A 0.4840 g sample was treated to separate Bi3* from other metals. The resulting

Question

A 0.4840 g sample was treated to separate Bi3* from other metals. The resulting solution was treated with an excess of MgY2 and, after adjusting the pH to 10, the liberated Mg2+ (BiY is more stable than Mgr2) requireo 22.61 mL of 0.0122 M EDTA for titration to the EBT endpoint, what is the %Bi in the sample? a.) 2.30 % b) 9.17% a) 10.5 % d) 11.9% 8) Cr3+ reacts slowly with EDTA (Y4) so it is usually analyzed using back-titration. A 100.0 mL treated water sample containing only Cr3+ was diluted to 1 L and 25.O mL 0.005 M Y4- was added to a 10 mL aliquot of the diluted sample. The excess Y4- required 23.0 mL 0.005 M CaCl2 to reach the EBT end-point. Find the ppm Cr3+ in the water sample. a.) 5.2 ppm b.) 52 ppm 9) c.) 520 ppm d.) 5200 ppm

Explanation / Answer

Question 8.

Formula: Molarity = no. of moles/volume (L)

For EDTA: 0.0122 mol/L = no. of moles of EDTA/22.61*10-3 L

i.e. The no. of moles of EDTA = 0.0122 * 22.61*10-3 mol = 0.275842*10-3 mol

i.e. The sample contains 0.275842*10-3 mol * 208.98 g/mol = 0.0576 g

Therefore, the percentage of Bi in the sample = (0.0576/0.4840)*100 = 11.9 % (option D)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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