Because someone got confused (widjaja), ionic refers to net ionic
Acid + Metal oxide/hydroxide (base) $\to$ Salt + water e.g.
$$\begin{align} HCl_{(aq)}+KOH_{(aq)} \to H_{2}O_{(l)} + KCl_{(aq)} \\ \text{This is WRONG! It's not an ionic equation!!!} \\ H_{3}O_{(aq)}^{+}+ OH_{(aq)}^{-} \to 2H_{2}O_{(l)} H_{3}O_{(aq)}^{+} \checkmark\\ \text{This is the preferred equation,} \\ \text{it is ionic and does not have} \\ \text{spectator ions.} \end{align} $$Note that $H_{3}O_{(aq)}^{+}$ is a strong acid, and the $OH_{(aq)}^{-}$ comes from a soluble hydroxide.
Here is an example of a weak acid reacting with a non-soluble hydroxide:
$$\begin{align} 3CH_{3}COOH_{(aq)}+Fe(OH)\_{3(aq)} \to (CH_{3}COO)\_{3}Fe_{(aq)}+3H_{2}O \\ \text{This is wrong, because the Iron (III) ethanoate} \\ \text{is separated in solution(it is dissociated).} \\ 3CH_{3}COOH_{(aq)}+Fe(OH)\_{3(aq)} \to 3CH_{3}COO\_{(aq)}^{-}+Fe_{(aq)}^{3+}+3H_{2}O \checkmark \end{align} $$If you aren't given the state of a carbonate, assume it is a solid!
These are what you should be thinking:
Example:
$$\begin{align} Al_{2}(CO_{3})\_{3(aq)} + 6HNO_{3(aq)} \to 2Al(NO_{3})\_{3(aq)}+3CO_{2(g)}+H_{2}O_{(l)} \\ \text{This is wrong, for same reason as above,} \\ \text{and because nitric acid is strong!} \\ Al_{2}(CO_{3})\_{3(aq)} + 6H_{3}O_{(aq)}^{+} \to 2Al_{(aq)}^{+}+ 6NO_{3(aq)}^{-}+3CO_{2(g)}+H_{2}O_{(l)} \end{align} $$Mr Norrie believes you get the whole distinguishing between strong and weak acid, solubility of reactants/products, so no more examples (thank god very hard to write latex)(i was wrong there are examples)
Acid + sulfide $\to$ $H_{2}S$ + Salt
Acid + sulfite ($SO_{3}^{2-}$) $\to$ $SO_{2}$ + Salt + water
Acid + metal $\to$ $H_{2}$ + salt (consider redox!!!)
Every acid BUT $HNO_{3}$, because it reacts to produce $NO_{2}/NO$!!!!!
$$\begin{align} Zn_{(s)}+2HCl \to ZnCl_{2(aq) }+ H_{2(g)} \\ \text{This is wrong.} \\ Zn_{(s)}+2H_{3}O_{(aq)}^{+} \to Zn_{(aq)}^{2+} 2Cl_{(aq)}^{-}+ H_{2(g)} + 2H_{2}O_{(l)} \end{align} $$19.6mL of $CH_{3}COOH$ reacts completely with 17.6mL of 0.106$molL^{-1} \ NaOH$. What is the concentration of $CH_{3}COOH?$
$$\begin{align} CH_{3}COOH_{(aq)}+NaOH_{(aq)}\to CH_{3}COONa_{(aq)}+H_{2}O_{(l)} \\ \text{(Note: apparently you need ionic equation if its not given? } \\ \text{maybe you can write just balanced if no eq. given)} \\ n(NaOH) = cV = 0.106 \times 0.0176 = 0.0018656mol \\ \text{Since full reaction occurs, and 1:1 ratio:} \\ n(CH_{3}COOH) = n(NaOH) = 0.0018656mol \\ \therefore c = \frac{n}{V}, [CH_{3}COOH] = \frac{0.0018656}{0.0196} \\ =0.0962molL^{-1} (3SF) \end{align} $$Round your final answer to least number of sig figs in the question, in the case above the least number of sig figs is 3 sig figs, hence answer will be in 3 sig figs.
