Question about solubility equilibria, precipitation problems

You're adding NaOH(s) to a soln of Mg(NO3)2 (aq) ... so you stop once the NaOH(s) is added so that NaOH(aq) concentration is 0.001M. Since Molarity is equal to

Molarity =moles/volume solvent (aka H2O in (aq) solns)

concentration of the solution Mg(NO3)2(aq) doesn't change, since adding a solid won't affect the concentration.

I sat here for like 15 minutes trying to figure it out, I mean I *knew* how to solve it, and I knew how to approach the problem... then it hit me. Lol. In the cases where you're taking, like, 0.5M 500mL Mg(NO3)2 (aq) to 0.5M 500mL NaOH (aq) ... then that's when you need to take into account the dilution of the solution concentration.

With the eqn

M1V1 = M2V2

This entry passed through the Full-Text RSS service - if this is your content and you're reading it on someone else's site, please read the FAQ at http://ift.tt/jcXqJW.

Question about solubility equilibria, precipitation problems