I have a PhD in chemistry but haven't done acid base chemistry in years but needed to help my daughter who is home from college during pandemic and this was an awesome refresher!
Good point-it can also be said of the first equivalence point that [H2A] is essentially zero there. Of course, where it's indicated at 3:24 in the central flat region of the curve, [H2A] is even smaller! It is important to keep in mind that at the equivalence point, some back reaction of HA- with water produces H2A and hydroxide.
Thank you so much. You have cleared my doubt on finding the pka2 from the titration curve, since I could not understand why the pH at half of neutralization point was not coming as pka2
Thank you! It's bery useful. But just wondering a point: why there is a steep increase at the first equivalence point? Should those two buffer regions connected together? Because there is no sudden increase in [H+] at the first equivalence point. I will quite appreaciate if anyone can answer me this question.
Oy, that should have read [HA-]/[H2A] inside the parentheses. Right idea, but the wrong molecules were written inside the brackets. Good catch, and thanks!
Arup, the key is to realize that at each equivalence point, all you have is a solution of the conjugate (acid/base) of whatever you started with. (Here, we have a solution of the conjugate bases at each equivalence point.) The volume at this point is the sum of the volume of titrant added and the volume of analyte solution. To calculate the pH, treat the solution as a weak acid/base solution (as the case may be).
Since all you have is your conjugate base of your acid, it'll be that reacting with water. You then make an ice table and get Kb = x^2/(conc of base -x), the x in the denominator we can usually ignore because Kb is usually very small. Since you only know the Ka of your acid, you will use this equation KaKb = Kw and solve for Kb. The x in the above equation is equal to the OH- concentration. Then you use your OH- concentration, calculate the pOH and use that to get your pH. (Sorry this is late.)
I have a PhD in chemistry but haven't done acid base chemistry in years but needed to help my daughter who is home from college during pandemic and this was an awesome refresher!
Thank you. It really helped. I have my A2 exams in a few hours and this was the area i wasnt quite confident about. Great explanation
Thanks! Hope your exams went well!
Michael Evans it did we got a question on the buffer region you explained where pKa=pH
At 3:24 shouldn't the point where [H2A] is practically zero be at the middle of the 1st steep section?
Good point-it can also be said of the first equivalence point that [H2A] is essentially zero there. Of course, where it's indicated at 3:24 in the central flat region of the curve, [H2A] is even smaller! It is important to keep in mind that at the equivalence point, some back reaction of HA- with water produces H2A and hydroxide.
Thank you so much. You have cleared my doubt on finding the pka2 from the titration curve, since I could not understand why the pH at half of neutralization point was not coming as pka2
Thank you! It's bery useful. But just wondering a point:
why there is a steep increase at the first equivalence point? Should those two buffer regions connected together? Because there is no sudden increase in [H+] at the first equivalence point.
I will quite appreaciate if anyone can answer me this question.
To the point and well-explained :). Thank you very much!
thank u so much 🙌💕💕💕💕baghdad
Thank you so much, This is the explanation realy I needed
Hi, how to calculate acid concentration from this titration?
what do you mean
[CoVo(for acid)-V(aded)C(NaOH)]/Vo+V(aded) . for one protonic acids
Great explanation, it really helped sort this out for me!
thank you Michael Evans :,)
How about diprotic weak base titration with strong acid (e.g Quinine, Piperazine)
At 3.24 minute there is some thing maybe wrong 🙈🌿 when you divide (A-/HA) TO Prove PH=PKa1 from handerson equation??
Thank you very much 🍂🌺
yeah I felt the same!
Oy, that should have read [HA-]/[H2A] inside the parentheses. Right idea, but the wrong molecules were written inside the brackets. Good catch, and thanks!
Very nice explanation. Can you help me getting pH at both the equivalent points??
Arup, the key is to realize that at each equivalence point, all you have is a solution of the conjugate (acid/base) of whatever you started with. (Here, we have a solution of the conjugate bases at each equivalence point.) The volume at this point is the sum of the volume of titrant added and the volume of analyte solution. To calculate the pH, treat the solution as a weak acid/base solution (as the case may be).
Michael Evans thanks a lot.
how to calculate the pH at equivalence point:(
Use an ice table
Since all you have is your conjugate base of your acid, it'll be that reacting with water. You then make an ice table and get Kb = x^2/(conc of base -x), the x in the denominator we can usually ignore because Kb is usually very small. Since you only know the Ka of your acid, you will use this equation KaKb = Kw and solve for Kb. The x in the above equation is equal to the OH- concentration. Then you use your OH- concentration, calculate the pOH and use that to get your pH. (Sorry this is late.)
from hydrolize reacion
Thank you