Magnesium in hydrochloric acid solution.

  1. 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 ?
  2. jcsd
  3. SteamKing

    SteamKing 9,971
    Staff Emeritus
    Science Advisor
    Homework Helper

    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?
  4. 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:
    $$KOH(aq) + H_3O^+(aq) \rightarrow K^+(aq) + 2H_2O$$
    Last edited: Oct 19, 2013
  5. SteamKing

    SteamKing 9,971
    Staff Emeritus
    Science Advisor
    Homework Helper

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

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

    Staff: Mentor

    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).
    1 person likes this.
  7. 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 ?
    Last edited: Oct 20, 2013
  8. Borek

    Staff: Mentor

    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.
  9. 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.
  10. Borek

    Staff: Mentor

    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.
  11. 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:
    $$pH = -log([NHO_2]) = 4.24$$
    Last edited: Oct 21, 2013
  12. Borek

    Staff: Mentor

    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
  13. 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)
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted