Can someone please fully work out this problem? A 32.23- ml. sample of was treat
ID: 1008557 • Letter: C
Question
Can someone please fully work out this problem?
A 32.23- ml. sample of was treated with excess Na_2C_2O_4 to precipitate La_2(C_2O_4)_3, which was washed, dissolved in acid, and titrated with 13.29 mL of 0.004455 M KMnO_4 to reach a purple end point. A black titration required 0.03 mL. KMnO_4 to produce enough color to be seen. Find the La^3 molarity in the unknown Reactions: 2La^3+. 3C_2O_4^2 rightarrow La_2(C_2O_4)(8) La_2(C_2O_4)_3 (s) + 6H^+ rightarrow 21.a^3- + 3H_2C_2O_4 5H_2C_2O_4 + 2MnO_4 + 6H rightarrow 10CO_2 + 2Mn^2+ 8H_zO (purple)Explanation / Answer
The reactions are
2La+3 + 3C2O4-2 --> La2(C2O4)3
La2(C2O4)3 + 6H+ --> 2La+3 + 3H2C2O4
5H2C2O4 + 2MnO4- + 6H+ --> 10|CO2 + 2Mn+2 + 8H2O
Volume of La+3 = 32.23mL
The moles of ppt of La2(C2O4)3 = 0.5 X moles of La+3 initially taken (as per balanced equation)
Moles of oxalic acid produced by acid treatement = 3 X moles of La2(C2O4)3
moles of KMnO4 used = 5/2 moles of oxalic acid
Moles of KMnO4 used = Molarity X volume
we should substract the excess volume of KMnO4 used = 0.03
Moles of KMnO4 used = 0.004455 X 13.26 = 0.059 millimoles
Now let us do the back calculation
Moles of Oxalic acid produced = 2/5 X moles of KMnO4 = 0.059 X 2 / 5 = 0.0236 milimoles
Moles of La2(C2O4)3 = 1/3 X moles of oxalic acid = 0.0236 / 3 = 0.00787 millmoles
Moles of La+3 = 2X moles of La2(C2O4)3 = 0.0157 millimoles
Molarity = Millimoles / Volume in mL = 0.0157 / 32.23 = 4.88 X 10^-4 molar