Question

3. A weak triprotic acid, H3Wa, has the following pKa's: pKal = 4. 75 p Ka2 = 7. 89 p Ka3 = 10.65 a. Calculate the pH of the resulting solution when 20.00 mLs of 0.200 M Hz Wa + 45.00 mLs of 0.200 M NaOH are mixed together. Please show all your work. b. Sketch an approximate titration curve for this titration. Show the general shape, and label the: axes equivalence points + who is present buffer regions + who is present where pH =pKa for each of the pKa's + who is present

Answer 1

Answer 3a)

For a weak triprotic acid, moles = 0.200 M * 20.00 ml = 4.00 millimole.

After addition of 2 millimole of base, half equivalence of first proton will be there, pH = pka1 = 4.75

At 4 millimole, first proton will be completely titrated.

pH = (pKa1 + pKa2)/2 = (4.75 + 7.89)/2 = 6.32

At 6 millimole, half equivalence of second proton, pH = pKa2 = 7.89

At 8 millimole, second proton will be completely titrated. pH = (pKa2 + pKa3)/2 = (7.89+10.65)/2 = 9.27

At 10 millimole, half equivalence of third proton, pH = pKa3 = 10.65.

At 12 millimole, third proton will be completely titrated. (End of titration)

Answer a)

moles of base added = 0.200 M * 45.00 ml = 9 millimole.

So, the curve is between complete second titration and half equivalence of third, buffer will be formed between HWa^2- and Wa^3-

Final volume = 20.00 + 45.00 = 65.00 ml

Henderson Hasselbalch equation will be

pH = pKa3 + log (Wa^3-)/(HWa^2-)

Now, in present condition, at 8 millimole of base, there will be 4 millimole of HWa^2-. After addition of 9th millimole, millimole of HWa^2- = 4 - 1 = 3 millimole.

[HWa^2-] = 3 millimole/65 ml = 0.0461 M

Millimole of Wa^3- = 1 millimole

[Wa^3-] = 1 millimole/65 ml = 0.0154 M

Substituting in Henderson Hasselbalch equation

pH = 10.65 + log(0.0154)/(0.0461)

= 10.17

Answer b)

TITRATION CURVE:

Initial pH = √ka₁*C

ka1= 10^-pka1 = 1.77 x 10​​​​​^-5

Initial pH = -log√(1.77 x 10^-5)*0.200 = 2.72

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