Calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point. Compare these theoretical values with the actual values found on the titration curves created in lab.

(Hint: Use total volume of titrant (the initial and final volume on the buret) in conjunction with the time required to deliver that volume to convert the volumes given above to time.) Show a sample of all necessary calculations. Calculate the percent error from the theoretical values.

molarity is 0.1 for all substances, volume of HCl and HC2H3O2 is 8.00ml each, both diluted in 100ml of water

PLESASE HELP!!

millimoles HCl = 8.00 mL x 0.1M=0.8mmols.

mmoles NaOH = 2.50 mL x 0.1M = 0.25 mmols.
mmoles NaOH = 9.50 x 0.1M = 0.95 mmoles.
mmoles HAc = 8.00 mL x 0.1M = 0.8 mmoles.
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..........HCl + NaOH ==> H2O + NaCl
initial...0.8....0........0......0
add.............0.25.................
change...-0.25..-0.25....+0.25..0.25
equil...0.55......0.......0.25..0.25
pH of the soln is determined by the excess HCl since there is no NaOH present and NaCl is not hydrolyzed.
(HCl) = 0.55mmoles/(100+8.00+2.50)mL = ?? and convert to pH.

initial......0.8..0......0.......0
add...............0.95.............
change.....-0.8...-0.8...+0.8..+0.8
equil........0....0.15....0.8..0.8
pH of the soln is determined by the excess NaOH since there is no HCl present and NaCl is not hydrolyzed.
(NaOH) = 0.15mmoles/(100+8+9.5)mL = ?, convert to pOH and pH.

...........HAc + NaOH ==> NaAc + H2O
initial...0.8.....0........0.......0
add..............0.25..............
change...-0.25..-0.25....+0.25...+0.25
equil.....0.55.....0.......0.25...0.25

Here you have a soln in HAc and NaAc so that is a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log (0.25/0.55) = ?

............HAc + NaOH ==> NaAc + H2O
initial....0.8.....0........0......0
add...............0.95.............
change...-0.8...-0.8......+0.8..+0.8
equil.......0.....0.15.....0.8....0.8

Here you have a soln in NaOH and NaAc but that isn't a buffer and the pH will be determined by the excess NaOH present. It is true that the Ac^- will hydrolyze but it is small enough to ignore it.
(NaOH) = mmoles NaOH/(100 + 9.5+8)mL = ?, convert to pOH and to pH.
Post your work if you get stuck.

To calculate the theoretical pH after adding NaOH in the titration of HCl and HC2H3O2, we need to use the concepts of stoichiometry and acid-base reactions. Here are the steps to determine the theoretical pH:

Step 1: Calculate the number of moles of HCl and HC2H3O2 used in the titration. Since the molarity and volume of the acids are given, we can use the equation:

Moles = Molarity * Volume

For HCl:
Moles of HCl = (0.1 mol/L) * (8.00 mL/1000 mL/L) = 0.008 mol

For HC2H3O2:
Moles of HC2H3O2 = (0.1 mol/L) * (8.00 mL/1000 mL/L) = 0.008 mol

Step 2: Determine the limiting reagent. In an acid-base titration, the limiting reagent is the one that is completely consumed and forms the equivalence point. In this case, both HCl and HC2H3O2 have the same number of moles, so both will be fully consumed.

Step 3: Calculate the number of moles of NaOH added in the titration. Since the volume of NaOH is given, we can use the equation:

Moles of NaOH = Molarity * Volume

For both cases:
Moles of NaOH = (0.1 mol/L) * (2.50 mL/1000 mL/L) = 0.0025 mol

Moles of NaOH = (0.1 mol/L) * (9.50 mL/1000 mL/L) = 0.0095 mol

Step 4: Determine the volume of the NaOH solution at the equivalence point. The equivalence point occurs when the moles of the acid equal the moles of the base. In this case, the moles of NaOH added is the same as the initial moles of the acid. Since the molarity of NaOH is also given (0.1 M), we can calculate the volume at the equivalence point using the equation:

Volume at equivalence point = Moles of acid / Molarity of NaOH

For HCl:
Volume at equivalence point = 0.008 mol / 0.1 mol/L = 0.08 L = 80 mL

For HC2H3O2:
Volume at equivalence point = 0.008 mol / 0.1 mol/L = 0.08 L = 80 mL

Step 5: Compare the volume of NaOH added with the volume at the equivalence point. From the given volumes, we can see that 2.50 mL and 9.50 mL are both less than the volume at the equivalence point (80 mL). Therefore, the volumes of NaOH added in both cases are before the equivalence point.

Step 6: Calculate the percent error between the theoretical and actual values found on the titration curves. To do this, you would need the actual pH values obtained from the lab.

Calculate the theoretical pH using the Henderson-Hasselbalch equation for both cases:

For HCl:
Theoretical pH = -log([HAcid] / [Salt]) = -log(0.008 mol / 0.08 L) = -log(0.1) = 1.00

For HC2H3O2:
Theoretical pH = -log([HAcid] / [Salt]) = -log(0.008 mol / 0.08 L) = -log(0.1) = 1.00

To calculate the percent error, you need to subtract the actual pH value obtained from the lab from the theoretical pH value, divide by the theoretical pH value, and multiply by 100:

Percent Error = (Actual pH - Theoretical pH) / Theoretical pH * 100

Remember to substitute the actual pH values obtained from the lab to complete the calculation.

I hope this explanation helps you understand how to calculate the theoretical pH and compare it with the actual values found in the lab.