The silver ion in a 25.00-mL sample was converted to dicyanoargentate (1) ion by the addition of an excess solution containing Ni(CN),²: Ni(CN)42 + 2 Ag → 2Ag(CN)21 + Ni2 The liberated nickel ion was titrated with 43.77-mL of 0.02408M EDTA. Calculate the molar concentration of the silver solution. Molar Masses: Ni = 58.69 Ag 107.9 %3!
The silver ion in a 25.00-mL sample was converted to dicyanoargentate (1) ion by the addition of an excess solution containing Ni(CN),²: Ni(CN)42 + 2 Ag → 2Ag(CN)21 + Ni2 The liberated nickel ion was titrated with 43.77-mL of 0.02408M EDTA. Calculate the molar concentration of the silver solution. Molar Masses: Ni = 58.69 Ag 107.9 %3!
Chapter17: Complexation And Precipitation Reactions And Titrations
Section: Chapter Questions
Problem 17.35QAP
Related questions
Question
Answer provided: 0.08432M Ag^+1
Please show your complete solution and write your answer clearly and readable. Thank you.
![The silver ion in a 25.00-mL sample was converted to dicyanoargentate (l) ion by the addition of an
excess solution containing Ni(CN),²:
Ni(CN)4? + 2 Ag → 2Ag(CN)21 + Ni2
The liberated nickel ion was titrated with 43.77-mL of 0.02408M EDTA. Calculate the molar
concentration of the silver solution.
Molar Masses: Ni
58.69
Ag
107.9
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff99cf4bf-ebe1-4ebf-8f00-91eabd7e6b5d%2Ff6bdb678-be6b-444a-b042-ec84028aa45b%2Fyiqws5i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The silver ion in a 25.00-mL sample was converted to dicyanoargentate (l) ion by the addition of an
excess solution containing Ni(CN),²:
Ni(CN)4? + 2 Ag → 2Ag(CN)21 + Ni2
The liberated nickel ion was titrated with 43.77-mL of 0.02408M EDTA. Calculate the molar
concentration of the silver solution.
Molar Masses: Ni
58.69
Ag
107.9
%3D
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