# A question about net ionic equations

I have a question regarding net ionic equations. In a solution, sodium fluoride and hydrochloric acid are mixed together. The "correct" net ionic equation is shown below. However, how can this actually BE a net ionic equation, since none of the ions change state (all aqueous)?

• There's no condition that there must be changes in state. Commented Jun 5, 2020 at 3:33
• Oh, perhaps what's confusing you is that $$\ce{Na+(aq) + Cl-(aq) -> NaCl(aq)}$$ (for example) is not valid. Well, HF is not the same as NaCl. HF nearly always sticks together, even in water. NaCl doesn't like sticking together. Commented Jun 5, 2020 at 3:35
• So, is it basically because it's a weak electrolyte? Commented Jun 5, 2020 at 3:44
• Yup, that's right. Commented Jun 5, 2020 at 3:55

To answer your question it is important to know which compound dissociate completely in water when in aqueous solution. For example, consider following reaction:

$$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}} \tag1$$

When ionic compound dissolves in water, each molecule separates to corresponding ions, which influenced by water. All of $$\ce{AgNO3}$$, $$\ce{NaCl}$$, and $$\ce{NaNO3}$$ are ionic compounds, which are soluble in water. $$\ce{AgCl}$$ is, onthe other hand, an ionic compound, but it is insoluble in water (refer to solubility rules). Hence the reaction $$(1)$$ can be rewrite as follows:

$$\ce{Ag+_{(aq)} + NO3-_{(aq)} + Na+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} + Na+_{(aq)} + NO3-_{(aq)}} \tag2$$

Like in algebra, you can canceled out common ions from each side of the equation $$(2)$$, which leaves:

$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} } \tag3$$

We called the equation $$(3)$$ a net ionic equation.

Now, let's look at your equation:

$$\ce{HCl_{(aq)} + NaF_{(aq)} -> HF_{(aq)} + NaCl_{(aq)}} \tag4$$

Now, $$\ce{NaF}$$ and $$\ce{NaCl}$$ are ionic compounds, which are soluble in water, hence they separated into ions in aqueous medium. How about $$\ce{HCl}$$, and $$\ce{HF}$$? They are soluble in water, but are they ionic? Since both are acids, to answer this question, we should look into their respective $$\mathrm{p}K_\mathrm{a}$$ values (or $$K_\mathrm{a}$$ values). Ref.1 is listed followings:

$$\begin{array}{c|ccc} \text{acid/base (chemical formula)} & \mathrm{p}K_\mathrm{a} & K_\mathrm{a} & \\ \hline \text{hydrogen iodide }(\ce{HI}) & -10.0 & 1.0 \times 10^{10} \\ \text{hydrogen bromide }(\ce{HBr}) & -9.0 & 1.0 \times 10^{9} \\ \text{hydrogen chloride }(\ce{HCl}) & -8.0 & 1.0 \times 10^{8} \\ \text{hydronium ion }(\ce{H3O+}) & -1.7 & 50.1 \\ \text{hydrogen fluoride }(\ce{HF}) & 3.2 & 6.3 \times 10^{-4} \\ \text{acetic acid }(\ce{CH3CO2H}) & 4.7 & 2.0 \times 10^{-5} \\ \hline \end{array}$$

Accordingly, $$\ce{HCl}$$ should be ionic in aqueous solutions and should dissolve in water according to following equation:

$$\ce{HCl_{(aq)} + H2O_{(l)} -> H3O+_{(aq)} + Cl-_{(aq)}} \tag5$$

I put forward arrow because it'd dissociate completely, since $$K_\mathrm{a} = \frac{\ce{[H3O+][Cl-]}}{\ce{[HCl]}} = 1.0 \times 10^{8}$$.

However, $$\ce{HF}$$ is a different story. Since its $$K_\mathrm{a} = 6.3 \times 10^{-4} \lt \lt 1$$, its dissociation is not complete, and most of $$\ce{HF_{(aq)}}$$ molecules stay intact (keep its covalent nature):

$$\ce{HF_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + F-_{(aq)}} \tag6$$

Note that $$K_\mathrm{a}^\ce{HF} = \frac{\ce{[H3O+][F-]}}{\ce{[HF]}} = 6.3 \times 10^{-4}$$, and hence $$\ce{[HF]} \approx 1.6 \times 10^{3}\ce{[H3O+][F-]}$$. Like sugar molecules, which are soluble in water but did not dissociate, $$\ce{HF}$$ stays as $$\ce{HF_{(aq)}}$$ in water. Thus, the equation $$(4)$$ can be rewritten as:

$$\ce{H+_{(aq)} + Cl-_{(aq)} + Na+_{(aq)} + F-_{(aq)} -> HF_{(aq)} + Na+_{(aq)} + Cl-_{(aq)}} \tag7$$

When you made common ions cancelled out, the net ionic equation renains as:

$$\ce{H+_{(aq)} + F-_{(aq)} -> HF_{(aq)} }$$

References:

1. K. Peter C. Vollhardt, Neil E. Schore, In Organic Chemistry: Structure and Function; Fifth Edition; W. H. Freeman & Co.: New York, NY, 2007 (ISBN: 0-7167-9949-9).