NaF(aq) + HCl(aq) --> ?

[rthmjohn]

Okay, so this equation appeared on my last chemistry test. I wrote the following:

 

NaF(aq) + HCl(aq) --> HF + NaCl(aq)

 

Apparently my answer was correct except that I didn't know what phase the HF would be. I asked my chem teacher and she said she didn't know (sad huh?). So I have to resort to a science forum for the answer. What would the phase of the HF be? I thought it would be a gas.

[woelen]

The HF will not appear as a gas. It will be dissolved in the water. When the liquid is heated, it will not be driven off as the pure gas. HF forms an azeotrope with water. When the solution is dilute, then first water will be driven away and when the concentration of the HF has risen to the azeotropic value, then a constant ratio mix of HF and H2O is boiled away.

 

Pure HF is borderline between gas an liquid. It boils just under 20C at normal atmospheric pressure.

[Cloud]

Okay, so this equation appeared on my last chemistry HF would be. I asked my chem teacher and she said she didn't know (sad huh.)

 

Wow.

 

I think Hydrogen Flouride would be liquid. (supposing your equation is correct).

 

But wouldn't the equation look like this:

 

2NaF (aq) + 2HCl (aq) -> 2NaCl(aq) + H2(g) + 2F(l)

 

I'm quute sure that this reaction would give off hydrogen gas and that flourine would be deposited as a liquid.

[RyanJ]

Wow.

 

I think Hydrogen Flouride would be liquid. (supposing your equation is correct).

 

But wouldn't the equation look like this:

 

2NaF (aq) + 2HCl (aq) -> 2NaCl(aq) + H2(g) + 2F(l)

 

I'm quite sure that this reaction would give off hydrogen gas and that flourine would be deposited as a liquid.

 

Not at all

 

1) Fluorine is a gas at room temperature' date=' you have it listed there as a liquid.

 

Maybe this will explain better:

 

[ce']Na^{+} + F^{-} + H^{+} + Cl^{-} -> HF + NaCl[/ce]

 

* All noted above occur in solution.

 

As you can see on dissolving the sodium fluoride disocciates into Sodium and Fluoride ions and the hydrochloric acid too chloride and hydrogen ions. These then react producing sodium chloride and hydrogen fluoride.

 

Make any sence?

 

Cheers,

 

Ryan Jones

[woelen]

Wow.

 

I think Hydrogen Flouride would be liquid. (supposing your equation is correct).

 

But wouldn't the equation look like this:

 

2NaF (aq) + 2HCl (aq) -> 2NaCl(aq) + H2(g) + 2F(l)

 

I'm quute sure that this reaction would give off hydrogen gas and that flourine would be deposited as a liquid.

Well, you are totally wrong. Suppose you would have a combination of fluorine and hydrogen gas, then it would explode at once. In fact, fluorine is so increadibly reactive, it even reacts with glass. With some difficulty you can water make burn in an atmosphere of fluorine.

[budullewraagh]

the reaction is just an application of acidity and basicity. H+ is more acidic than Na+ and F- is more basic than Cl-, so those ions react more. the weaker acid, Na+ and weaker base, Cl- stay in solution completely. Much H+ and F- stays in solution, but some covalency is observed and we see some HF

[woelen]

This is quite a good explanation, however, we see quite some HF. The majority of all H(+) ions and F(-) ions are combined to HF. HF only is a weak acid.

[Cloud]

Ok - I admit - chemistry is not my forte(e.g. . . .whatever)

 

I also liked how bulldull exlpained it.

I was thinking it was a simple ionic swap - But I couldn't get my head round the HF as being a product of an ionic reaction (since its covalent) I guess I haven't learnt chemistry to that degree.

 

I figured from the knowledge that I do have that it would output in the same way as an ionic reaction. As Woelen and RyanJ mentioned - Flourine is a gas at room temperature. Duh. Simple chemistry.

 

Sorry about the inconvenience.

[RyanJ]

I was thinking it was a simple ionic swap - But I couldn't get my head round the HF as being a product of an ionic reaction (since its covalent) I guess I haven't learnt chemistry to that degree.

