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A sol of 100 mL HOCl of unknown conc was titrated with 40 mL of .1 M NaOH. Ka = 3E-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 = 3.33e-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?
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 = 3.33e-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?
