Question: Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni2+ solution (containing excess Ni2+) to convert the cyanide into tetracyanonickelate(II): 4CN- + Ni2+

Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni2+ solution (containing excess Ni2+) to convert the cyanide into tetracyanonickelate(II):
4CN- + Ni2+ → Ni(CN)2-4
Excess Ni2+ was then titrated with 10.15 mL of 0.01307 M EDTA. Ni(CN)42- does not react with EDTA. If 39.35 mL of EDTA were required to react with 30.10 mL of the original Ni2+ solution, calculate the molarity of CN- in the 12.73 - mL sample.

Step by Step Solution

3.39 Rating (155 Votes )

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

3010 mL Ni 2 reacted with 3935 mL 001307 M EDTA so Ni 2 molarity is Ni ... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Document Format (1 attachment)

Word file Icon

878-E-C-E-E-C (1960).docx

120 KBs Word File

Students Have Also Explored These Related Chemical Engineering Questions!