PRELAB QUESTIONS (RESPOND ON A SEPARATE SHEET OF PAPER): 1. Assume that a drop h

Question

PRELAB QUESTIONS (RESPOND ON A SEPARATE SHEET OF PAPER): 1. Assume that a drop has a volume of 0.05 mL. If a titration requires 30.00 mL for completion, what % error will each extra drop over 30.00 mL cause? 2. For a tablet containing 500 mg of vitamin C, calculate how many mL of 0.095 M NaOH are required (molar mass of ascorbic acid is 176.12 g/mol). a tablet containing 1000 mg of vitamin C, calculate the minimum concentration of the NaOH solution so that the titration would take no more than 45.00 mL of the NaOH from the buret. 3. For 4. A student collected the following data in the laboratory: Mass of flask Mass of flask and sample Concentration of NaOH Initial buret reading Final buret reading 52.8741 g 53.4293 g 0.1010 M 0.50 mL 23.78 mL Calculate the mass percent of ascorbic acid in the sample.

Explanation / Answer

Ans. #1. % error = (Excess volume added / Actual required volume) x 100

                                    = (0.05 mL / 30.00 mL) x 100

                                    = 0.167 %

#2. Moles of vitamin C = Mass in grams / Molar mass

                                    = 0.500 g / (176.12 g/ mol)

                                    = 0.002839 mol

Balanced reaction: C6H7O6 (aq) + NaOH (aq) ----> H2O (l) + NaC6H7O6 (aq)

Stoichiometry: 1 mol Vitamin C / ascorbic acid (C6H7O6) is neutralized by 1 mol NaOH.

So, at equivalence point, total number of moles of NaOH consumed is equal to total number of moles of ascorbic acid present in solution.

So, minimum number of moles of NaOH required = 0.002839 mol

Now,

            Volume of NaOH required = Required number of moles / Molarity

                                                = 0.002839 mol / 0.095 M

                                                = 0.002839 mol / (0.095 mol/ L)

                                                = 0.02988 L

                                                = 29.88 mL

#3. Moles of vitamin C = 1.0 g / (176.12 g/ mol) = 0.005678 mol

Following stoichiometry of balanced reaction, required moles of NaOH = 0.005678 mol

Given, maximum volume of NaOH solution = 45.00 mL = 0.045 L

Now,

            Desired molarity of NaOH = Moles / Volume of solution in liters

                                                            = 0.005678 mol / 0.045 L

                                                            = 0.1262 M

#4. Mass of sample = Mass of flask plus sample – Mass of flask

                                    = 53.4293 g – 52.8741 g

                                    = 0.5552 g

# Volume of NaOH consumed = Final burette reading – Initial burette reading

                                    = 23.78 mL – 0.50 mL

                                    = 23.28 mL

                                    = 0.02328 L

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

                                    = 0.1010 M x 0.02328 L

                                    = 0.00235128 mol

Following stoichiometry of balanced reaction, moles of ascorbic acid present in sample must be equal to the moles of NaOH consumed.

So,

            Moles of ascorbic acid in sample = 0.00235128 mol

            Mass of ascorbic acid = 0.00235128 mol x 176.12 g mol-1 = 0.4141074336 g

Now,

            % Ascorbic acid = (Mass of ascorbic acid / Mass of sample) x 100

                                                = (0.4141074336 g / 0.5552 g) x 100

                                                = 74.59 %

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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