# 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 weren't correct.

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

1. 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. $\ce{Mg+(aq) + NO3 (aq) -> MgNO3(s) }$

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

For the second, 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. – Brinn Belyea Aug 7 '14 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 $OH^-$.