# What is the correct molecular, total ionic, and net ionic reaction equation of reaction between magnesium nitrate and sodium chromate? [closed]

From the question, I think that the reactants are $$\ce{Mg(NO3)2(aq) + Na2CrO4(aq)}$$

But the problem is, I am confused about the result of the reaction. Is it $$\ce{MgCrO4}$$ and/or $$\ce{NaNO3}$$?

How is the state of the product? In my opinion, there should be at least $$\ce{NaNO3(aq)}$$ as the product, but what about the magnesium chromate? Is it really magnesium chromate or it is something else?

• Please put the question (which you ask in the title) also in the body of the post along with a citation where it is taken from. Sep 15 at 21:53

There is no solvated molecule of either of 4 ionic compounds. All interactions ( or lack of ) happen on hydrated ionic level, including eventual precipitation or formation of ionic pairs.

All four salts are soluble, including magnesium chromate with the solubility $$\pu{137 g/100 mL}$$ at $$\pu{20^{\circ}C}$$ (solubility table).

When dissolved, salts form independent hydrated ions:

$$\ce{Mg(NO3)2(s) + Na2CrO4(s) ->[H2O] Mg^2+(aq) + 2 NO3-(aq) + 2 Na+(aq) +CrO4^2-(aq)}$$

If the solution is being evaporated, the salt with lowest solubility starts precipitating the first, with the repective ions recombining into the solid, stripping off their hydration cover.

The same solution is formed, if we start with equivalent amounts of the other two salts:

$$\ce{2 NaNO3(s) + MgCrO4(s) ->[H2O] Mg^2+(aq) + 2 NO3-(aq) + 2 Na+(aq) +CrO4^2-(aq)}$$

Ions have no memory which ions they were paired with in the solid state.

This question shows up a couple times in a google search, e.g. The carbonates, phosphates, borates, sulfites, chromates and arsenates of all metals except sodium, potassium and ammonium are insoluble in water but soluble in dilute acids. $$\ce{MgCrO4}$$ is soluble in water; $$\ce{MgSO3}$$ is slightly soluble in water.
So if you miss the statement about $$\ce{MgCrO4}$$ and $$\ce{MgSO3}$$ being an exception, you might think otherwise.