You can derive this equation with a little work but the equilibrium constant is, in reality, the hydrolysis constant for OCl. That is Kh = Kw/Ka where
Ka is for HOCl.
A 1.2M NaOCl solution is prepared by dissolving solid NaOCL in distilled water at 298K. The hydrolysis reaction
OCl-(aq) + H2O(l) --> HOCl(aq) + OH-(aq) occurs.
a. Calculate the value of the equilibrium constant at 298K for the hydrolysis reaction.
b. Calculate the value of [OH-] in the 1.2M NaOCl solution at 298K.
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So far I have the expression:
K= {[HOCl][OH-]}/[OCl-]
...but I don't understand how to find K if you're only given the concentration of NaOCl. Help???
Thanks!
5 answers
Okay, then. I think I know what you're getting at, but is it ok if you point out where I can start? (Not a little sure on deriving equations...)
To be honest about it, all of the derivations I've seen were done in books used 30 years ago when we called that constant the hydrolysis constant and had to derive that it was Kw/Kb. In modern texts, the word hydrolysis constant is never used. They simply write the equation for the hydrolysis, as you have done, and say OCl is acting as a base (note that it pulls a hydrogen away from HOH to make HOCl so it is a Bronsted-Lowry base). Then these same modern texts note that Ka*Kb = Kw, so obviously, Kb = Kw/Ka. Voila! And that's as close as the modern texts get to deriving it BECAUSE modern students have no clue as to what a hydrolysis constant is. It's simply not used in the lingo. WHY? Mostly because the idea is that OCl is acting as a base and everyone KNOWS that KaKb=Kw so we can find Kb by Kb = Kw/Ka.Said another way, Kb IS the hydrolysis constant.
But I can show you how to do it if you wish.
But I can show you how to do it if you wish.
Yeah, that would be nice. Thanks much!
I have tried several times to post this but the site has been having some problems. Maybe this time.
We take a salt NaA and throw it into some water. The Na^+ doesn't hydrolyze but the A^- does. I will omit all of the charges to make this simpler to type.
A + HOH ==> HA + OH
Kh = (HA)(OH)/(A)
Now multiply that expression by (H)/(H) which gives
(H)*(HA)(OH)
--------------- = Kh
(H)*(A)
Note that the numerator contains (H)(OH) which is Kw. What is left?
(HA)
-----
(H)(A)
but that is just 1/Ka; therefore,
Kh = Kw/Ka
We take a salt NaA and throw it into some water. The Na^+ doesn't hydrolyze but the A^- does. I will omit all of the charges to make this simpler to type.
A + HOH ==> HA + OH
Kh = (HA)(OH)/(A)
Now multiply that expression by (H)/(H) which gives
(H)*(HA)(OH)
--------------- = Kh
(H)*(A)
Note that the numerator contains (H)(OH) which is Kw. What is left?
(HA)
-----
(H)(A)
but that is just 1/Ka; therefore,
Kh = Kw/Ka
2 answers