Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: escapeartistq on May 13, 2005, 01:45:40 PM

I was solving a titration problem (titrating a solution of KHO with a solution of H_{2}SO_{4}).
And the final question is the number of molecules of water produced during the titration. The problem is that the neutralization net equation is given by:
HO^{}(aq) + H^{+}(aq) > H_{2}O(l) (1)
and that gives me a correct result since I have the number of moles of HO^{} (and H_{3}O^{+}).
However if I use the equation
H_{3}O^{+}(aq) + HO^{}(aq) > 2H_{2}O(l) (2)
I get the double number of molecules which is obvious since the stoichiometry of this equation is different.
My question is: when should I use equation (1) and when should I use equation (2)? And why do chemists use both since it can be a source of confusion?

1. It's KOH, not KHO. So it's OH^{}, not HO^{}.
2. It's not a source of confusion. Actually, writing something like H_{3}O^{+}(aq) is confusing because the structure already suggests it's an aqueus solution. Anyway, both of those equations are correct, but I can't tell you which you should use, because I always wrote: "H^{+} + ..."

Technically, the second is correct. I've never encountered a case where it mattered whether I used one or the other, the Kw value is still the same.

H+ refers to the proton. In fact, H+ never actually exist in solution. H3O+ is one of the forms that H+ exist. In fact, the most common form of H+ is H_{9}O_{4}^{+}
We write H+ to reflect the transfer of the proton.

I understand that the second one is correct because the H^{+} associates with a water molecule to form H_{3}O^{+} but if you use this equation to calculate the number of molecules formed at the end of the titration you will get the double value of the real one because the equation says that for each HO^{} two H_{2}O are formed, which is incorrect!

I understand that the second one is correct because the H^{+} associates with a water molecule to form H_{3}O^{+} but if you use this equation to calculate the number of molecules formed at the end of the titration you will get the double value of the real one because the equation says that for each HO^{} two H_{2}O are formed, which is incorrect!
So instead of writing H_{3}O^{+} write it down (for yourself only!) as H_{2}O.H^{+} to denote the fact that water particle is only accompanying proton and is not taking part in the reaction. This way if you write down the reaction equation water will cancel out and you will get right amount of water from the neutralization.

Thanks. What I did was to write the neutralization equation, in this case it was H_{2}SO_{4}(aq) + 2KHO(aq) > Na_{2}SO_{4}(aq) + 2H_{2}O(l) , and from it I wrote the ionic equation and removed the spectator ions to obtain the net equation and obtained HO^{}(aq) + H^{+}(aq) > H_{2}O(l)
What was bugging me was the discrepancy between the two equations for the water formation. Maybe because I am new to Chemistry and my background is in Computer Science, I tend to see chemical equations symbolically. But as I am learning now, in Chemistry you always have to see the "hidden" entities behind the symbols :)

I don't see the problem of whether you have 2 molecules formed versus 1, in reality, 2 molecules of water are actually formed. Why don't you give me an example of where the difference affects a calculation result in the long run.

the net production is 1 mole of water molecule per mole of neutralisation reaction.

As I said in the first post, the problem I was trying to solve asked for the number of molecules of water produced in the titration.
let's say n(HO^{}) = 5 mol
according to the stoichiometry of equation (1) n(H_{2}O) = n(HO^{})= 5
so the number of molecules of water formed is N = N_{A} * n(H_{2}O) = 6.022 * 10^{23} * 5 = 3.011 * 10^{24}
according to the stoichiometry of equation (2) n(H_{2}O) = 2 * n(HO^{}) = 10
so the number of molecules of water formed is N = N_{A} * n(H_{2}O) = 6.022 * 10^{23} * 10 = 6.022 * 10^{24}

if your reaction consume 1 water molecule and produce 2 water molecule subsequently, what is the overall production?
1. H2O + H+ > H3O+
2. H3O+ + OH> 2H2O
net reaction:
H+ + OH => H2O

As I said in the first post, the problem I was trying to solve asked for the number of molecules of water produced in the titration.
let's say n(HO^{}) = 5 mol
according to the stoichiometry of equation (1) n(H_{2}O) = n(HO^{})= 5
so the number of molecules of water formed is N = N_{A} * n(H_{2}O) = 6.022 * 10^{23} * 5 = 3.011 * 10^{24}
according to the stoichiometry of equation (2) n(H_{2}O) = 2 * n(HO^{}) = 10
so the number of molecules of water formed is N = N_{A} * n(H_{2}O) = 6.022 * 10^{23} * 10 = 6.022 * 10^{24}
well, in that case, you'll need to use the second equation. In my chemistry courses we always had to write H30+, H+ was somewhat unacceptable...

1. H2O + H+ > H3O+
2. H3O+ + OH> 2H2O
net reaction:
H+ + OH => H2O
Seems to me that's the most elegant explanation :)