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Topic: How to know the products of NH3 + H2O? (Read 161808 times)

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Offline bongalez

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How to know the products of NH3 + H2O?
« on: July 16, 2011, 05:07:22 am » Hey guys,How can we know the products of an equation? E.g. NH3 + H2O ::equil:: NH4+ + OH-. But how do you know this. I was just staring blankly at NH3 + H2O now knowing where to go. Please help. Thanks in advance. Logged

Offline Professor 0110

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Re: How to know the products of NH3 + H2O?
« Reply #1 on: July 16, 2011, 10:14:16 am » What I think is happening is a mixture of a Bronsted-Lowry acid base reaction (the acid being the hydrogen ion donor and the base being the hydrogen ion receiver) and the Lewis Acid Base theory (where the base is the electron donor and the acid is the electron receiver). So ammonia is the base in the reaction and the water is the acid (it receives an electron and donates a hydrogen). Hope this helps clear things up. Google Bronsted-Lowry and Lewis Acid Base theory if you are still unsure (or ask here, and I'll be happy to explain in further detail). Logged Attempting to be a Chemistry teacher as best I can. :)

Offline bongalez

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Re: How to know the products of NH3 + H2O?
« Reply #2 on: July 16, 2011, 10:33:23 am » Hey Professor,Thanks for the reply. We touched on bronsted lowery theory last year, but i don't think its in our course any more, so i don't think i should worry about it. I have asked my teacher if we need it and am waiting for a reply. Thanks, see you around! Logged

Offline vmelkon

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Re: How to know the products of NH3 + H2O?
« Reply #3 on: July 17, 2011, 08:36:11 am » We know that water molecules break up and recombine all the timeH2O <=> H+ + OH-Quote
But how do you know this
I read it and I remember it.Quote
How can we know the products of an equation?
In the general sense? To start out, look at all sorts of reaction and get a "feel" for it. Logged

Offline fledarmus

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Re: How to know the products of NH3 + H2O?
« Reply #4 on: July 18, 2011, 11:45:20 am » Mostly you know the products of the reaction by knowing what types of reactions the reactants can undergo, and what is required for them to undergo that type of reaction.For example, we know that ammonia (NH3) is a base. That means it can pick up a proton to form an ammonium cation (NH4+). What else is in the reaction that might give up a proton? Water, which can act as an acid or a base. If the ammonia takes a proton from water, what is left behind? A hydroxide ion (OH-).But maybe you don't remember that ammonia is a base (or never knew it to begin with). You can look at it's structure - it is forming three bonds to hydrogen atoms, and since it has five electrons in it's outer shell, that leaves a lone pair of electrons. Lone pairs of electrons can be shared with protons, so you know the molecule at least has the potential to act as a base. From there, you can look up the pKb to see if it is a strong enough base to react with water in an acid-base reaction.You might also realize that nitrogen is more electronegative than hydrogen, and that the nitrogen-hydrogen bonds in ammonia are polar, with the hydrogens being somewhat more positive than the nitrogen. So possibly the ammonia could act as an acid as well. Then you would look up the pKa to see if it was a strong enough acid to react with water, and you will find that it isn't.The key to predicting reaction products is knowing what types of reactions can occur, what types of structures will undergo those reactions, and what conditions are required. After that, you just need to practice to recognize those structures. Logged

Offline adam456

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Re: How to know the products of NH3 + H2O?
« Reply #5 on: July 28, 2011, 05:29:53 pm » Hi dear friend,I found this equation can be solve by doing this.kookiekim77 is getting there, but is not completely right. Ammonia is a weak base so will be only partially ionized in solution..NH3 + H2O → NH4+ + OH -[initial]..0.100... 0 .. 0[change].. .-x .. +x . +x[final]..0.100-x .. x .. xthen the equilibrium constant, Kb is given byKb = x^2/(0.100 - x)now [OH-] = 1.34 x 10^-3 = xso substitute for x and calculate KbTo calculate pHfirst find pOH = -log(1.34 x 10^-3)thenpH + POH = 14 sopH = 14 - pOH Logged
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