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Standard addition. An unknown sample of Ni2 gave a current of 2.36 A in an
Chapter 5, Problem 5-B(choose chapter or problem)
Standard addition. An unknown sample of Ni2 gave a current of 2.36 A in an electrochemical analysis. When 0.500 mL of solution containing 0.028 7 M Ni2 was added to 25.0 mL of unknown, the current increased to 3.79 A. (a) Denoting the initial, unknown concentration as [Ni2 ]i, write an expression for the final concentration, [Ni2 ]f, after 25.0 mL of unknown were mixed with 0.500 mL of standard. Use the dilution factor for this calculation. (b) In a similar manner, write the final concentration of added standard Ni2 , designated as [S]f. (c) Find [Ni2 ]i in the unknown. 5
Questions & Answers
QUESTION:
Standard addition. An unknown sample of Ni2 gave a current of 2.36 A in an electrochemical analysis. When 0.500 mL of solution containing 0.028 7 M Ni2 was added to 25.0 mL of unknown, the current increased to 3.79 A. (a) Denoting the initial, unknown concentration as [Ni2 ]i, write an expression for the final concentration, [Ni2 ]f, after 25.0 mL of unknown were mixed with 0.500 mL of standard. Use the dilution factor for this calculation. (b) In a similar manner, write the final concentration of added standard Ni2 , designated as [S]f. (c) Find [Ni2 ]i in the unknown. 5
ANSWER:Step 1 of 3
(a)
The volume of the unknown solution is 25.0 mL.
The volume of the standard is 0.500 mL.
Therefore,
Concentration is inversely proportional to volume, therefore, the ratio of initial concentration of the unknown to its final concentration is given as:
