Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nanorx on December 30, 2007, 03:10:32 PM
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A sol of 100 mL HOCl of unknown conc was titrated with 40 mL of .1 M NaOH. Ka = 3x10-8.
Q1) what is the OCl- conc at equivalence point?
A1) [OCl-] = .029 M
;They reason that at equivalence point, all initial HOCl has been converted to OCl- so they calc for all the initial mole of HOCl which is .004 and equate that to the value of OCl-.
My question #1: I thought HOCl is a weak acid and so it can never completely dissociate all of HOCl to all of H+ & OCl-? If so then their assumption and calc is wrong right?
Q2) what is the pH of sol at equiv point?
A2) pH = 10
; They found 10 by solving for Kb which = 3.33x10-7 then use the Kb = ([HOCl] [OH-]) / [OCl-] to solve for [OH-]. Thereafter, they do the pOH and subtract from 14 to get the pH = 10.
My question #2: since Ka is given and = ([H+] [OCl-])/[HOCl-], why not just use Ka to find out the [OCl-] instead of using the Kb? Do they have to use the Kb route to figure out the pH? It doesn't quite match up to the 10 with the Ka but close. What logic or understanding am I missing here?
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NaOH is a strong base.
Strong with Weak react completely.
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My question #1: I thought HOCl is a weak acid and so it can never completely dissociate all of HOCl to all of H+ & OCl-? If so then their assumption and calc is wrong right?
My question #2: since Ka is given and = ([H+] [OCl-])/[HOCl-], why not just use Ka to find out the [OCl-] instead of using the Kb? Do they have to use the Kb route to figure out the pH? It doesn't quite match up to the 10 with the Ka but close. What logic or understanding am I missing here?
End point is when you add stoichiometric amount of titrant. What you have then is a solution of NaOCl - so starting point for the calculation of pH is [OCl-] = 0.029M
What happens then? Hydrolysis, which mean OCl- reacts with water:
OCl- + H2O = HOCl + OH-
This in turn means you have 0.029 solution of weak base. pH calculations are much easier now - you calculate pOH for weak base with known Kb and convert it to pH. It can be done in many different ways, but the only easier one I know is the use of BATE ;)
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NaOH is a strong base.
Strong with Weak react completely.
Are you saying because OH- is strong base, due to Le Chatlier, it eventually extracts off every H+ out of HOCl thereby making OCl- at the same time (basically convert every HOCl to OCl-) by equiv point time?
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My question #1: I thought HOCl is a weak acid and so it can never completely dissociate all of HOCl to all of H+ & OCl-? If so then their assumption and calc is wrong right?
My question #2: since Ka is given and = ([H+] [OCl-])/[HOCl-], why not just use Ka to find out the [OCl-] instead of using the Kb? Do they have to use the Kb route to figure out the pH? It doesn't quite match up to the 10 with the Ka but close. What logic or understanding am I missing here?
End point is when you add stoichiometric amount of titrant. What you have then is a solution of NaOCl - so starting point for the calculation of pH is [OCl-] = 0.029M
So initially, HOCl autoionize to H+ & OCl- until equiv point. But by adding OH-, it disrupts the would be equilibrium and neutralize the H+. Le Chatelier law's kick in and makes more H+ and eventually consume all of HOCl. Hence, you get Na+ OCl- solution. Is that logic correct?
What happens then? Hydrolysis, which mean OCl- reacts with water:
OCl- + H2O = HOCl + OH-
This in turn means you have 0.029 solution of weak base. pH calculations are much easier now - you calculate pOH for weak base with known Kb and convert it to pH. It can be done in many different ways, but the only easier one I know is the use of BATE ;)
Because hydrolysis occurs, what's left is just Na+ and OCl-, hence it's better to start using pOH instead of pH since there are no H+ left after equiv point. Is this logic correct?
Hydrolysis always occurs in every problem right? Whether acidic or basic. BATE seems cool software. Thanks.
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So initially, HOCl autoionize to H+ & OCl- until equiv point. But by adding OH-, it disrupts the would be equilibrium and neutralize the H+. Le Chatelier law's kick in and makes more H+ and eventually consume all of HOCl. Hence, you get Na+ OCl- solution. Is that logic correct?
Almost. In reality all of HOCl is never consumed, as the hydrolysis starts immediately when there is OCl- base present in the solution.
Because hydrolysis occurs, what's left is just Na+ and OCl-, hence it's better to start using pOH instead of pH since there are no H+ left after equiv point. Is this logic correct?
Hard to say if it is correct or not. There is always some H+ because we are in water. However, final solution is identical with the one obtained by dissolving solid NaOCl - thus you may treat it as containing (initially, before dissolution) only base OCl-. If so, pOH calculation is much easier.
Hydrolysis always occurs in every problem right? Whether acidic or basic.
As long as we are in water, hydrolysis occurs always.