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What is a strong Acid?

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 -. Strong Acids: 100% ionized (completely dissociated) in water.

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What is a strong Acid?

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  1. What is a strong Acid? An Acid that is 100% ionized in water. Strong Acids: 100% ionized (completely dissociated) in water. HCl + H2O  H3O+ + Cl- often written as: HCl  H+ + Cl-

  2. Strong Acids: 100% ionized (completely dissociated) in water. HCl + H2O  H3O+ + Cl- Strong Acids: Chloric, HClO3 Hydrobromic, HBr Hydrochloric, HCl Hydroiodic, HI Nitric, HNO3 Sulfuric, H2SO4 Perchloric HClO4

  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

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

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

  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 Molarity 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? Yes, since the contribution from the HCl is much, much larger than 1 x 10-7. Work with someone next to you to solve. 2. Find new Molarity(of H+ ion) and the pH. 10.0mL ______________________________= mol H+ 0.10molHCl 1molH+ 0.033 L 1molHCl 1000mL 0.030L 3. pH = -Log[0.033] = 1.48

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

  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 + H2O ex. HCl + NaOH  NaCl + H2O or: H+ + OH-  H2O 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  H2O + NaCl + XS Net Ionic: H+ + OH-  H2O 0.10 molHCl 1 H+ Moles of H+ : _______________________=molH+ 25.0mL 0.0025 1000mL 1 HCl 0.15molNaOH 8.00mL 1 OH- Moles OH-: ________________________ = mol OH- 0.0012 1 NaOH 1000mL

  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  H2O + NaCl + XS Net Ionic: H+ + OH-  H2O Moles of H+ : __________________ = molH+ 25.0mL 0.10 mol 0.0025 1000mL Moles OH-: ____________________ = mol OH- 8.00mL 0.15mol 0.0012 1000mL Moles XS: Subtract 0.0025 - 0.0012 = 0.0013molH+ New Volume = 25.0 + 8.00 = 33.0 mL = 0.033L pH = -Log[0.0013/0.033] = 1.404

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

  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? @ 25oC [H+]= 1 x 10-7 so pH = 7 Question: What is conc. of OH- ion?

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

  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  H2O + NaCl + XS Net Ionic: H+ + OH-  H2O 1. Find moles H+: 0.10molHCl 1molH+ 25.00mL= molH+ 0.0025 1000mL 1molHCL 2. Find moles OH-: 0.15molNaOH 1molOH- 18.00mL = molOH- 0.0027 1000mL 1molNaOH

  19. Net Ionic: H+ + OH-  H2O + XS 1. Find moles H+: 0.10molHCl 1molH+ 25.00mL= molH+ 0.0025 1000mL 1molHCL 2. Find moles OH-: 0.15molNaOH 1molOH- 18.00mL = molOH- 0.0027 1000mL 1molNaOH 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

  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?

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