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Question

You have prepared a solution by diluting 10.00mL of 0.961M copper(II) nitrate to a total volume of 50.00mL. What is the concentration of Cu2+ ions in the resulting solution? Report your answer to 3 decimal places.

Answer 1

Cu(NO₃)₂ dissociates in water as follows:

Cu(NO₃)₂ <==> Cu²⁺ + 2NO₃^-

For every mole of Cu(NO₃)₂ you get 1 mole of Cu²⁺ and 2 mols of NO₃^-

(10.00 ml)(0.961 M) = (50.00 ml)(x M)

x = 0.1922 M Cu(NO₃)₂

[Cu²⁺] = 0.192 M (to 3 decimal places)

Answer 2

The concentration of Cu²⁺ initially is .010 liters * .96 mole/l of Cu(NO₃)₂ * 1 mole of Cu²⁺/ mole Cu(NO₃)₂

You dilute the solution to a V of 5 x, therefore the concentration will be 1/5 as high.

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