A 20.00 mL sample of a .1000M unknown acid solution is titrated with .1000M NaOH. Given that the acid is diprotic and its pKa's are 1.90 and 6.70
a.) Estimate the pH after 10 mL of base are added
b.) estimate the pH after 20 mL of base are added
c.) estimate the pH after 30 mL of base are added
d.) sketch the titration curve ( i do not need the sketch just some help on what would be some critical points on the curve)
Thanks!
5 answers
The critical points on the curve are given by a,b,c. How much of this do you know how to do. Explain the parts you don't know how to do. I don't want to do a lot of work on what you already understand.
I have no clue how to start this any help would be appreciated!
I have no clue how to start this any help would be appreciated!
H2A + OH^- ==> HA^- + H2O k1 = 0.0126
HA^- + OH^- ==> A^2- + H2O k2 = 2E-7
Convert pKa values to k1 and k2 as above. pk1 = 1.9; pk2 = 6.7
You start with 20 mL of 0.1M H2A and you titrate it half way with 10 mL of the NaOH; i.e., starting with 20 mL of 0.1M H2A you have used up half of it by adding 10 mL of the 0.1M NaOH. That leaves half of the H2A untitrated. If you use the Henderson-Hasselbalch equation this means that the salt formed (the conjugate base) = acid untitrated and
pH = pKa + log (base)/(acid)
pH = pK1 + log (1/1) = pKa
b. The pH at 20 mL is the pH determined by the the NaHA. That is done, I'm sure in your text, and is
(H^+) = sqrt[(k2*CHA^- + Kw)/1+(CHA^-/k1)]
c. At 30 mL you have neutralized the first H (it took 20 for that) and you have now used half of HA^- and formed A^2- so HA^- is 1 millimols and A^2- is 1 millimole so pH = pK2.
Here is an approximate view of how that will look.
https://www.google.com/search?q=image+titration+diprotic+acid&ie=utf-8&oe=utf-8
HA^- + OH^- ==> A^2- + H2O k2 = 2E-7
Convert pKa values to k1 and k2 as above. pk1 = 1.9; pk2 = 6.7
You start with 20 mL of 0.1M H2A and you titrate it half way with 10 mL of the NaOH; i.e., starting with 20 mL of 0.1M H2A you have used up half of it by adding 10 mL of the 0.1M NaOH. That leaves half of the H2A untitrated. If you use the Henderson-Hasselbalch equation this means that the salt formed (the conjugate base) = acid untitrated and
pH = pKa + log (base)/(acid)
pH = pK1 + log (1/1) = pKa
b. The pH at 20 mL is the pH determined by the the NaHA. That is done, I'm sure in your text, and is
(H^+) = sqrt[(k2*CHA^- + Kw)/1+(CHA^-/k1)]
c. At 30 mL you have neutralized the first H (it took 20 for that) and you have now used half of HA^- and formed A^2- so HA^- is 1 millimols and A^2- is 1 millimole so pH = pK2.
Here is an approximate view of how that will look.
https://www.google.com/search?q=image+titration+diprotic+acid&ie=utf-8&oe=utf-8
I assume you made that entry by error and put me as the author since the IP address for your original question and that answer attributed to me is the same. If my answer still needs clarification I can help but let's not have any funny business.
2 answers