Step 1 is to calculate the concn of the stock solution so you will know what you're starting with.
mols Sn(NO3)4 = grams/molar mass
Step 2. The M of the stock soln is mols/L soln or mol/0.250. Let's call this y molar.
Then dilutions are handled with the dilution formula which is
c1v1 = c2v2 where
c = concn
v = volume.
In this problem you will have
15.0mL x yM = ?M x 90 mL and you solve for ?M. This is all true IF volumes are additive. Technically they are not (and they aren't here) but the difference is so small it won't make any difference. I'm sure the asker means for you to add the 75 mL to 15 mL to end up with a solution volume of 90 mL.
Now you have M of the diluted solution of Sn(NO3)4.
Since there is 1 mol Sn^4+ per 1 mol Sn(NO3)4 the M Sn^4+ is the same as M of the solution.
M of the NO3^- is 4x M of Sn(NO3)4 because there are 4 mols NO3^- for each mol Sn(NO3)4.
1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.
This is my first time learning this so please do the above as an example. I have more to do so this can be an example.
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