Sample Exam lll.docx 100% as Charts Styles 5. (0 points) A solution contains La\

Question

Sample Exam lll.docx 100% as Charts Styles 5. (0 points) A solution contains La" and Scy" ions. A50.00 mL aliquot is taken, 10.00 mL ofpH-8.00 buffer is added, and the solution is titrated with 0,003142 M disodium EDTA. It required 21.37 mL of the EDTA solution to reach the first equivalence point and a total of45.88 mL to reach the second. (Hint: Think of the sequence.) a) What are the concentrations of lanthanum and scandium ions in the original solution? c) If a total of50.00 mL of the EDTA solution is added, what is the concentration of the EDTA ion in the resulting solution? words: 205 of 697 HL 1 Pages: 3 of 6

Explanation / Answer

La3+ and SC3+ make a solution.

Aliquot = 50 ml

Buffer = 10 ml, pH = 8

Solution titrated with = 0.003142 M disodium EDTA

I equivalent solution = 21.37 ml

II equivalent solution = 45.88 ml

1) La3+ = 0.001540 M

  SC3+ = 0.001343 M

Atomic Mass of La = 138.90 amu

Atomic Mass of Sc = 44.955 amu

Moles of EDTA added to the solution = (0.003142 M) x 21.37 mL (or 0.02137 L)

= 0.003142 M x 0.02137 L = 0.000067144 mol or 6.714 x 10-5 mol

Excess moles of EDTA = (0.003142 M) x 45.88 mL (or 0.0458 L)

= 0.003142 M x 0.04588 L = 0.00014415 mol or 1.44 x 10-4 mol

Concentration of La3+ = 1.44 x 10-4 mol / 0.010 L

= 0.00154 M

Concentration of Sc3+ = 6.714 x 10-5 mol / 0.050 L

= 134.28 x 10-5 mol or 0.0013428 M

2) 1.1 x 10-12

Concentration of EDTA which remains free = (0.003142 M) x (45.88 mL - 21.37 mL) (or 0.0458 L - 0.02137 L)

= 0.003142 M x 24.51 mL (or 0.02451 L)

=0.07701042 M/L or 0.00007701042 M or 7.7 x 10-5 M

So, concentration of Sc3+ which remains free = (7.7 x 10-5 M) - 6.714 x 10-5 M

= 0.0000113342 M

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