Chemistry Solving Ionic Equilibrium Problem: Volume of HCl & Reaction Explained

AI Thread Summary
The discussion centers on a titration problem involving a weak base (BOH) and HCl, where the volume of HCl needed to reach the endpoint is determined to be 20 ml. Participants clarify that at the endpoint, the solution contains BH+ and Cl- ions, and when NaOH is added, the resulting species include BOH and NaCl. The Henderson-Hasselbalch equation is suggested for calculating pH after NaOH addition, emphasizing the importance of understanding the chemical species present in the solution. Confusion arises regarding the buffer formation and the role of various ions, but clarity is achieved through examples and corrections. Ultimately, the calculated pOH value aligns with the book's answer, highlighting the need for careful reading and comprehension of the problem.
Prabs3257
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Homework Statement
When a 20 ml of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 ml of 0.04M NaOH is added to the resulting solution?
Relevant Equations
concept of hydrolysis
i am having a problem in understanding the question , what will be the volume of hcl ? and how is the reaction going to take place ? please help me i am stuck on this problem for quite a long time.
 

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Prabs3257 said:
Homework Statement:: When a 20 ml of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 ml of 0.04M NaOH is added to the resulting solution?
Relevant Equations:: concept of hydrolysis

i am having a problem in understanding the question , what will be the volume of hcl ? and how is the reaction going to take place ? please help me i am stuck on this problem for quite a long time.

For your question about the HCl, what do you understand as meant by "titrated"? Have you done any titrations?
 
epenguin said:
What do you understand as meant by "titrated"? Have you done any titrations?
i think it means when the equivalents of acid and base are equal we get the end point
 
Prabs3257 said:
i think it means when the equivalents of acid and base are equal we get the end point

Good enough.So how many equivalents of HCl do you need to add to reach that? - or, rather directly considering the HCl and base concentrations are equal, what volume of HCl?
 
its 20 ml hcl right and the reaction would be boh + hcl == bcl +h2o and in the resulting solution we will have only bcl and h2o right ?? i have done a few calculations and i got the concentration of bcl as 0.04 in the mixture and then with the concept of hydrolysis i calculated the pkb of boh which came out to be 5.4 is it right ?? now i don't know what to next
 
I agree with your result. Now it is as if you had taken a solution of the buffer salt BH+Cl- (often written on the bottle label "B.HCl" - check some out if you get a chance when next in a laboratory, or else check out a catalogue), and added NaOH. To calculate the pH on adding base to this salt you would normally go through the Henderson-Hasselbalch equation, but here the NaOH has been added in a special ratio, which saves you trouble.
 
that is the problem i don't understand why is their a buffer ?? can you please write the reactions on how the buffer is formed what are its constituants
 
That is what you have to know or be able to reason out yourself, qualitatively before quantitatively.
When you are at the equivalence point (or when you have dissolved buffer salt) what chemical species are present as major components? Then when you add to that NaOH, but not so much as to totally neutralise the acid, what other major components are present?
 
please correct me if i am wrong is it that before adding there is b+ ions(1.6 moles) from boh and now by adding 0.4 moles of naoh they both react to form boh (which acts as the weak base in the buffer) and as naoh is limiting reagent we have b+ ions left which are now 1.2 moles which act as the salt part (conjugate) of the buffer is it correct ??
 
  • #10
Have another try. There are no B+ ions and I almost told you the ions present in the salt in #6.
 
  • #11
epenguin said:
Have another try. There are no B+ ions and I almost told you the ions present in the salt in #6.
isnt there b+ from bcl which was formed ?? sorry i am confused right now
 
  • #12
Check #6 again, and, for example if the base is ammonia, or trimethylamine or pryidine, and you add equivalent HCl, what have you got in solution?
 
  • #13
by adding ammonia and hcl you should
epenguin said:
Check #6 again, and, for example if the base is ammonia, or trimethylamine or pryidine, and you add equivalent HCl, what have you got in solution?
by adding ammonia and hcl you should get nh4cl
 
  • #14
Do you know, in aqueous solution of ammonium chloride what chemical species are present?
 
  • #15
epenguin said:
Do you know, in aqueous solution of ammonium chloride what chemical species are present?
nh4+ and cl-
 
  • #16
Yes, which is also what you would get if you titrate ammonia with an equivalent amount of HCl. So if the base is instead of ammonia, just general B, what species will be present?
 
  • #17
ohh i got it now thanks alot
 
  • #18
Plus then what other species are present after you have added less than equivalent NaOH? Better see it through, giving whole argument for your solution, see my sig.
It could be a bit treacherous as I think the book answer is wrong. (Accordingly, other interventions welcome.)
 
  • #19
i think after adding naoh only bcl boh and nacl should remain in the solution according to the equation bcl + naoh == boh + nacl
 
  • #20
by using the Henderson-Hasselbalch equation i got the correct answer
 
  • #21
Prabs3257 said:
i think after adding naoh only bcl boh and nacl should remain in the solution according to the equation bcl + naoh == boh + nacl
Prabs3257 said:
by using the Henderson-Hasselbalch equation i got the correct answer
You seem reluctant to use, don't give much evidence of knowing, what chemical species are present in solution.You hopefully do know that after titrating with HCl, the solution will contain BH+ and Cl- - actually both at 0.04 M. After adding a less than equivalent amount of NaOH it will contain some of those and also B and Na+ - each of those, to good approximation, in equal molarities.

You will be in difficulties for this kind of calculation if you cannot readily call to mind these things.
However I now agree with the book about the calculated pOH value.
(The added moles of NaOH are a quarter of those of base. If they had been a half (as I was mistakenly calculating in#6) the pOH would have equalled pKb, not needing HH, so maybe this was a trap for superficial reading of the problem!)
 
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