Question

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, indicating a reaction took place. Write the net ionic equation for this reaction.

Answer 1

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, indicating a reaction took place. The net ionic equation for this reaction is;

$\mathbf{Mg^{2+}_{(aq)} + 2OH^- _{(aq)} \to Mg (OH)_2_{(s)}}$

From our inorganic chemistry, we need to understand that $\mathbf{NH_{(aq)} }$ forms an aqueous ammonium hydroxide $\mathbf{NH_4OH_{(aq)}}$ when reacted in solution.

• The molecular reaction can be well represented as follows:

$\mathbf{Mg(NO_3)_{2(aq)} \to 2NH_4OH_{(aq)} \to Mg (OH)_2{(s)} + 2NH_4NO_{3}_{(aq)}}$

The ionic reaction for the reaction between 0.1 M Mg(NO3)2 and  a limited amount of aqueous ammonia can be expressed as:

$\mathbf{Mg^{2+}_{(aq)} + 2NO_3^-_{(aq)} + 2NH_4^+_{(aq)}+ 2OH^-_{(aq)} \to Mg (OH)_2_{(s)}+ 2 NH^+_4_{(aq)} + 2NO_3^-_{(aq)}}$

Now, the net ionic equation of the chemical reaction that displays only the elements, compounds, or ions that are involved directly in the chemical process can be determined from the ionic reaction. This is done by eliminating similar species of ions, elements, or compounds from both sides of the reaction;

i.e.

The net ionic equation is:

$\mathbf{Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \to Mg (OH)_2_{(s)}}$

Therefore, we can conclude that we have understood the concept of how to clearly write the net ionic equation of a given ionic reaction.

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Answer 2

The net ionic reaction for given reaction will be

$\rm \boldb {\rm \bold{ Mg^2^++ 2(NO_3)^-\rightarrow Mg(OH_2(s) }$

Net ionic equation:

It is the chemical reaction that displays only the elements, compounds, or ions that are involved directly in the chemical process can be determined from the ionic reaction.

The given reaction is,

$\rm \bold{ Mg (NO_3)_2(aq) + NH_4OH(aq)\rightarrow Mg(OH)_2(s) + NH_4NO_3(aq)}$

The ionic reaction will be,

$\rm \bold{ Mg^2++ 2(NO_3)^- + NH_4^++ OH^-\rightarrow Mg(OH_2(s) + NH_4^++NO_3^-}$

The net ionic reaction can be calculated by deleting the similar chemical species from both side.

Hence, we can conclude that the net ionic reaction will be

$\rm \boldb {\rm \bold{ Mg^2^++ 2(NO_3)^-\rightarrow Mg(OH_2(s) }$

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