# Balancing ionic equations and determining number of ions

There were two questions that I've listed below that I've had troubles on. I got an answer in both, but both of them were incorrect.

For the first question, they list the answer to be D and then for the second they list the answer as A.

6.Determine the net ionic equation for the reaction between $$\ce{HNO3}$$ and $$\ce{Mg(OH)2}$$.

a. $$\ce{2H+(aq) + (OH)2^{2-}(aq) -> H2(OH)2(aq) }$$
b. $$\ce{2HNO3(aq) + Mg(OH)2(aq) -> Mg(NO3)2(aq) + 2H2O(l) }$$
c. $$\ce{Mg2+(aq) + 2NO3 (aq) -> Mg(NO3)2(s) }$$
d. $$\ce{H+(aq) + OH(aq) -> H2O(l) }$$
e. $$\ce{Mg+(aq) + NO3 (aq) -> MgNO3(s) }$$

8.Determine the number of ions that are present in $$38.8~\mathrm{mL}$$ of $$0.113~\mathrm{M}$$ $$\ce{(NH4)2CO3. }$$

a. $$7.92\cdot 10^{21}~\text{ions }$$
b. $$2.64\cdot 10^{21}~\text{ions }$$
c. $$2.07\cdot 10^{23}~\text{ions }$$
d. $$6.20\cdot 10^{23}~\text{ions }$$
e. $$1.32\cdot 10^{23}~\text{ions }$$

For the first question, I did this: $$\ce{2 H+ + 2NO3- +Mg(OH)2 -> 2H2O + Mg+2 + 2NO3-}$$ The question I have is why $$\ce{Mg(OH)2}$$ splits up. I don't get it because things you split up are things that are aqueous or soluble. $$\ce{Mg(OH)2}$$ isn't soluble so I listed it as a solid and thought it wouldn't break down. All I need to know is why I have to break down $$\ce{Mg}$$ and the $$\ce{OH2}$$, because they aren't soluble.

For the second problem I got an answer of $$2.64\cdot 10^{21}~\text{ions}$$ but the review said it was A. I set it up as $$.113~\mathrm{M} = \text{moles} / 0.0388~\mathrm{L}$$ and then multiplied $$.113$$ by $$0.0388$$ and then multiplied that amount by $$6.022\cdot 10^{23}~\text{ions}$$ which got me the answer B.

• You are correct on the net ionic equation. For the other, think about how many particles each ammonium carbonate generates when it dissolves. Aug 7, 2014 at 3:25

Actually $Mg(OH)_2$ has (low) solubility ($K_{sp}=1.5\cdot 10^{-11}$). Since it is asking for the net ionic equation, the only ions reacting (since $NO_3^-$ stays in solution) are $H^+$ and $OH^-$.