Solubility or such (1 Viewer)

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Hola! Does anyone know how to do this?

Equal volumes of Silver Nitrate Solution and Magnesium chloride solution, each 0.20 mol L-1 are mixed. Determine the concentration of the ions remaining in the resulting solution.

Thanks in advance!:sun:
 

Kurosaki

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by inspection silver chloride will precipitate out, while magnesium and nitrate will remain in solution.
now it doesn't give us any volume, so lets just say we mixed 1 L of each solution for simplicity.
then 0.2 mol AgNO3 is in the initial solution, and 0.2 mol MgCl2 in the other solution.
by inspection of mole ratios from dissociation equations, there are:

0.2 mol Ag+
0.2 mol NO3-
0.2 mol Mg2+
0.4 mol Cl-

then by inspection of the precipitation equation, you see that 0.2 mol Cl- is used in forming AgCl, since Ag+ is the limiting reagent.
Assuming that the resultant solution has a volume of 2L, then there are
0.2 mol NO3-
0.2 mol Mg2+
0.2 mol Cl- (since 0.2 mol of it got used in precitation)
then just use c=n/v
 

HeroicPandas

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by inspection silver chloride will precipitate out, while magnesium and nitrate will remain in solution.
now it doesn't give us any volume, so lets just say we mixed 1 L of each solution for simplicity.
then 0.2 mol AgNO3 is in the initial solution, and 0.2 mol MgCl2 in the other solution.
by inspection of mole ratios from dissociation equations, there are:

0.2 mol Ag+
0.2 mol NO3-
0.2 mol Mg2+
0.4 mol Cl-

then by inspection of the precipitation equation, you see that 0.2 mol Cl- is used in forming AgCl, since Ag+ is the limiting reagent.
Assuming that the resultant solution has a volume of 2L, then there are
0.2 mol NO3-
0.2 mol Mg2+
0.2 mol Cl- (since 0.2 mol of it got used in precitation)
then just use c=n/v
 

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