Section : Name: PRE-LABORATORY QUESTIONS (WEEK I) EXPT. 7A: GROUP PROJECT: QUANT

Question

Section : Name: PRE-LABORATORY QUESTIONS (WEEK I) EXPT. 7A: GROUP PROJECT: QUANTITATIVE ANALYSIS OF SEA WATER To be completed and handed in prier to the start of the laboratory period. Show all work. 1. (6 points) Chloride ion is determined by a potentiometric titration using 0.200 M AgNO, ast titrant. AgCI (8) is formed during the titration by the following reaction: Ag, (aq) + cr (aq) AgCl (s) A 1:10 dilution of an unknown water sample is made and a 25 mL aliquot of the diluted samples is placed into a 100-mL beaker. Two electrodes (Ag wire and Ag/AgCI reference electrode) are immersed, the voltage reading between the electrodes is allowed to stabilize, and the initial voltage recorded. Nineteen titration aliquots of 0.100 mL of 0.200 M AgNO, are made with an automatic pipettor. After each aliquot, the solution is stirred, the voltage reading stabilizes, and the voltage in mV recorded. The following graph of Potential (m V) versus volume of AgNO, titrant is made. Potentiometric titration of chloride 380.0 360.0 340.0 320.0 300.0 280.0 260.0 240.0 220.0 180.0 140.0 120.0 1000 5 8 8 Volume of AgNO, (mL) a) The endpoint of this titration is the midpoint of the steepest vertical rise in potential. Estimate the volume of titrant added to reach the endpoint. b) How many moles of AgNO, have been added at the endpoint? How many moles of CI have reacted at the endpoint?

Explanation / Answer

Potentiometric titration of chloride ion with AgNO3 solution

From the graph shown above,

a) The end point volume of titrant (AgNO3) added = 1.30 ml

b) mols of AgNO3 added = mols of Cl- present in diluted sample

                                        = 0.200 M x 0.0013 L = 0.00026 mol

c) molar concentration of Cl- in diluted sample = mols/L of solution

                                                                           = 0.00026 mol/0.025 L

                                                                           = 0.0104 M

d) molar concentration of Cl- in original solution = 0.0104 M x 10 = 0.104 M

e) moles of BaSO4 formed = 0.068 g/233.38 g/mol = 0.0003 mol

moles of Sulfate present in water sample = 0.0003 mol

average molar concentration of sulfate in unknown water sample = 0.0003 mol/0.01 L

                                                                                                        = 0.03 M

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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