Magnesium in hydrochloric acid solution.

[DarthRoni]

I am confused as to why this happens when magnesium is introduced into a hydrochloric acid solution:
$$2HCl_(aq) + Mg(s) \rightarrow H_2(g) + MgCl_2(aq)$$
I understand that the magnesium will form an ionic bond with chlorine and then quickly dissolve. However, I don't understand why the hydrogen is affected at all. Furthermore the hydrogen ions are positively charged so how could they possibly bond ?

 

[SteamKing]

What ions? Free hydrogen gas molecules are formed by the reaction. Where do you think that the chlorine to bond with the magnesium is coming from?

 

[DarthRoni]

But hasn't the hydrogen already separated from the chlorine?
I was thinking the equation should look something like this
$$H_3O^+(aq) + 2Cl^-(aq) + Mg(s) \rightarrow MgCl_2(aq) + H_3O^+(aq)$$

I have one more question, when calculating the pH of a solution after neutralizing it using titration, can we simply determine the pH in the following way:
$$KOH(aq) + H_3O^+(aq) \rightarrow K^+(aq) + 2H_2O$$

 

[SteamKing]

The hydronium ions (H3O+) are unstable. Magnesium is higher in the reactivity series than hydrogen and replaces it in the HCl.

http://en.wikipedia.org/wiki/Reactivity_series

Second question, no. There is no longer any hydrogen ion activity in a potassium hydroxide solution.

 

[Borek]

But hasn't the hydrogen already separated from the chlorine?
I was thinking the equation should look something like this
$$H_3O^+(aq) + 2Cl^-(aq) + Mg(s) \rightarrow MgCl_2(aq) + H_3O^+(aq)$$

The way you wrote it H₃O⁺ cancels out as a spectator and you are left with

Mg(s) + 2Cl^-(aq) → MgCl₂(aq)

which is not (and can't be) balanced. Properly balanced equation is balanced both in terms of atoms and charge, and you can't balance charge having only negatively charged Cl^- on one side of the reaction. So there is apparently something missing here.

And what is missing is the fact Mg is getting oxidized by H⁺, and the real reaction is

Mg(s) + 2H⁺ → Mg²⁺(aq) + H₂(g)

(you can put H₃O⁺ in place of H⁺ and add water to the RHS if you prefer).

 

[DarthRoni]

Thanks Borek ! Always great help. However about the titration question. When we add a base (such as KOH) in an acidic solution. Is the base reacting with the acid or simply the hydrogen ions to create water ?

 

[Borek]

You have to decide what your question is. First, you asked about pH (and to be honest, I have no idea what the question was). Now, you are asking about neutralization mechanism (which I don't know, and I don't remember ever seeing it). Problem with neutralization and dissociation is that these are typically pretty fast and both possible mechanisms (dissociation first, neutralization second, or direct neutralization) yield exactly the same products.

 

[DarthRoni]

I will try to be more clear. Take a solution with a certain concentration of acetic acid ($C_2H_4O_2$). We are attempting to neutralize the solution using a another solution containing potassium hydroxide ($KOH$). My textbook suggests I use the following formula in order to find the molar quantity of acetic acid:
$$KOH(aq) + C_2H_4O_2(aq) \rightarrow KC_2H_3O_2(aq) + H_2O$$
In this case 1 mole of KOH reacts with 1 mole of acetic acid. I could then calculate the concentration of acetic acid. They then suggest that I should use this concentration to find the pH. This does not make much sense to me because acetic acid does not ionize completely in a solution due to the fact that it is a weak acid. Upon further research, I discovered that there is a way to calculate the quantity of hydrogen ions which will be released using something called the Acid dissociation constant. This leads me to believe that we are not actually determining the amount of acetic acid that the KOH is reacting with but instead the amount of Hydrogen ions that are being reacted with to create water. Which is why I wrote the following formula:
$$KOH(aq) + H_3O^+(aq) \rightarrow K^+(aq) + 2H_2O$$

Is my assumption correct? Should I apply the dissociation constant to find my pH or can I just use the second formula and determine my pH using a direct relation between the potassium hydroxide and the hydrogen ions.

