Calculate concentration of ions?

Question

I have a question about calculating molarities.

You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 2.50 M solution of AgNO3(aq). Calculate the concentrations of Na+, CrO4(2-), Ag+ and NO3- when the two solutions are added together.

I have no clue how to start. Please help. Thanks!

Answer 1

this mix reacts, producing a precipitate:

Na2CrO4  & AgNO3 --> Ag2CrO4  &  NaNO3

Na+  & NO3- are "spectator ions" tha stay in solution

========================

Na+

2.50 Molar Na2CrO4 releases twice the Na+ :  5.00 Molar Na+

5.00 Molar Na+ diluted @ 75.0 ml / 200 ml total = 1.875

first answer Na+ = 1.88 MOlar

============

NO3- :

2.50 Molar NO3- diluted @ 125 ml / 200 ml = 1.563

second answer: NO3- = 1.56 Molar

==================

find moles involved in the reaction:

0.075 L @ 2.50 mol / litre = 0.1875 moles CrO4)-2

0.125 L @ 2.50 mol / litre = 0.3125 moles Ag+

since 2 Ag+ reacts with 1 CrO4)-2 --> 1 Ag2CrO4 ,...

0.3125 moles of Ag+ reacts with, 1/2 it, 0.1563 moles CrO4)-2

which leaves un-precipitated,...

0.1875 - 0.1563 = 0.03125 moles of CrO4)-2 in the 200 ml of solution

find molarity:

0.03125 moles of CrO4)-2 / 0.200 litres = 0.156 Molar

third answer: 0.156 Molar (CrO4)-2

========================

finding [Ag+] needs the Ksp

Ag2CrO4 --> 2 Ag+  &  CrO4-2

Ksp = [Ag+]^2 [CrO4]

1.2e-12 = [Ag+]^2 [0.156 Molar]

[Ag+]^2  =  1.2e-12 / [0.156 Molar]

[Ag+]^2  = 7.7 e-12

Ag+ = 2.77e-6

Your last answer is : Ag+ = 2.8e-6

(the Ksp had only 2 sig figs)