Complex chemical equations

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I'm having trouble with sample AP Equation writing questions such as; solid zinc hydroxide is treated with concentrated sodium hydroxide solution. The sheet my teacher provided does not cover this reaction and many others is there any material out there that tells what will happen in most reactions
 

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  • #2
Monique
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Well, NaOH is a very strong base, ZnOH is also a base, but a weaker one (I think it is the PKa that determines this).

Since ZnOH is the weaker base, it will act as an acid. As you know, acids donate H+ and bases accept H+, this is the basics the reaction depends upon.
 
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that wasn't my question I don't want to know the answer to the sample I provided I just want to know how to do problems like that in general
 
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Monique
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Well, look at the back of your chem book which should have a listing of the ionization constants of acids and bases (Ka)
 
  • #5
Zargawee
Mind that concentration is a basic factor.
but still , the NaOH will act like a base (higher Kb than ZnOH)

So , If you wanted to see which one of them behaves like a base or not , the one with higher pH than the other will act like an base , but remeber than concentration *IS* a basic factor .
 
  • #6
This is not an acid base reaction. This is just a solvation. Zn(OH)n (s) + -OH(aq) + -> Zn(+n)(aq) + -OH (aq)

Or something like that. Zinc, I belive is Zn(II) in this case.

JKLM, to answer your question, no, there's not one convenient source that you can just look up any given reaction. Check your textbook though. You might have missed this when you did your reading assignment.
 
  • #7
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Originally posted by Chemicalsuperfreak
This is not an acid base reaction. This is just a solvation. Zn(OH)n (s) + -OH(aq) + -> Zn(+n)(aq) + -OH (aq)

(snip)
Cotton & Wilkinson say acid-base; Greenwood & Earnshaw say acid-base; Hampel & Hawley say acid-base --- let's go with the mainstream thinking on this.
 
  • #8
Originally posted by Bystander
Cotton & Wilkinson say acid-base; Greenwood & Earnshaw say acid-base; Hampel & Hawley say acid-base --- let's go with the mainstream thinking on this.
How is a hydroxide ion going to abstract a proton from another hydroxide ion? Zn(OH)2 is two hydroxide ions coordinated with Zn+2. In the same way that NaOH is one hydroxide coordinated with Na+1. Are Cotton, Wilkinson, et. al. claiming that this forms O-2? Because that would be a pretty neat trick.
 
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Originally posted by Chemicalsuperfreak
How is a hydroxide ion going to abstract a proton from another hydroxide ion? Zn(OH)2 is two hydroxide ions coordinated with Zn+2. In the same way that NaOH is one hydroxide coordinated with Na+1. Are Cotton, Wilkinson, et. al. claiming that this forms O-2? Because that would be a pretty neat trick.
How? Think wwaaayyyy back to freshman chem --- "amphoteric properties of metal oxides" --- oxy-anions, that sort of "stuff."

You wanta fight city hall on this, be my guest --- whatever happens, you've done it to yourself.
 
  • #10
Originally posted by Bystander
How? Think wwaaayyyy back to freshman chem --- "amphoteric properties of metal oxides" --- oxy-anions, that sort of "stuff."

You wanta fight city hall on this, be my guest --- whatever happens, you've done it to yourself.
See, now that's why I didn't go into inorganic chemistry. Does this form a transient Zn(I) before kicking off the hydroxide? Or is it concerted? Also, will all group II or greater metals do this?
 
  • #11
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Without writing a damned textbook on the subject, there are two "mainstream" rationalizations: 1) hydration (solvation) of the cation followed by charge transfer ionization(s) of the solvating molecules; 2) hydrolysis of the metal-oxide bonds just as occurs for anhydrides of the oxy-acids. Neither is entirely satisfying, and various combinations dressed up with arguments including IP, EA, a little QM, and a lotta Kentucky windage have been presented here and there to "predict" which metals do, and which don't (after the fact) --- it's a case of over-detailing trends through the periodic table.

Do a "yoohoo" on "amphoteric metal oxides" for lists.

Let's see if this works ---

M.(H2O)m+n = M.(H2O)m-1OH +(n-1) + H+

Okay, and,

MO + H2O = M(OH)2 = MO(OH)- + H+ (generalization of this to other metal-oxide stoichiometries will be left as an exercise to the reader.)
 
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