A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni 2+ s

Question

A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
excess Ni2+) to convert the cyanide totetracyanonickelate(II):
4CN- + Ni2+ ------Ni(CN)42-
The excess Ni2+ was then titrated with 10.15 mL of0.01307 M ethylenediaminetetraacetic acid
(EDTA). One mole of this reagent reacts with 1 mol ofNi2+:
Ni2+ + EDTA4- ----Ni(EDTA)2-
Ni(CN)42- does not react with EDTA. If 39.35mL of the 0.01307 M EDTA was required to react
with 30.10 mL of the original Ni2+ solution, calculatethe molarity of CN- in the 12.73 mL cyanide
sample.

Explanation / Answer

39.35 mL *(1 L /1000 mL ) *(0.01307 mol EDTA/ 1 L) = 0.0005143045moles EDTA => moles Ni2+ moles Ni2+ = 0.0005143045 mol EDTA *( 1 mol Ni2+ / 1 molEDTA) = 0.0005143045 moles Ni2+ concentration of Ni2+ = 0.0005143045 moles Ni2+ / 30.10mL   *(1000 mL/ L) = 0.0170865282 M Ni2+ of this 30.10 mL, only 25.00 mL was used to react with CN- moles CN- = 0.0170865282 M Ni2+ *( 25 mL *(1 L /1000mL)) =0.000427163205 moles CN- concentration CN- = 0.000427163205 moles CN- / (12.73 mL *(1 L/1000mL)) = 0.0335556328 M concentration CN- = 0.03356 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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