How would I go about on calculating the volume of stock required to prepare my d

Question

How would I go about on calculating the volume of stock required to prepare my desired concentrations of [Cu2+]? I believe that I would need to use the M1V1=M2V2 formula in this situation, but I'm not sure what goes what into the formula.

1. Reagents CU(NO3)2.3H20 salt 0.1 M HCI 2. Procedures This week, you are only going to prepare your solutions for the next two weeks of lab experiments. You will need to make 10 solutions in total . You will only need 50 mL of each solution. Each solution should use 0.1 M HCI as the solvent The final Cu concentrations of each solution should be: 10, 20, 30, 40, 60, 100, 200, 400, 600, 800, and . 1000 ?M .You should first prepare a stock solution of appropriate concentration that can be diluted for each of the following standard solutions. You can also dilute one of the other solutions .Your dilutions need to be made with volumes that we can easily measure with volumetric pipettes (5, 10, 15 20, 25 mL, etc.) Feel free to work together on your calculations. A good place to automate some of this would be in Excel, it will save you an awful lot of time... Table 1: Helpful suggestions for calculating solution volumes Volume Stock Total stock required, volu Desired concentration, mM Solution [Cu2+], ?? mL 1000 800 600 400 200 100 80 60 40 20 4 50 mL 6 7 9 10

Explanation / Answer

Solution (1) can be prepared and used as stock to prepare the rest of the solutions by dilution.

[Cu2+]1=1000uM=1000*10^-6 mol/L=10^-3 mol/L=0.001mol/L

So, V1=volume of solution 1=1L (say)

molar mass of Cu(NO3)2.3H2O=241.602 g/mol

mass of Cu(NO3)2.3H2O required to prepare 1L of 0.001M solution=0.001mol *(241.602 g/mol)=0.2416 g

So, 0.2416g salt is to be taken in a volumetric flask of 1L and dissolved in little0.1M HCl and then the HCl solution added upto the mark.

50 ml of this solution is our solution 1.Rest can be used as stock to prepare the rest.

for the table:

Concentration of stock =1*10^-3 mol/L=1mM

volume of stock=50ml

total volume=50ml

Solution 2)

[Cu2+]stock=1mM

[Cu2+]sol2=800uM=800*10^-6 M=8*10^-4 M=0.8mM

Volume of stock=Vstock

Volume of sol 2=50ml

Vstock*[Cu2+]stock=Vsol2*[Cu2+]sol2

Vstock=(Vsol2*[Cu2+]sol2)/[Cu2+]stock=50ml*(0.8mM/1mM)=40ml

sol 3)

[Cu2+]stock=1mM

[Cu2+]sol3=600uM=600*10^-6 M=6*10^-4 M=0.6mM

Volume of stock=Vstock

Volume of sol 3=50ml

Vstock*[Cu2+]stock=Vsol3*[Cu2+]sol3

Vstock=(Vsol3*[Cu2+]sol3)/[Cu2+]stock=50ml*(0.6mM/1mM)=30ml

soln 4)

[Cu2+]stock=1mM

[Cu2+]sol4=400uM=400*10^-6 M=*10^-4 M=0.4mM

Volume of stock=Vstock

Volume of sol 4=50ml

Vstock*[Cu2+]stock=Vsol4*[Cu2+]sol4

Vstock=(Vsol4*[Cu2+]sol4)/[Cu2+]stock=50ml*(0.4mM/1mM)=20ml

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