Calculating and Mixing NiAc and ZnAc Solutions for Molar Proportion

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To achieve the desired Ni/Zn molar proportions of 1%, 3%, and 5% using 1M NiAc and ZnAc solutions, the calculations must reflect the correct ratios of the two components. For a 1% ratio, the calculation suggests using a total of 5 mmol of the mixture, resulting in 0.05 mmol of Ni and 4.95 mmol of Zn. However, this approach leads to a solution that does not maintain the original 1M concentration of Zn, thus failing to meet the requirement for a 1% molar proportion accurately. The correct method is to mix the solutions in a straightforward ratio based on the desired proportion. For a 1% ratio, mix the solutions in a 100:1 ratio, ensuring that the final concentrations are consistent with the original molarities. This principle applies similarly for the 3% and 5% ratios, where the mixing should reflect the specified proportions without altering the initial concentrations significantly.
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I have been given NiAc and ZnAc solutions of 1M concentration. And being asked to mix the two solutions together in Ni/Zn nominal molar proportions 1%, 3% and 5%. How can I calculate and mix the two solutions to get the required molar proportion.
 
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bilalarif said:
calculate and mix the two solutions to get the required molar proportion
Ni/Zn = 0.01, 0.03, and 0.05. Calculate it and do it.
 
Bystander said:
Ni/Zn = 0.01, 0.03, and 0.05. Calculate it and do it.

Molar fraction = (number of moles of component of interest) ÷ (total number of moles of mixture)

For Ni/Zn nominal proportion 1%
0.01= number of moles of Ni/ total number of moles of Ni+Zn
Suppose we have total moles of (Ni+Zn) solution= 5 mmol
0.01= moles of Ni/ 5mmol
Moles of Ni = 0.05 mmol
Moles of Zn= total moles of (Ni+Zn)- moles of Ni
Moles of Zn= 4.95 moles
As both NiAc and ZnAc concentration is 1M.
M=n/V
V (ZnAc solution) = 4.95mmole/1 mol/l = 4.95ml

V (NiAc solution) = 0.05mmole/1 mol/l = 0.05ml

is it correct calculation?
 
Last edited:
That's one. Two to go.
 
No it isn't, though the error is small. You have made a solution which is 0.01 M Ni, but that is not what was asked for. It is not quite 1M Zn any more, so 1 % is not exactly the molar proportion.
If the starting molarities are equal an you are asked for a 1 % ratio you just mix them in the ratio 100:1.
You're probably so used to problems of achieving a given final molarity that you have overthught this one
 
Divalent ions, one to one.
 
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