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!
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
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.
To estimate the pH at different points during the titration, you need to consider the stoichiometry of the reaction and the dissociation constants of the acid. Let's break down the steps to find the answers to each part of the question.
a.) Estimate the pH after 10 mL of base are added:
1. Calculate the moles of acid initially present:
Moles of acid = volume (in L) x concentration (in mol/L)
Moles of acid = 20.00 mL x 0.1000 M / 1000 = 0.00200 moles
2. Calculate the moles of base added:
Moles of base = volume (in L) x concentration (in mol/L)
Moles of base = 10.00 mL x 0.1000 M / 1000 = 0.00100 moles
3. Determine which pKa is relevant for this calculation. Since the acid is diprotic, the first pKa (1.90) will be relevant for the first deprotonation step.
4. Calculate the moles of acid remaining after the reaction with 10 mL of base:
Moles of acid remaining = Moles of acid initially present - Moles of base added
Moles of acid remaining = 0.00200 moles - 0.00100 moles = 0.00100 moles
5. Calculate the moles of conjugate base formed:
Moles of conjugate base formed = Moles of base added
Moles of conjugate base formed = 0.00100 moles
6. Calculate the concentration of the conjugate base:
Concentration of conjugate base = Moles of conjugate base formed / volume (in L)
Concentration of conjugate base = 0.00100 moles / 0.03000 L = 0.0333 M
7. Calculate the pOH of the solution:
pOH = -log10(concentration of conjugate base)
= -log10(0.0333 M)
≈ 1.48
8. Finally, calculate the pH of the solution using the relation pH + pOH = 14:
pH + 1.48 = 14
pH ≈ 12.52
So, the estimated pH after adding 10 mL of NaOH is approximately 12.52.
b.) Follow the same steps as in part a, but this time consider that 20 mL of base is added.
c.) Again, follow the same steps, but now consider that 30 mL of base is added.
To sketch the titration curve, you need to plot the pH on the y-axis against the volume of base added on the x-axis. The critical points on the curve correspond to the equivalence points and midpoint between them. The first equivalence point occurs when the moles of base added are equal to the moles of acid initially present, and the second equivalence point occurs when the moles of base added are equal to twice the moles of acid initially present. The midpoint between the two equivalence points occurs when half of the acid has been neutralized.