LAB REPORT WORKSHEETS and/or is not considered a major species. To decide which

Question

LAB REPORT WORKSHEETS and/or is not considered a major species. To decide which are the major species, consider what reactants were put into molecules, include the spectator ions. Since water is the solvent, it the titration beaker at each stage, and what, if any, chemical reactions between them have occurred. i. Before any NaOH is added (at the start of the titration). Carefully think about what you have put into your solution at this point i In the buffer region. Carefully think about what species must be present to make the buffer. Are there any spectator ions? At the equivalence point. What is equivalent at this point? What has been used up? What is left? Are there any spectator ions? iv. Beyond the equivalence point. What has been added since the equivalence point? Has anything reacted further since the equivalence point? Concentration of stock NaOH: (from your brown bottle) Volume of NaOH corresponding to the equivalence point: (from your graph) Calculate the Concentration of "unknown" acetic acid using the equivalence point volume from your graph: Complete any additional questions on the online General Lab Marker System (ULAB) for Part 1. 164

Explanation / Answer

Ans. Balanced reaction: CH3COOH(aq) + NaOH(aq) --------> CH3COONa(aq) + H2O(l)

Stoichiometry: One mol NaOH neutralizes 1 mol acetic acid, CH3COOH.

# Step 1: Calculate Moles of NaOH consumed to reach equivalent point:

Moles of NaOH consumed = Molarity x Volume of NaOH consumed in liters

                                                = 0.1208 M x 0.01605 L

                                                = 0.00193884 mol

Step 2: Calculate moles of CH3COOH using #step 1:

Following the stoichiometry of neutralization, the number of moles of NaOH consumed to reach equivalence point must be equal to the moles of acetic acid taken in sample being titrated.

So,

            Moles of CH3COOH in sample = moles of NaOH consumed

            Hence, Moles of CH3COOH in sample = 0.00193884 mol

# Calculate [CH3COOH] in sample: It would depend on the way you prepared the acetic acid solution and the way you want to solve it. Two approaches are presented (they don’t affect the result, though)

I. Some unknown was dissolved in water to prepare sample: Let the final volume of acetic acid solution 10.0 mL = 0.010 L. It would serve as standard solution which was titrated against NaOH.

Now,

[CH3COOH] in standard solution = moles of acetic acid / Volume of solution in liter

                                                            = 0.00193884 mol / 0.010 L

                                                            = 0.193884 M

Therefore, [CH3COOH] in standard solution = 0.193884 M

II. If an unknown solution, 10.0 mL was provided:

At equivalence point,           (M1V1), NaOH = (M2V2), acetic acid

            Or, 0.1208 M x 0.01605 L = M2 x 0.010 L

            Or, M2 = (0.1208 M x 0.01605 L) / 0.010 L

            Hence, M2 = 0.193884 M

Therefore, [CH3COOH] in unknown solution = 0.193884 M

# Note: Use the actual value of unknown solution to get the correct value.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.