NEED HELP A 50.50-mL sample containing La3+ was treated with sodium oxalate to p

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NEED HELP

A 50.50-mL sample containing La3+ was treated with sodium oxalate to precipitate La2(C2O4)3, which was washed, dissolved in acid, and titrated with 18.19 mL of 0.006352 M KMnO4. Write the titration reaction. (Use the lowest possible whole number coefficients. Omit states-of-matter from your answer.)



Find [La3+] in the unknown.

VA 502215 Assignment 13 h C ponses/submit?dep 4204284 012 points Previous Answers My Notes Ask Your Teacher A 50.50-ml sample containing La3+ was treated with sodium oxalate to precipitate La2(C204)3, which was washed, dissolved in acid, and titrated with 18.19 ml of 0.006352 M KM 4. Write the titration reaction. n se the lowest possible whole number roefficients. Omit states-of-matte from Your answer.) g Help chen Pad Greek 2+ 6 +5C20 2Mn +10CO2 8H20 2+ GH 5C 20 2Mn 2 OCO 218H 20 ur answer uonudirls arribiguous ol plete reac equation. Check all the components on the product side of the equation. rind [La3+] n the unknown. 3 816-3 Supporting Materlals Ea Physical Constants Supplemental Periodic Table Materials 7, 72 points Frewious Answer My Notes Ask Your Teacher In which technique, iodimetry or iodometry, is starch indicator not added until just betore the end poin Why? Starch is not added until just betore the end point in iodometry Other starch will irreversibly to iodine rvise, Supporti ng Materials 28 PM Ask me any

Explanation / Answer

Reaction is

5 C2O42- + 2 MnO4- + 16 H+ --> 2 Mn2+ + 10 CO2 + 8 H2O

Moles of MnO4- = volume * concentration of KMnO4

= 18.19/1000 * 0.006352 = 0.000115543 mol

Moles of C2O42- = 5/2 * moles of MnO4-

= 5/2 * 0.000115543 = 0.0002888575 mol

Moles of La3+ = 2/3 * moles of C2O42-

= 2/3 * 0.0002888575 = 0.00019257166 mol

Molarity = moles/volume of La3+

= 0.00019257166/0.05050

= 0.0038133 M = 3.813 * 10^(-3) M = 3.813 mM

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