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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)?

Ionic equation

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    $\begingroup$ There's no condition that there must be changes in state. $\endgroup$ Commented Jun 5, 2020 at 3:33
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    $\begingroup$ 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. $\endgroup$ Commented Jun 5, 2020 at 3:35
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    $\begingroup$ So, is it basically because it's a weak electrolyte? $\endgroup$
    – Evan Ewing
    Commented Jun 5, 2020 at 3:44
  • $\begingroup$ Yup, that's right. $\endgroup$ Commented Jun 5, 2020 at 3:55

1 Answer 1

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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).
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