Redox equation and ionic form #1

Rewrite the following equation showing all substances in their ionic form in solution (the ionic equation) and balance it:
$$Hg(NO_3)_2+KI \rightarrow K_2HgI_4+KNO_3$$

SOLUTION:
The ionic equation is:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+K^++NO_3^-$$

Note that in reactants there are two $NO_3^-$ and only one $NO_3^-$ in products. So multiply by two $KNO_3$ in products:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+2K^++2NO_3^-$$

Now sum the same species -$K^+$- and simplify -$2NO_3^-$-:
$$Hg^{2+}+K^++I^- \rightarrow 4K^++HgI_4^{2-}$$

Now there are $4K^+$ in products. So multiply by four the $KI$ in reagents:
$$Hg^{2+}+4K^++4I^- \rightarrow 4K^++HgI_4^{2-}$$

Now simplify $K^+$:
$$Hg^{2+}+4I^- \rightarrow HgI_4^{2-}$$
This is the net ionic equation. Note that che charges are balanced.

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PROBLEM 14:
Balance the following redox equation and put it in the ionic form:
$$Na+H_2O \rightarrow NaOH+H_2$$