1. Titration of weak Triacid Instructor : Dr.Amer ghannoum Student name : Mona hefnawy File number: 6004 1Titration of Polyprotic Acid 12/23/2020

2. Terminology A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it. An ampholyte is a molecule containing both acid and base functionality. A Titrant is solution of a known concentration. An Analyte or titrand is the solution whose concentration has to be determined. A Triacid is an acid containing 3 acidic H +. 2Titration of Polyprotic Acid 12/23/2020

3. Dissociation constants CouplesKaPka H 3 PO 4 /H 2 PO 4 - 7.9x10 -3 2.1 H 2 PO 4 - /HPO 4 2- 6.3x10 -8 7.2 HPO 4 2- /PO 4 3- 3.98x10 -13 12.4 Kw =Ka 1 x Kb 3 = ka 2 x kb 2 =ka 3 x kb 1 = [H 3 O + ][OH - ] =10 -14 3Titration of Polyprotic Acid 12/23/2020

4. Titration of phosphoric acid with sodium hydroxide Phosphoric acid is triprotic ; it reacts with sodium hydroxide in 3 steps : H3PO4 + NaOH  NaH2PO4 + H2O NaH2PO4 + NaOH  Na2HPO4+ H 2 O Na2HPO4 + NaOH  Na3PO4+ H 2 O 4 3 buffer solutions 2 ampholytes Titration of Polyprotic Acid 12/23/2020

5. Titration of Polyprotic Acid5

6. First Equivalence 1 st step : H 3 PO 4 + NaOH  NaH 2 PO 4 + H 2 O 6 VolumeRuleReason V=0mlka 1 >10 3 ka 2 H 3 PO 4 is considered as behaving as weak monoacid. 0<V<10mlBuffer solution pH=pka 1 1 st half equivalence V=10ml V=V E1 NaH 2 PO 4 “ampholyte” Titration of Polyprotic Acid 12/23/2020

7. Second Equivalence 2 nd step :NaH 2 PO 4 + NaOH  Na 2 HPO 4 + H 2 O 7 VolumeRuleReason 10<V<20mlbuffer solution pH=pka 2 2 nd half equivalence V=20ml V=V E2 Na 2 HPO 4 “ampholyte” Titration of Polyprotic Acid 12/23/2020

8. Third Equivalence 3 rd step : Na 2 HPO 4 + NaOH  Na 3 PO 4 + H 2 O 8 VolumeRuleReason 20<v<30 ml Buffer solution pH=pka 3 half Veq3 V=30ml V=V E3 Na 3 PO 4  weak base v>30 mlExcess NaOH Titration of Polyprotic Acid 12/23/2020

9. V E1 V E2 V E3 pH=pka 1 pH=pka 2 pH=pka 3 H 3 PO 4 H 2 PO 4 - H 2 PO 4 - HPO 4 2- PO 4 3- H 2 PO 4 - HPO 4 2- PO 4 3- Ampholyte Basic salt V base PO 4 3- OH - 9Titration of Polyprotic Acid 12/23/2020

10. Titration of Polyprotic Acid10 H 3 PO 4 + NaOH  NaH 2 PO 4 InitialCaVaCbVb Change-CbVb CbVb EndCaVa-CbVb _CbVb NaH 2 PO 4 + NaOH  Na 2 HPO 4 InitialCaVaCb(Vb-V E1 ) Change-Cb(Vb-V E1 ) Cb(Vb-V E1 ) EndCaVa-Cb(Vb-V E1 ) _Cb(Vb-V E1 ) Na 2 HPO 4 + NaOH  Na 3 PO 4 InitialCaVaCb(Vb-V E2 ) Change-Cb(Vb-V E2 ) Cb(Vb-V E2 ) EndCaVa-Cb(Vb-V E2 ) _Cb(Vb-V E2 ) Divide by V total

11. Calculation of ka 1 ka 2 ka 3 11Titration of Polyprotic Acid 12/23/2020

12. Application Consider the titration of H 3 PO 4 with a strong base NaOH Given : Ca=0.2M Cb=0.2M Va=10ml Calculate pH after adding: 12 V NaOH ml 05710151820223040 pH Titration of Polyprotic Acid 12/23/2020

13. Solution 13Titration of Polyprotic Acid 12/23/2020

14. Solution 14Titration of Polyprotic Acid 12/23/2020

15. Solution 15Titration of Polyprotic Acid 12/23/2020

16. Solution 16 V NaOH ml 05710151820223040 pH 1.42.12.464.657.27.89.811.7912.5512.6 Titration of Polyprotic Acid 12/23/2020

17. Exercise Phosphoric acid, H3P04,is a triprotic acid with Kal= 7.5 X 10 -3,Ka2= 6.2 X 10 -8 and Ka3= 4.8 X 10 -13 o Consider the titration of 50.0 mL of 1.0 M H 3 P0 4 by 1.0 M NaOH and o answer the following questions. A. Write the reactions associated with Ka 1, Ka 2 'Ka 3 ' B. Calculate the pH after the following total volumes of NaOH have been added: 1) 0.0 mL of NaOH (pH =1.08) 2) 25.0 mL of NaOH 3) 50.0 mL of NaOH 4) 75.0 mL of NaOH 5) 100.0 mL of NaOH 6) 125.0 mLof NaOH 7) 150.0 mL of NaOH (pH =12.80) C.Sketch the titration curve for this titration 12/23/2020 Titration of Polyprotic Acid17

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