-3 moles

Total volume of solution = volume at equivalence point + volume of excess NaOH added
Total volume = 20.00 mL + 1.19 mL = 26.19 mL = 0.02619 L

Molarity of excess KOH = moles of excess KOH/total volume of solution
Molarity = (1.19 x 10^-3 moles)/(0.02619 L) ≈ 0.045 M

Using pH + pOH = pKw:
pH = pKw - pOH
pH = 14 - (-log(0.045))

Solving for pH:
pH ≈ 14 - (-log(0.045)) ≈ 14 + 1.35 ≈ 15.35

So, the pH for (h) 25.00 mL NaOH is approximately 15.35.

I hope this helps you in finding the pH for the remaining parts of the titration. Let me know if you need further assistance!