Question

If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?

Answer 1

`[Cu^(2+)]=0.041 M`

Step-by-step explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

`[Cu^(2+)]=0.041(molCu(NH_3)_4^(2+))/(L)*(1molCu^(2+))/(1molCu(NH_3)_4^(2+)) =0.041 (molCu(NH_3)_4^(2+))/(L)`

`[Cu^(2+)]=0.041 M`

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