confuesd about neutralization

[iwilliwill]

I am a grade 10 student studying AP chemistry

what factors determine something's ability to neutralize?

does it have somrthing to do with concentration? or pH? I'm confused

[insane_alien]

its all about pH. which is basically the concentration of H+ ions. if it restores the H+:OH- ion concentrations to 1:1 (pH 7) then it neutralized it.

[ecoli]

exactly... so a two different solutions, one with concentration X of H+ ions and one with concentration X of OH- ions will neutralize each other to form pH 7.

[iwilliwill]

oh..ok thanks!!

so does that mean the stronger an acid the better it is at neutralizing a base?

but weak electrolytes can neutralize too right?..does it have to do with equlibrium?

[aj47]

The stronger the acid, the more molecules disscoiate into H+ ions, so a smaller volume is required to netralise a base compared to a weak acid e.g. ethanoic.

 

Not sure what you mean about electrolytes.

[RyanJ]

Not sure what you mean about electrolytes.

 

Maybe the charged isons such as H+ or OH-?

 

Cheers,

 

Ryan Jones

[woelen]

exactly... so a two different solutions, one with concentration X of H+ ions and one with concentration X of OH- ions will neutralize each other to form pH 7.

No, unfortunately you are not right. Things are quite a lot more complicated. What you are stating here only is valid for a strong acid and a strong base, which both are fully ionized in water (e.g. HCl and NaOH).

 

A counterexample:

 

Suppose we have 1 liter of dilute acetic acid (vinegar) at a concentration of 1 mol/l. Such a solution contains much less than 1 mol/l of H(+) ions, somewhat less than 0.005 mol/l. Now, suppose we add 0.005 mol of NaOH to this solution. When all is mixed well, the concentration of H(+) ions still is around 0.005 mol/l, it only has changed around 0.5% and the solution definitely is not neutralized.

 

Computing the amount of NaOH, needed to make this solution have a pH equal to 7 is amazingly complex, but of course it can be done. If you add 1.00 mol of NaOH, then the pH will be well above 7 (somewhere between 11 and 12) and the liquid is quite alkaline. So, 0.005 mol of NaOH does not neutralize, nor does 1.00 mol of NaOH.

 

The same holds for solutions of equal volume. Suppose we have half a liter of 1 M acetic acid and we have half a liter of 0.005 M NaOH (which contain approximately the same amount of H(+) and OH(-) ions, with slight excess of OH(-) ions), and then we mix both liquids, then the resulting liquid of 1 liter of volume still contains H(+) ions in a concentration not far from 0.005 mol/l.

If on the other hand we mix 1 M acetic acid and 1 M NaOH at equal volumes, then the resulting liquid will be alkaline with a pH between 11 and 12 (conc. of OH(-) around 0.005 mol/l).

[budullewraagh]

eh, woelen, i have to disagree.

 

ecoli said: "one with concentration X of H+ ions and one with concentration X of OH- ions will neutralize each other to form pH 7."

 

you said: "No, unfortunately you are not right. Things are quite a lot more complicated. What you are stating here only is valid for a strong acid and a strong base, which both are fully ionized in water (e.g. HCl and NaOH)."

 

the key words are boldfaced. if you have, say, 1.1 moles of a hydroxide, but only 1 mole ionizes, the mole of hydroxide anions will neutralize 1 mole of hydronium cations. of course, there will be an equilibrium and more hydroxide will ionize as it reacts with the hydronium, but that varies from case to case and a correction can be made stating that we have, say, slightly less hydroxide in this example. but back to the point, ecoli was talking about available ions, and he is right- if you have, say, 80 moles of a monoprotic acid and that yields only 1 mole of hydronium cations, it will neutralize the mole of hydroxide anions from a mole of, say, cesium hydroxide.

[woelen]

but back to the point, ecoli was talking about available ions, and he is right- if you have, say, 80 moles of a monoprotic acid and that yields only 1 mole of hydronium cations, it will neutralize the mole of hydroxide anions from a mole of, say, cesium hydroxide.

No, budullewraagh, you're not right. Your 80 moles of monoprotic acid will indeed neutralize the 1 mole of OH(-) ions of your CsOH, but after this, the solution definitely will NOT be neutral with a pH equal to 7. Still, the solution will be quite acidic (in fact, the amount of H(+) ions still will be fairly close to 1 mol in this example).

 

On the other hand, if you want to introduce the concept of potentially available H(+), then you could say that there is 80 mol of potentially available H(+). Now, if you add 80 mol of CsOH to this, then again, the solution will not have a pH, equal to 7. In that case, the solution will be quite alkaline and you'll have 1 mol of free hydroxide in solution (disregarding the fact of dilution, but that does not change the story very much).

 

Try to perform the computations with the Ka of the monoprotic acid and you'll see what I mean. Also try something similar in reality with e.g. vinegar and sodium hydroxide.

[jdurg]

No' date=' budullewraagh, you're not right. Your 80 moles of monoprotic acid will indeed neutralize the 1 mole of OH(-) ions of your CsOH, but after this, the solution definitely will NOT be neutral with a pH equal to 7. Still, the solution will be quite acidic (in fact, the amount of H(+) ions still will be fairly close to 1 mol in this example).

 

On the other hand, if you want to introduce the concept of potentially available H(+), then you could say that there is 80 mol of potentially available H(+). Now, if you add 80 mol of CsOH to this, then again, the solution will not have a pH, equal to 7. In that case, the solution will be quite alkaline and you'll have 1 mol of free hydroxide in solution (disregarding the fact of dilution, but that does not change the story very much).

 

Try to perform the computations with the Ka of the monoprotic acid and you'll see what I mean. Also try something similar in reality with e.g. vinegar and sodium hydroxide.[/quote']

 

Woelen, I think there's just a misinterpretation of what is being said. If you are talking about the molar quantities of the ACID, then yes, you are correct. Budullewraagh was just pointing out that ecloi was talking about H+ and OH- ions already in solution, not bound to a molecule. Therefore, the Ka/Kb values can be ignored since those relate to the formation of the H+ and OH- values.

 

One mole of H+ is one mole of H+. It doesn't matter where it comes from. One mole of a strong acid, however, is indeed quite different than one mole of a weak acid in regards to available H+.

 

If you have a solution with X moles of H+ ions in it and one with X moles of OH- ions in it, they WILL cancel each other out. Woelen is right in that if you don't have completely ionized acids that the pH will not be 7, but if we are talking hypothetically about a solution with only the H+ ions and one with only the OH- ions, the solution will wind up being at pH 7.