 

You can think of is that way - exclude the ionic part and your right on! Actualy HF is one of those gray areas. F having an electronegativity of 4.0 and Hydrogen having an electronegativity of 2.1 the difference between those determins the bond type. If greater then 2 then its ionic - in this case its slightly more covalent but still pretty close to the mark

 

http://www.chemguide.co.uk/atoms/bonding/electroneg.html#top

 

May provide some insight for you.

 

Cheers,

 

Ryan Jones

[WillTheNewf]

Ok - I admit - chemistry is not my forte(e.g. . . .whatever)

 

I also liked how bulldull exlpained it.

I was thinking it was a simple ionic swap - But I couldn't get my head round the HF as being a product of an ionic reaction (since its covalent) I guess I haven't learnt chemistry to that degree.

 

I figured from the knowledge that I do have that it would output in the same way as an ionic reaction. As Woelen and RyanJ mentioned - Flourine is a gas at room temperature. Duh. Simple chemistry.

 

Sorry about the inconvenience.

 

 

Well the thing is, the reaction IS an ionic reaction. However you must take it one step further. The Flourine from the dissociated NaF will react with H20 as follows:

F- (aq) + H20(l) <--> HF(aq) + OH-(aq)

[jdurg]

Well the thing is' date=' the reaction IS an ionic reaction. However you must take it one step further. The Flourine from the dissociated NaF will react with H20 as follows:

F- (aq) + H20(l) <--> HF(aq) + OH-(aq)[/quote']

No it won't. The fluorine will react with the free H+ ions as opposed to ripping apart a water molecule. Energetically, it costs energy to break bonds and releases energy when you form bonds. Therefore, it would be MUCH more energetically favorable to form a bond with H+ then it would be to break apart an H-O bond.

[WillTheNewf]

No it won't. The fluorine will react with the free H+ ions as opposed to ripping apart a water molecule. Energetically, it costs energy to break bonds and releases energy when you form bonds. Therefore, it would be MUCH more energetically favorable to form a bond with H+ then it would be to break apart an H-O bond.

 

Yes your right haha sorry, i forgot about the HCL as I was writing what i said. (that and ive got a lot of buffer calculations recently punched into my brain.)

[woelen]

No it won't. The fluorine will react with the free H+ ions as opposed to ripping apart a water molecule. Energetically, it costs energy to break bonds and releases energy when you form bonds. Therefore, it would be MUCH more energetically favorable to form a bond with H+ then it would be to break apart an H-O bond.

Well, in fact, formally you are right, but the reaction of WillTheNewf also can be regarded as right.

 

If you dissolve NaF in plain water (without added acid), then the solution becomes quite alkaline. Enough to make litmus blue and if sufficient NaF is dissolved, phenolphtalein becomes red.

 

So, you get free OH(-) ions in solution at quite a high concentration.

 

There are multiple equlibria:

 

H2O <--> H(+) + OH(-) (very weak, mostly at the left)

F(-) + H(+) <--> HF (mostly at the right)

 

The acid HF is so weak, that indeed F(-) takes up most H(+) and this drives the equlibirium for water more to the right. The H(+) is consumed and the OH(-) remains behind.

 

The net equlibrium indeed is:

 

H2O + F(-) <--> OH(-) + HF, being mostly to the right. So, Jdurg and WillTheNewf, you both are right.

[jdurg]

Well' date=' in fact, formally you are right, but the reaction of WillTheNewf also can be regarded as right.

 

If you dissolve NaF in plain water (without added acid), then the solution becomes quite alkaline. Enough to make litmus blue and if sufficient NaF is dissolved, phenolphtalein becomes red.

 

So, you get free OH(-) ions in solution at quite a high concentration.

 

There are multiple equlibria:

 

H2O <--> H(+) + OH(-) (very weak, mostly at the left)

F(-) + H(+) <--> HF (mostly at the right)

 

The acid HF is so weak, that indeed F(-) takes up most H(+) and this drives the equlibirium for water more to the right. The H(+) is consumed and the OH(-) remains behind.

 

The net equlibrium indeed is:

 

H2O + F(-) <--> OH(-) + HF, being mostly to the right. So, Jdurg and WillTheNewf, you both are right.[/quote']

But in an acidic solution, as the OP had mentioned, the free H+ ions will "react" with the F- as opposed to the neutral H2O. The fluoride ion forming HF due to interactions with water will only predominate in a solution which began as neutral.