 

[Borek]

You aren't correct - one you add KOH to neutralize acid you are just shifting dissociation. It doesn't mean you removed all H⁺ from the solution.

Could be you are misreading your textbook, but it is hard to say not knowing what the problem was and how it was solved/explained.

 

[DarthRoni]

This is their process:
71.14ml of $NaOH$ with concentration $2.00*10^-5 mol/L$ is used to neutralize 25.00ml of $HNO_2$ solution
They then state the following
$$NaOH(aq) + HNO_2(aq) \rightarrow NaNO_2(aq) + H_2O$$
they then find the moles of NaOH
$$n_{NaOH} = 2.00*10^-5 * 0.07114 = 1.423*10^-6mol$$
and since one mole of $NaOH$ reacts with one mole of $NHO2$ then that means there's 1.423*10^-6 moles of $NHO_2$.
Next they determine the concentration of acid.
$$[NHO_2(aq)] = (1.423 * 10^-6)mol / 0.0250L = 5.69*10^-5mol/L$$
Now this is the part that confuses me. They say the pH is equal to the following:
$$pH = -log([NHO_2]) = 4.24$$

 

[Borek]

This is their process:

OK, this is the process, but it won't hurt to know the question they are trying to solve. It is not clear to me at all.

71.14ml of $NaOH$ with concentration $2.00*10^-5 mol/L$ is used to neutralize 25.00ml of $HNO_2$ solution
They then state the following
$$NaOH(aq) + HNO_2(aq) \rightarrow NaNO_2(aq) + H_2O$$
they then find the moles of NaOH
$$n_{NaOH} = 2.00*10^-5 * 0.07114 = 1.423*10^-6mol$$
and since one mole of $NaOH$ reacts with one mole of $NHO2$ then that means there's 1.423*10^-6 moles of $NHO_2$.
Next they determine the concentration of acid.
$$[NHO_2(aq)] = (1.423 * 10^-6)mol / 0.0250L = 5.69*10^-5mol/L$$

So far so good, just a titration calculation. Slightly stupid if you ask me, but technically correct.

Now this is the part that confuses me. They say the pH is equal to the following:
$$pH = -log([NHO_2]) = 4.24$$

If their plan was to calculate pH, this is wrong, HNO₂ is not 100% dissociated, even that diluted. About 11% would be in the not dissociated form and pH would be much closer to 4.30

 

[DarthRoni]

Their plan WAS to calculate the pH. You can see why I was thrown off. Thanks again! You've once again helped me understand chemistry a little bit more. (I assume the acid dissociation constant would be applied at the end)

 

[megmog4]

2HCl ( aq)+Mg(s)→H2 (g)+MgCl2 (aq)
In this equation, hydrochloric acid is reacting with magnesium to form hydrogen gas and magnesium chloride. The reason this is formed is because magnesium is higher up in the reactivity series than hydrogen, causing the chloride to "transfer" from hydrogen to magnesium. This leaves hydrogen as a single element rather than a compound. The reason that hydrogen has formed a bond with another hydrogen atom is because the outer shell of electrons is not whole and therefore unstable, and the atoms "feel" an "attraction" with each other to fill these outer shells. (hydrogen atoms have one electron in their outer (and only) electron shell). To fill these outer shells, the element must form a bond with another atom of the same element in order to fill this outer shell (which needs two electrons to fill). In conclusion, the chloride "joins" the magnesium to form magnesium chloride (MgCl2) due to the fact that magnesium is higher up in the reactivity series, leaving hydrogen as a lone element, which forms a bond with another atom of itself in order to become more stable and fill its outer electron shell.

 

[Borek]

2HCl ( aq)+Mg(s)→H2 (g)+MgCl2 (aq)
In this equation, hydrochloric acid is reacting with magnesium to form hydrogen gas and magnesium chloride. The reason this is formed is because magnesium is higher up in the reactivity series than hydrogen, causing the chloride to "transfer" from hydrogen to magnesium. This leaves hydrogen as a single element rather than a compound. The reason that hydrogen has formed a bond with another hydrogen atom is because the outer shell of electrons is not whole and therefore unstable, and the atoms "feel" an "attraction" with each other to fill these outer shells. (hydrogen atoms have one electron in their outer (and only) electron shell). To fill these outer shells, the element must form a bond with another atom of the same element in order to fill this outer shell (which needs two electrons to fill). In conclusion, the chloride "joins" the magnesium to form magnesium chloride (MgCl2) due to the fact that magnesium is higher up in the reactivity series, leaving hydrogen as a lone element, which forms a bond with another atom of itself in order to become more stable and fill its outer electron shell.

