For the titration of 10.00mL of 0.05116M phosphoric acid (triprotic) with 0.1M N

Question

For the titration of 10.00mL of 0.05116M phosphoric acid (triprotic) with 0.1M NaOH, and given that pKa1=2.89, and pKa2=6.15, please calculate pKa3. Please explain in full detail, thank you.

Table 3.5 pKa values for phosphoric acid pKat pKa: pKa Calculate the Ka values for phosphoric acid 1.29 10 pH ot lst half-equt Glerce po.nt 2.8? fkat, kat. IO pH at 2nd half-equivalence point-6.5-P!? ) ka? :10-pker:08-10-7 Table 3.6 Ks values for phosphoric acid Experimental values Literature values Kai Ka2 Kas Reference for literature values Compare the experimental and literature Ka valucs Revised Summer 2016 4 Espt 3 Potentiometric Titrations

Explanation / Answer

The pKa2 calculated by you is also incorrect.

The half equivalence point is at 6.15. The pH at 6.15 ml is 7.17.

So, pka2 = 7.17.

Ka2 = 10-7.17 = 6.76*10-8

Now to calculate pKa3:

First equivalence point is when you added 4.10 ml of base.

Second equivalence point is when you added 8.20 ml of base, which is double of 4.10.

So, third equivalence point will be when you added (3*4.10) ml = 12.20 ml of base.

Now 3rd half-equivalence point will be at (12.20+8.20)/2 = 10.20 ml

The pH at 10.20 ml is approximately 11.1.

So, pKa3 = 11.1

Ka3 = 10-11.1 = 7.94*10-12

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.