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 -

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 -

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 ]

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

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

mL 0.10 M HCl 0.10 M NaOH

mL 0.10 M HCl 0.10 M NaOH Region 1: only acid and water

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.

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 Find new Molarity(of H + ion) and the pH. ______________________________= mol H + L 10.0mL 0.10mol HCl 1000mL 1mol HCl 1molH L pH = -Log[0.033] = 1.48

mL 0.10 M HCl 0.10 M NaOH Region 2: some base has been added

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.

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 mL 0.10 mol HCl 1000mL Moles OH - : ________________________ = mol OH mL 0.15mol NaOH 1000mL HCl 1 H + 1 NaOH 1 OH -

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 mL0.10 mol 1000mL Moles OH - : ____________________ = mol OH mL0.15mol 1000mL Moles XS: Subtract = mol H + New Volume = = 33.0 mL = 0.033L pH = -Log[0.0013/0.033] = 1.404

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

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). 25 o C [H + ]= 1 x so pH = 7 Question: What is conc. of OH - ion?

mL 0.10 M HCl 0.10 M NaOH Region 4: XS Base

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.

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 mL 0.10mol HCl 1mol HCL 1molH Find moles OH - : 18.00mL = molOH mol NaOH 1000mL1mol NaOH 1mol OH

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

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?