[Wollmy]

I have a question about the preparation of a solution. I have a solid Cu(NO3)2 · 3H2O and I want to make solutions with differente Concentration [M] (0,005; 0,010; 0,015 and 0,020) using NH3 30% (density 0,89) as a solvent.
My first though was to simply Multiply the desired Molar Concentration with the Volume of NH3 and the Molmass of the Cu(NO3)2 · 3H2O and so I would obtain the amount of solid I needed to weight. For example: to obtain a Molar Concentration of 0,010: 0,010 mol.L^-1 × 0,020 L × 241,62 g.mol^-1 and thus obtain the amount of 0,048324g needed to be weight to obtain a 0,010 M solution. For me this doesnt seem right.
My other though was to make a solution i.e. 0,05M of Cu(NO3)2 · 3H2O with water and a 1M solution of NH3 with water. The desired Molar concentrations I could achieve with dilution, right? If this is doable, im kinda lost and its kinda frustrating because im not seeing how hahahah.

So to summarize it:

- Can I work with the solid beeing dissolved directly in the Solvent and calculate the Concentration by just weighing the Mass of the Cu(NO3)2 · 3H2O and adding the Volume of NH3? (If yes, it would be great if you could show how to calculate it)

- Is it better to prepare 2 solutions. One solution dissolving Cu(NO3)2 · 3H2O in water and prepare i.e. 0,05M solution and prepare another solution of NH3 with water i.e. 1M and then mixture both to the disired concentration? (If yes, it would be great if you could show how to calculate it)


Many many thanks for any input

[mjc123]

What's the molar concentration of a 30% ammonia solution?

[Wollmy]

What's the molar concentration of a 30% ammonia solution?

On the bottle it doesnt give the information, it only says: Molar mass (M) 17,03 + aq g/mol. I found a table and also calculated it as 15,67 M
0,30(g NH3 / g aq. sol.) × 0,89(g aq. sol. / mL) × (1000 mL/L)/(17.03 g NH3/mol NH3) = 15,67 M (as NH3).

[mjc123]

What does that say about the possibility of using 1M ammonia to prepare your solutions?

[Wollmy]

What does that say about the possibility of using 1M ammonia to prepare your solutions?

Sorry, didnt understand your question. Is it if I could work with a 1M solution or how I would prepare the 1M Solution?

If its referring to dilute the probe, I did it by adding 3,189 mL of the NH₃ 30% in a 50mL Volumetric flask and adding Water until the 50 mL menisco.

As I mentioned, im kinda lost because im overlooking something or not seeing how to proceed in preparing my solutions

[mjc123]

How would you use a 1M solution to prepare a solution in 30% NH₃?

[Wollmy]

I think I managed it, would be great if someone could check and see If this makes logic.

Side note before I write down what I calculated: The Ammonia Volume and Water I want to keep constant, because the purpose is to do a calibration curve with a UV-VIS. For Ammonia I calculated I would need 8,33mL of NH 30% to have 2,5mL of Ammonia. For water I chose 20 mL.

I started with fixating the concentration of Cu(NO₃)₂, since I want to measure the Cu in further solutions. Here I chose 0,005M.
With this I calculated the needed mass of Cu(NO₃)₂ to achieve this concentration.
 
m_Cu(NO3)2=M_Cu(NO3)2×MM_Cu(NO3)2×V_NH3×(1L/1000mL)

With the obtained mass of Cu(NO₃)₂, I calculated the amount of mols of Cu(NO₃)₂.

n_Cu(NO3)2=m_Cu(NO3)2/MM_Cu(NO3)2

With the amount of mols I can calculate the amount of Molecules of Water in my solute. Since my Copper(II) nitrate is trihydrated  I will have 3 mols of H₂O for every mol of Cu(NO₃)₂. And with the number of mols I can obtain the mass and Volume of Water.

n_H2O=3×n_Cu(NO3)2

m_H2O=n_H2O×MM_H2O

V_Cu,H2O= m_H2O / ρ_H2O  I assumed the density (ρ) as 0,997 g/mL

Adding the m_Cu(NO3)2 and m_H2O I'll have the amount of mass Cu(NO₃)₂ · 3H₂O (m_{Cu(NO3)2 · 3H2O}) I have to weight.
m_{Cu(NO3)2 · 3H2O} = m_Cu(NO3)2 + m_H2O

Now to the Volume of Ammonia and Water. To obtain the total of Water I have to add to my system, I first calculated the amount of Water that is add when I want 2,5 mL of Ammonia.
V_NH3,H2O: Volume of Water contained in the Volume of NH₃ 30%

V_NH3,H2O = (2,5×100)/30 - 2,5

The total amount of Water I have to add to have constant 20 mL is:
V_T = V_H2O - V_Cu,H2O - V_NH3,H2O The value of V_H2O is 20mL

I got to a Value of:
m_{Cu(NO3)2 · 3H2O} = 0,00302 g
V_T = 14,16598 mL

Thanks again for any help.

[mjc123]

For Ammonia I calculated I would need 8,33mL of NH 30% to have 2,5mL of Ammonia.

Why do you need a value for volume of ammonia? 30% NH₃ means 30% by mass, not volume. And volumes are not additive anyway. Why not just use a constant concentration of ammonia (same volume of 30% NH₃ in same total solution volume)?

For water I chose 20 mL.

Why do you want to specify (as appears from below) total amount of water? It makes more sense to specify a constant total solution volume (and is practically easier, since as mentioned, volumes are not additive).

Here I chose 0,005M.

Is this the intended Cu concentration in the 30% ammonia, or in the final diluted solution?

mCu(NO3)2=MCu(NO3)2×MMCu(NO3)2×VNH3×(1L/1000mL)

I assume m means mass, M is molarity and MM is molar mass, but it would be as well to make this explicit.

The total amount of Water I have to add to have constant 20 mL is:
VT = VH2O - VCu,H2O - VNH3,H2O The value of VH2O is 20mL

See above; the answer is not right because of non-additivity of volumes, and the sensible thing is to have a fixed total volume. Add the requisite amounts of Cu salt and ammonia solution to a volumetric flask, and add water to make up to 20 mL.