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!
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.
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.
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
Ah, the chemistry of hydrolysis. Well, it seems like you're on the right track with the equilibrium expression. To find the value of K, you would need the equilibrium concentrations of all the species involved.
But fear not, my inquisitive friend, for I will guide you through this perplexing path of chemistry! To find the value of K, you need to consider the initial and equilibrium concentrations of the species.
Let's break it down step by step:
Step 1: Write the balanced equation for the hydrolysis reaction.
OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-(aq)
Step 2: Identify the initial concentrations of the species.
In this case, you are given the concentration of NaOCl, which is 1.2 M. Since NaOCl is a strong electrolyte, it completely dissociates to give OCl- and Na+ ions. So, the initial concentration of OCl- is also 1.2 M.
Step 3: Assume that x M of OCl- reacts, which results in the formation of x M of HOCl and OH-. Since the stoichiometric coefficient of OCl- in the balanced equation is 1, the change in concentration will also be x M for HOCl and OH-.
Step 4: Determine the equilibrium concentrations.
The initial concentration of OCl- is 1.2 M, and it reacts to form HOCl and OH-. Therefore, the equilibrium concentration of OCl- will be (1.2 - x) M. Since OCl- is the concentration of a weak electrolyte, we can assume that it does not change significantly during the reaction.
The equilibrium concentrations of HOCl and OH- will both be x M, as they are formed from the hydrolysis of OCl-.
Step 5: Substitute the equilibrium concentrations into the equilibrium expression.
K = {[HOCl][OH-]}/[OCl-]
K = (x)(x)/(1.2 - x)
Now, solving for x may require some fancy footwork with quadratic equations, depending on the value of x. But once you determine the value of x, you can plug it back into the equilibrium expression to find the value of K.
As for calculating the value of [OH-], at equilibrium, the concentration of OH- will be equal to x. So, once you find the value of x, you can simply substitute it into the equation to get [OH-].
I hope this helps, my chemically curious companion! If you have any more questions or need a bit more clownish assistance, feel free to ask!
To find the value of the equilibrium constant (K) for the hydrolysis reaction, you can use the given concentration of NaOCl and the balanced chemical equation. Here's how you can calculate it:
a. Calculate the value of the equilibrium constant (K) at 298K for the hydrolysis reaction:
First, let's assign variables to the concentrations of the species involved in the equation. Let [OCl-] represent the concentration of OCl-(aq), [HOCl] represent the concentration of HOCl(aq), and [OH-] represent the concentration of OH-(aq).
Since NaOCl is a strong electrolyte, it dissociates completely in water, so the concentration of OCl-(aq) in the solution is equal to the concentration of NaOCl, which is given as 1.2M.
The balanced chemical equation shows that the stoichiometric coefficient of OCl-(aq) is 1, and the stoichiometric coefficient of OH-(aq) is 1. Therefore, the concentration of OH-(aq) would also be equal to the concentration of NaOCl, which is 1.2M.
Now, let's consider the concentration of HOCl(aq). According to the equation, for every 1 mole of OCl-(aq) that hydrolyzes, 1 mole of HOCl(aq) is formed. Therefore, the concentration of HOCl(aq) would be 0M initially.
Now, substitute these concentrations into the equilibrium constant expression:
K = [HOCl][OH-] / [OCl-]
Since [HOCl] = 0M, the value of K will be 0.
b. Calculate the value of [OH-] in the 1.2M NaOCl solution at 298K:
As mentioned earlier, the concentration of [OH-] in the 1.2M NaOCl solution is also 1.2M. Therefore, the value of [OH-] is 1.2M.
So, to summarize:
a. The value of the equilibrium constant (K) for the hydrolysis reaction is 0.
b. The value of [OH-] in the 1.2M NaOCl solution is 1.2M.