Ion Concentration/Dissociation

Question

What is the concentration of each ion of the resulting solution when 18.5 mL of 0.25 M Al(NO₃)₃ is mixed with 22.4 mL of 0.50 M Cu(NO₃)₃?

Answer 1

The given compounds are ionic so they dissociate completely. For your information copper exists either as Cu+ or Cu2+. There is no Cu3+. Thus Cu(NO3)3 doesn't exist. It should be Cu(NO3)2.   

 

1st. Find the number of moles = Molarity x Volume (in L)

 

 Al(NO3)3 = (0.25 mol/L) x 0.0185 L = 0.0046 mol

 

Cu(N)3)2 = (0.5 mol/L) x 0.0224 L = 0.011 mol

 

Then find the mol of ions from each compound:

 

Al(NO3)3 produces 0.0046 mol Al3+ and 3 x 0.0046 mol = 0.014 mol NO3-

 

Cu(NO3)3 produces 0.011 mol Cu2+ and 3 x 0.011 mol = 0.033 mol NO3-

 

To find the concentration of each ions: divide the total number moles for each ions by total volume. make sure to

 

convert the total volume ( 18.5 ml + 22.4 ml 40.9 ml) to Liter before you do the calculation.

 

I will let you complete the calculation.