2. 1 What is a strong Acid? An Acid that is 100% ionized in water. Strong Acids: 100% ionized (completely dissociated) in water. HCl + H 2 O  H 3 O + + Cl - often written as: HCl  H + + Cl -

3. 2 Strong Acids: Chloric, HClO 3 Hydrobromic, HBr Hydrochloric, HCl Hydroiodic, HI Nitric, HNO 3 Sulfuric, H 2 SO 4 Perchloric HClO 4 Strong Acids: 100% ionized (completely dissociated) in water. HCl + H 2 O  H 3 O + + Cl -

4. 3 What is a strong Base? A base that is completely dissociated in water (highly soluble). NaOH(s)  Na + + OH - Strong Bases: Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy Group 2A metal hydroxides [Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 ]

5. 4 pH Calculations Along a Titration Curve of a Strong Acid Being Titrated with a Strong Base

6. 5 Calculations Summary: Strong-Strong 1. Find moles of acid (H + ) and base (OH - ) 2. Subtract to find XS 3. Find new volume 4. Find Molarity of XS. 5. Find pH or pOH

7. 6 25.0 mL 0.10 M HCl 0.10 M NaOH 1 2 3 4

8. 7 25.0 mL 0.10 M HCl 0.10 M NaOH Region 1: only acid and water

9. 8 Region 1: Only the Strong acid and water are present and no base has been added. General Solution: Since strong acid is completely ionized. The pH is found by the expression: pH = -Log[H + ] but the total volume must be taken into account in calculating the M olarity of the hydronium ion.

10. 9 Problem: Calculate the pH of a solution prepared by combining 10.0mL of 0.10 M HCl with 20.0mL of water. Question: Can we ignore the H + contribution from water in this case? Work with someone next to you to solve. Yes, since the contribution from the HCl is much, much larger than 1 x 10 -7. 2. Find new Molarity(of H + ion) and the pH. ______________________________= mol H + L 10.0mL 0.10mol HCl 1000mL 1mol HCl 1molH + 0.030L 0.033 3. pH = -Log[0.033] = 1.48

11. 10 25.0 mL 0.10 M HCl 0.10 M NaOH Region 2: some base has been added

12. 11 Region 2: Some base has been added to the Strong acid solution. What has changed? 1. The number of moles of acid is reduced. HA + OH -  salt + H 2 O ex.HCl + NaOH  NaCl + H 2 O or: H + + OH -  H 2 O 2. The total volume is also changed and must be taken into account when calculating the new Molarity of the acid.

13. 12 Sample problem: Calculate the pH of a solution after 8.00 mL of 0.15 M NaOH is added to 25.0 mL of 0.10 M HCl. Solution: Reaction: HCl + NaOH  H 2 O + NaCl + XS Net Ionic: H + + OH -  H 2 O Moles of H + : _______________________= mol H + 25.0mL 0.10 mol HCl 1000mL 0.0025 Moles OH - : ________________________ = mol OH - 8.00mL 0.15mol NaOH 1000mL 0.0012 1 HCl 1 H + 1 NaOH 1 OH -

14. 13 Sample problem: Calculate the pH of a solution after 8.00 mL of 0.15 M NaOH is added to 25.0 mL of 0.10 M HCl. Solution: Reaction: HCl + NaOH  H 2 O + NaCl + XS Net Ionic: H + + OH -  H 2 O Moles of H + : __________________ = mol H + 25.0mL0.10 mol 1000mL 0.0025 Moles OH - : ____________________ = mol OH - 8.00mL0.15mol 1000mL 0.0012 Moles XS: Subtract 0.0025 - 0.0012 = 0.0013 mol H + New Volume = 25.0 + 8.00 = 33.0 mL = 0.033L pH = -Log[0.0013/0.033] = 1.404

15. 14 25.0 mL 0.10 M HCl 0.10 M NaOH Region 3, the equivalence pt.

16. 15 Region 3: The equivalence pt. All of the Strong acid has been neutralized and no XS base is present. Since the solution has a pH, what furnishes the H + ions? Only the ionization of water. Ions from strong acids and strong bases do not cause any ionization of water (no hydrolysis). So? @ 25 o C [H + ]= 1 x 10 -7 so pH = 7 Question: What is conc. of OH - ion?

17. 16 25.0 mL 0.10 M HCl 0.10 M NaOH Region 4: XS Base

18. 17 Region 4: All the Strong acid has been neutralized and the XS base is the dominating factor influencing pH. General Solution: 1. Find moles of XS base (OH - ion). 2. Use total volume in Liters to find Molarity of XS base.

19. 18 Problem: Find the molarity of a solution that was prepared by titrating 25.00mL of 0.10 M HCl with 18.00mL of 0.15M NaOH. Solution: Reaction: HCl + NaOH  H 2 O + NaCl + XS Net Ionic: H + + OH -  H 2 O 1. Find moles H + : 25.00mL = molH + 1000mL 0.10mol HCl 1mol HCL 1molH + 0.0025 2. Find moles OH - : 18.00mL = molOH - 0.15mol NaOH 1000mL1mol NaOH 1mol OH - 0.0027

20. 19 Net Ionic: H + + OH -  H 2 O + XS 1. Find moles H + : 25.00mL = molH + 1000mL 0.10mol HCl 1mol HCL 1molH + 0.0025 2. Find moles OH - : 18.00mL = molOH - 0.15mol NaOH 1000mL1mol NaOH 1mol OH - 0.0027 3. Find moles of XS: (subtract) 0.0027 -0.0025 = 0.0002 mol OH - 4. New Volume: 25.00 + 18.00 = 43.00mL = 0.043L 5. pOH = -Log[0.0002/0.043] = 2.33 and since pH + pOH = 14 : 14 - 2.33 = 11.67 = pH

21. 20 Any Questions?? Summary: Strong-Strong 1. Find moles of acid and base 2. Subtract to find XS 3. Find new volume 4. Find Molarity of XS. 5. Find pH and/or pOH Quiz anyone?