how many grams of solid magnesium nitrate 148.3g/mol should be added to 500ml of 0.118M of sodium nitrate in order to produce aqueous solution .25M of nitrate i

Question

I have been stuck on this question and I'm required to show work 

Answer 1

Hi Taylor,

 

This is a pretty complex problem that you essentially have to work backwards. So let's start with what we know:

 

*We have 500mL of a 0.118M solution of sodium nitrate and we will be adding an unknown mass of Mg(NO₃)₂ to the solution without changing the volume in order to increase the concentration of nitrate essentially.

 

*From the given information, we know that we have a certain amount of nitrate already in the solution, so what this problem is really asking, is how much more nitrate do we need to add to get to a certain concentration of nitrate. BUT FIRST, we need to figure out how much nitrate we already have.

1.) By multiplying 0.500L by 0.118M (moles/L) of NaNO₃, we see that we currently have 0.059 moles of NaNO₃, which translates to 0.059 moles of NO₃^- because for every 1 mole of NaNO₃, we have 1 mole of nitrate

2.) Also remember that our volume is not changing, so if we multiply 0.25M (moles/L) of nitrate (the desired concentration) by 0.500L, we will get the amount of moles needed to create this concentration in the amount of solution we have, which turns out to be 0.125 moles of NO₃^-.

3.) Now we know how many moles of NO₃^- we have to work with, and how many moles we need. So now we subtract 0.059 moles from 0.125 moles to determine how many moles of nitrate we need to add to the solution to increase the concentration to 0.25M, which equals 0.066 moles of NO₃^-.

4.) However, we are not done yet. Remember that for every 1 mole of Mg(NO₃^-)₂, there are 2 moles of NO₃^-.Therefore, we only need half as many moles of magnesium nitrate due to this ratio, and thus we need

5.) Lastly, we convert moles of magnesium nitrate to grams using its molar mass of 148.3g/mol:

0.033 moles Mg(NO₃^-)₂ x 148.3 g/mol = 4.89g Mg(NO₃^-)₂

 

Hope this helps! Let me know if you have any questions!

 

Shane E., MPAP, PA-C