Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

-Titration experiment: Please help. Help is greatly appreciated. (Strong acid- S

ID: 1035314 • Letter: #

Question

-Titration experiment: Please help. Help is greatly appreciated.

(Strong acid- Strong base)--->HCl +NaOH

(weak-acid - strong base)-->CH3COOH + NaOH

Please help me to calculate the following:

- Theoretical Volume of NaOH needed to reach endpoint for both HCl and CH3COOH

The Acidity Constant (Ka) of HCl

– Percent Error of HCl Ka ?

Theoretical Value = 1.3 x 106

– The Acidity Constant (Ka) of CH3COOH

? HINT: You can use pKa to find Ka

– Percent Error of CH3COOH Ka

? Theoretical Value = 1.8 x 10-5

Concentration of HCl: 0.0969M

Concentration of NaOH: 0.107M

Concentration of CH3COOH: 0.103M

Volume of HCl and CH3COOH in flask are both: 10 mL

---Here is the titration of HCl with NaOH

---Here is the titration of CH3COOH with NaOH

Questions:

Why is it not necessary to use phenolphthalein indicator to detect the endpoint in this titration?

? Explain why the weak acid-strong base titration equivalence point is not at pH= 7.0.

? What would the titration curve look like if the NaOH was being titrated by the CH3COOH?

– Explain your answer.

? Discuss the % Error. What is the major source of error?

Thank you

Volume NaOH added (mL) pH 1.0 1.02 2.0 1.30 3.0 1.45 4.0 1.34 5.0 1.54 6.0 1.60 7.0 1.90 8.0 2.16 9.0 9.63 10.0 11.65 11.0 12.25 12.0 12.50 13.0 12.48 14.0 12.55 15.0 12.60 16.0 12.61 17.0 12.30

Explanation / Answer

HCl + NaOH

theoretical volume of NaOH to reach end point = 0.0969 M x 10 ml/0.107 = 9.0 ml

pKa = -log[Ka] = 1.40

Ka = 3.98 x 10^-2

Percent error = (1.3 x 10^6 - 3.98 x 10^-4) x 100/1.3 x 10^6 = 99%

Is the Ka reported for HCl is correct, we see a huge error in measurement. This could be due to incorrect measurement of the amount of NaOH added to reach end point, or incorrect measurement of initial volume of HCl taken

CH3COOH + NaOH

Theoretical volume of NaOH to reach end point = 0.103 M x 10 ml/0.107 M = 9.6 ml

pKa from data = 4.5

Ka = 3.2 x 10^-5

percent error = (3.2 x 10^-5 - 1.8 x 10^-5) x 100/1.8 x 10^-5 = 78%

The veru high percent error could be due to incorrect measurement of the amount of NaOH added to reach end point, or incorrect measurement of initial volume of CH3COOH taken

--

Phenolphthalein is not needed here as the pH change is very different when all the acid is neutralized as compared to no base added and easily identifiable.

The weak acid at equivalence point is present in conjugate base form, which hydrolyzes and produces OH- in solution.Therefore, the pH at equivalence point is greater than pH 7.

The titration curve would be inverted S-shaped if NaOH is titrated with CH3COOH.

Major sources of error is due to inccorect measurement of end point of titration.