Sadly, this is a combination of truths, half-truths, and wrongs, and doesn't even mention the most important fact: that the metallic magnesium is being oxidized by hydronium ions.

 

[James Pelezo]

The way you wrote it H₃O⁺ cancels out as a spectator and you are left with

Mg(s) + 2Cl^-(aq) → MgCl₂(aq)

which is not (and can't be) balanced. Properly balanced equation is balanced both in terms of atoms and charge, and you can't balance charge having only negatively charged Cl^- on one side of the reaction. So there is apparently something missing here.

And what is missing is the fact Mg is getting oxidized by H⁺, and the real reaction is

Mg(s) + 2H⁺ → Mg²⁺(aq) + H₂(g)

(you can put H₃O⁺ in place of H⁺ and add water to the RHS if you prefer).

Exactly right! I think the H₃O⁺ should be used more often. Too many students think H⁺ exists as an independent entity when using it as in the above 'abbreviated' symbology when, in fact, it does not. H⁺ is a proton (p⁺) and does not float about in solution unattached. I am not saying it wrong using the abbreviated symbology, but it is - in my humble opinion - over used.

Just to add support may I suggest ...
2H₃O⁺(aq) + 2Cl^-(aq) + Mg^o(s) => Mg⁺²(aq) + 2Cl^-(aq) + H₂(g) + 2H₂O(l)
<or>
In half reaction form:
Mg^o(s) => Mg⁺²(aq) + 2e^- (Oxidation)
2H₃O⁺(aq) + 2Cl^-(aq) + 2e^- => 2Cl^-(aq) + H₂(g) + 2H₂O(l) (Reduction)
(Ionic Rxn):
Mg^o(s) + 2H₃O⁺(aq) + 2Cl^-(aq) => Mg⁺²(aq) + 2Cl^-(aq) + 2H₂(g) + 2H₂O(l) (Chloride is the only spectator ion)
(Net Ionic Rxn)
Mg^o(s) + 2H₃O⁺(aq) => Mg⁺²(aq) + 2H₂(g) + 2H₂O(l) (Charge & Mass are balanced as required)

 

[James Pelezo]

OK, this is the process, but it won't hurt to know the question they are trying to solve. It is not clear to me at all.
So far so good, just a titration calculation. Slightly stupid if you ask me, but technically correct.
If their plan was to calculate pH, this is wrong, HNO₂ is not 100% dissociated, even that diluted. About 11% would be in the not dissociated form and pH would be much closer to 4.30

I think what may be the question is ... 'What is the pH of the final mixed solution (titration) when 71.14 ml of 2 x 10^-5 Molar NaOH is added into 25 ml of Nitrous Acid solution?' The statements are assuming the system is neutralized by the weak acid addition and is at the equivalence point. They should note that only NaNO₂ remaining as all of the HNO₂ was consumed in neutralization of the NaOH. Looking at the calculation listed, 2 errors are noticed, 1) their calculation of the concentration of HNO₂ after wk acid addition is incorrect. It should be NaNO₂ and not HNO₂ because that is all that's in solution at the equivalence point. 2) they should divide by .09614 Liter and not 0.025 L. The mix is 71.14 ml + 25.00 ml = 96.14 ml = 0.09614 L. I calculated [NaNO₂] = 1.48 x 10^-5M. The NaNO₂ should then be applied as a salt acting to hydrolyze water producing an alkaline pH at the equivalence point. NaNO₂ => Na⁺(aq) + NO₂^-(aq). The Na⁺(aq) will not react with water, but NO₂^-(aq) will react causing the pH at the equivalence point to be alkaline due to anion hydrolysis. I calculate pH = 12.43 using the data given. I suggest to those that are interested, read (study) Hydrolysis of Salts in aqueous solution. It may also be titled pH of salt solutions. Most college level gen chem texts have good discussions.