did i do this right

Question

calculate the con of cu(nh4)^2+ of solution. Mass of cuso4*5h20=.6000g

First i dissolved the sample into 50.00ml of water. I filled up volumetric flask to 100.00 ml mark

I then, pipetted 10ml of this sample into a 50.00ml volumetric flask. I then topped it to the mark. Then I added 5ml of 6M NH3

I just did it this way.

.6000g x m/249.71g=2.403 x10^-3

2.403 x10^-3/.100l= 2.403 x 10^-2M

M1v1=m2v2

M2=m1v1/v2

=(2.403 x 10^-3M)(100.00ml)/50.00ml

i got 4.806 x 10^-3

Not sure about the NH3 that was added though ?

Answer 1

6000g x m/249.71g=2.403 x10^-3
This is ok.
2.403 x10^-3/.100l= 2.403 x 10^-2M
This is ok

M1v1=m2v2

M2=m1v1/v2

=(2.403 x 10^-3M)(100.00ml)/50.00ml
Didn't you use 10 mL of the 2.403 x 10^-2 M and make it to 50? Then
2.403 x 10^-2 x 10 = M x 50 and solve for M.
I would do this.
0.6/249.71 = 0.002403 moles in 100 mL. Then you take 10 mL of that and make it to 50, then add 5 mL 6M NH3.
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
moles Cu^+2 = 0.002403 moles x 0.1 (the 10 mL is 1/10 of the 100 mL sample) = 0.0002403 moles.
That reacts with 6M x 0.005 L = 0.03 moles NH3 to form 0.0002403 moles Cu(NH3)4^+2. So the concn of the NH3 complex is 0.0002403/0.055 = ??M
Check my thinking. Check my arithmetic. By the way, I think the complex is Cu(NH3)4^+2 and not Cu(NH4)4^+2.

Answer 2

k thanks

it says cu(nh4)2+ in my pre lab.

Answer 3

k thanks

it says cu(nh4)2+ in my pre lab.

yep. And then we added 5ml of 6M NH3

That step messed me up :)

Answer 4

k thanks

it says cu(nh4)2+ in my pre lab.

yep. And then we added 5ml of 6M NH3. Then duluted to 50.00ml mark.

Answer 5

Sorry, my browswer is going crazy.

Here's exactly what it says for part B

Use 10ml pipet to transfer 10.00ml into a clean 50.00ml flask. In the fumehood add about 5ml of 6M Nh3. Dilute this to 50.00ml using distilled water.

Answer 6

thanks.

your method works out.

Ill stop when i have questions writing it up :)

Answer 7

It may say Cu(NH4)^2+ but what is formed is Cu(NH₃)₄²⁺