A 25.00mL sample of 0.100M HC2H3O2 is titrated with 0.100M NaOH. What is the pH after 25.00mL of NaOH have been added? Ka HC2H3O2 = 1.8x10-5.

All of these look like homework dump to me. You show no idea of your own, how you think you could solve the problem yourself or even how you might procede. I've worked all of the problems anyway. In the future you should give me some indication of the work you have done and exactly what you don't understand about the problem.Remember that this is a homework HELP with the emphasis on HELP part. It is not my intention to take this chemistry course again. I already know how to work these problems. So I want to HELP but I don't want to DO them all.

I will call HC2H3O2 HA to save typing. NaA will be sodium acetate.
millimoles HAc = 25 mL x 0.1 M = 2.5 mmols
mmols NaOH added....25 mL x 0.1 M = 2.5 mmols
......................HAc + NaOH --> NaAc + H2O
I..................... 2.5..........0............0............0
add...............................2.5..............................
C.....................-2.5......-2.5...........+2.5................
E......................0...........0...............2.5.............
So after 25.00 are added you have aolution of Ac^- (from NaAc) and the pH depends upon the hydrolysis of the Ac^- as follows:
(Ac^-) = millimoles/mL = (2.5 mmols/50 mL) = 0.05M
.........................Ac^- + HOH ==> HAc + OH^-
I.......................0.05........................0...........0
C.......................-x............................x............x
E......................0.05-x........................x...........x
Kb for Ac^- = Kw/Ka for HAc = (x)(x)/(Ac^-)
Solve for OH, convert to pOH and then to pH as I've shown in one or two of the problems below.
Or there is a shortcut way of doint it directly
(H^+) = sqrt (KwKa/C) where C = (Ac^-) = 0.05 M

Unlike DrBOB222 I WILL SPOON FEED YOU ANSWER PLEASE LET ME KNOW ANYTHING WRONG I AM WEEK AT ENGLISH SO PLEASE LET ME KNOW...NO REQUIREMENT CALCULUS BCDEF.

Well, pH is determined by the concentration of hydrogen ions in a solution. In this case, we need to figure out how many moles of HC2H3O2 are left after the titration.

Given that you have a 0.100M HC2H3O2 solution and you added 25.00mL of 0.100M NaOH, we can figure out the number of moles of NaOH added using:

moles NaOH = concentration NaOH × volume NaOH added
= 0.100M × (25.00mL / 1000mL)

Knowing that the stoichiometry of the reaction is 1:1, the number of moles of HC2H3O2 left will be equal to the moles of NaOH added. So, we have:

moles HC2H3O2 = moles NaOH

Now, we need to find the remaining volume of the HC2H3O2 solution to calculate its concentration:

volume HC2H3O2 = initial volume - volume NaOH added
= 25.00mL - 25.00mL
= 0mL

Since the volume is now 0mL, we can conclude that all the HC2H3O2 has reacted with the NaOH, and we're left with only the NaC2H3O2 (sodium acetate) solution.

Now, we need to find the concentration of the resulting solution. However, since the volume is 0mL, it becomes undefined, just like the way I feel about that one time I tried to juggle chainsaws.

So, in conclusion, after adding 25.00mL of NaOH, we don't technically have a solution of HC2H3O2 anymore, and the pH is undefined.

To calculate the pH after 25.00 mL of NaOH have been added to the 25.00 mL sample of 0.100 M HC2H3O2, we need to determine the amount of moles of HC2H3O2 and NaOH that react.

Step 1: Determine the moles of HC2H3O2
Moles of HC2H3O2 = Volume (L) x Concentration (M)
Moles of HC2H3O2 = 0.025 L x 0.100 M
Moles of HC2H3O2 = 0.0025 moles

Step 2: Determine the moles of NaOH
Since NaOH is a strong base, it reacts with HC2H3O2 in a 1:1 ratio.
Moles of NaOH = Moles of HC2H3O2 = 0.0025 moles

Step 3: Calculate the concentration of HC2H3O2 after the reaction
Total volume = Volume of HC2H3O2 + Volume of NaOH
Total volume = 0.025 L + 0.025 L
Total volume = 0.050 L

Concentration of HC2H3O2 after the reaction = Moles of HC2H3O2 / Total volume
Concentration of HC2H3O2 after the reaction = 0.0025 moles / 0.050 L
Concentration of HC2H3O2 after the reaction = 0.050 M

Step 4: Calculate the pKa of HC2H3O2
pKa = -log(Ka)
pKa = -log(1.8x10^-5)
pKa = 4.74

Step 5: Calculate the concentration of OH- after the reaction
Since NaOH is a strong base, it dissociates completely into Na+ and OH- ions.
Concentration of OH- = Moles of NaOH / Total volume
Concentration of OH- = 0.0025 moles / 0.050 L
Concentration of OH- = 0.050 M

Step 6: Calculate the concentration of C2H3O2- after the reaction
Since NaOH reacts with HC2H3O2 in a 1:1 ratio, the concentration of C2H3O2- after the reaction is equal to the concentration of HC2H3O2 after the reaction, which is 0.050 M.

Step 7: Calculate the concentration of H+ (protons)
[H+] = √(Ka x [C2H3O2-] / [HC2H3O2])
[H+] = √((1.8x10^-5) x (0.050 M) / (0.050 M))
[H+] = √1.8x10^-5
[H+] = 4.24x10^-3 M

Step 8: Calculate the pH
pH = -log[H+]
pH = -log(4.24x10^-3)
pH = 2.37

Therefore, the pH after 25.00 mL of NaOH have been added to the 25.00 mL sample of 0.100 M HC2H3O2 is approximately 2.37.

To find the pH after 25.00mL of NaOH have been added, we need to determine the concentration of the remaining HC2H3O2 and then calculate the pH based on the dissociation of that acid.

Let's go step by step:

1. Start by calculating the initial number of moles of HC2H3O2 in the 25.00mL sample. To do so, use the formula:

Moles = Molarity x Volume (L)

Moles of HC2H3O2 = 0.100 mol/L x 0.02500 L
= 0.0025 mol

2. Next, calculate the number of moles of NaOH that have reacted with the HC2H3O2. Since the balanced chemical equation between HC2H3O2 and NaOH is 1:1, the number of moles of NaOH that have reacted will be the same as the number of moles of HC2H3O2. Therefore, 0.0025 mol of NaOH has reacted.

3. Now, determine the moles of HC2H3O2 that remain after the reaction with NaOH. Subtract the moles of NaOH that have reacted from the initial moles of HC2H3O2:

Moles of HC2H3O2 remaining = Moles of HC2H3O2 - Moles of NaOH
= 0.0025 mol - 0.0025 mol
= 0 mol

Since all the HC2H3O2 has reacted with the NaOH, there are no moles of the acid remaining.

4. Now, we need to calculate the concentration of the remaining HC2H3O2. Since we know the volume of the solution is 25.00 mL (0.02500 L) and the moles of HC2H3O2 is 0 mol, the concentration is:

Remaining HC2H3O2 concentration = Moles / Volume
= 0 mol / 0.02500 L
= 0 mol/L

Therefore, the concentration of HC2H3O2 is 0 mol/L after the NaOH has been added.

5. Since HC2H3O2 is a weak acid, it will only partially dissociate in water. To determine the pH, we need to calculate the concentration of H+ ions using the acid dissociation constant (Ka) of HC2H3O2.

Ka = [H+][C2H3O2-] / [HC2H3O2]

Since the concentration of HC2H3O2 is 0 mol/L, we can ignore it in the equation.

Ka = [H+][C2H3O2-] / 0

Therefore, [H+] = 0.

The concentration of H+ ions is 0 mol/L, which means the solution is neutral, and the pH is 7.

So, the pH after 25.00mL of NaOH have been added is 7.