1. Acids and Bases, Definitions, Ionization and pH, Neutralization and Molarity Calculations Chapter 10

2. Homework Due Problem from last lecture slide show. Homework problems assigned from syllabus and last lecture.

3. Acids and bases Properties and Definitions page 323; Table 10.2 page 323; Table 10.2 CharacteristicsAcidsBases Taste Feel pH LitmusRedBlue

4. Acids and Bases Acids (page 320) Give up protons Give up protons H → H + + H → H + + HCl (Hydrogen Chloride = Hydrochloric Acid) HCl (Hydrogen Chloride = Hydrochloric Acid) Bases (page 321) Accept protons OH - containing compounds (Hydroxide bases). NaOH (Sodium Hydroxide) NaOH (Sodium Hydroxide)

5. Naming Acids; Table 10.1, pg 320 Acid Formula Name of Acid Acid Anion Name of Acid Anion HCl Hydrochlori c acid Cl- HBr HNO 3 HNO 2 H 2 SO 4 H 2 SO 3 H 2 CO 3 H 3 PO 4 HClO 3 HClO 2 CH 3 COOH

6. Acids; Table 10.1 and Bases Acids (page 320) Give up protons Give up protons H → H + + H → H + + HCl (Hydrogen Chloride = Hydrochloric Acid) HCl (Hydrogen Chloride = Hydrochloric Acid) Bases (page 321) Accept protons OH - containing compounds (Hydroxide bases). NaOH (Sodium Hydroxide) NaOH (Sodium Hydroxide)

7. Hydrogen Ion Hydrogen Ion; Proton Review Review Hydrogen Atom H has 1 e - & 1 p + (Col1, Period 1) has 1 e - & 1 p + (Col1, Period 1) Hydrogen Ion H + Hydrogen Ion H + Hydrogen atom looses an e -, left with 1 p +

8. Ionization in Water H H \ / \ / O - +- + δ-δ- δ ++ δ Ionic compounds will dissolve in water to make ions. This process is called ionization (to make ions). Ions will be attracted to the partial charges of water. Several partial charges equal one whole charge. O / \ / \ H H δ-δ- δ + + δ Na Na Cl Cl NaCl (s) NaCl (s) → (in water; aqueous) δ-δ- + δ O / \ / \ H H δ + + (aq) (solid) (aq)

9. Ionization of Water H H \ / \ / O - +- + δ-δ- δ ++ δ Just like ionic compounds will dissolve in water to make ions. Water can ionize in itself. Ions will be attracted to the partial charges of water. Several partial charges equal one whole charge. O / \ / \ H H δ-δ- δ + + δ H H OH OH HOH (l) HOH (l) → (in water; aqueous) δ-δ- + δ O / \ / \ H H δ + + (aq) (liquid) (aq)

10. Hydronium Ion Hydronium Ion forms when a hydrogen ion is attracted to a water molecule. +- H Cl HCl (s) HCl (s) → + (aq) + H δ-δ- + δ O / \ / \ H H δ + H3O+H3O+H3O+H3O+ This is the state for hydrogen ions in water (ionized acids). Written as H 3 O +.

11. Ionization of Water + - H OH HOH (l) HOH (l) → + (aq) H3O+H3O+H3O+H3O+ Hyrodgen ions in water will also exist as hydronium ions. They are also written as H 3 O +. (aq)

12. Solution pH (Weak Acid/Base Solution) Neutral Solutions have no [H 3 O + ] pH =7 pH =7 Water Water Salts Salts Acid Solutions have [H 3 O + ] Weak acid solution (<1M) have a pH = 0 to 7 Weak acid solution (<1M) have a pH = 0 to 7 Base Solutions have no [H 3 O + ] but have [OH - ] Weak basic solution (<1M) have a pH = 7 to 14 Weak basic solution (<1M) have a pH = 7 to 14 pH measures the amount of H 3 O +. pH = log [H 3 O + ] [ ] means concentration in Molarity

13. Reaction of acids and bases Neutralization Reactions When an acid and a base react, they make a salt and water. they make a salt and water. HCl + NaOH → NaCl + H 2 O Hydrochloric Sodium Sodium Hydrogen Acid Hydroxide Chloride Hydroxide Because the acid reacted with the base to form a neutral solution of salt and water, we call this a neutralization reaction.

14. Titration in Neutralization Reactions Because acids and bases react to form a neutral solution, we can use the balanced equation to determine the exact amount of acid required to neutralize a know amount of base. If we know when the pH changes to neutral, we can determine the exact amounts of acid and base. We know the amounts because we can measure them. We first measure the amount of base. We first measure the amount of base. We then add measured amounts of acid (titrate) until the solution is neutral. We then add measured amounts of acid (titrate) until the solution is neutral. Too little acid and the solution is still basic. Too little acid and the solution is still basic. Too much acid and the solution will be acidic. Too much acid and the solution will be acidic.

15. Indicators in Titration Indicators are added to determine (indicate) when a pH change has occurred. Different indicators are used to determine different pH changes. Phenolphathalein is a different color in acid and base. In acid it is colorless, in base it is a dark redish purple. It changes color as you add acid to base. At pH = 7.0, it just turns pink. This is a good indicator that your reaction is neutral (balanced). We can use this information to calculate amounts and concentrations of solids and solutions.

16. Titration of Acids and Bases When a known volume of base is used to titrate (added to so it reacts) a known volume of acid, until the resulting solution is neutral (water and salt), we call this a titration reaction. Did you see the indicator in the solution in the flask change color when it became neutral? Did you see the indicator in the solution in the flask change color when it became neutral? The acid in the flask was neutralized with base. Known volume

17. Titration of Acids and Bases The acid in the flask was neutralized with base. Did you notice that some base remained in the buret? If we continue to add the base, the solution will be basic. Known volume

18. Titration of Acids and Bases – Calculations, Molarity You can use M and volume to calculate moles (same compound).

19. Review of Molarity Calculations KOH H 3 PO 4 K 3 PO 4 + → + H2OH2OH2OH2O Potassium hydroxide Hydrogen phosphate Potassium phosphate Hydrogen hydroxide (Water) 3 113 KOH moles (KOH) H 3 PO 4 moles (H 3 PO 4 ) K 3 PO 4 moles (K 3 PO 4 ) H 2 O moles (H 2 O) KOH Molarity (KOH) H 3 PO 4 Molarity (H 3 PO 4 ) K 3 PO 4 Molarity (K 3 PO 4 ) H 2 O Molarity (H 2 O) KOH vol,ml (KOH) H 3 PO 4 vol,ml (H 3 PO 4 ) K 3 PO 4 vol,ml (K 3 PO 4 ) H 2 O vol,ml (H 2 O) moles (compound in solution) M (compound in solution), moles/l = ---------------------------------, volume (compound in solution), l

20. Titration of Acids and Bases – Calculations, M and grams You can use M and volume to calculate moles (same compound). You can use moles to calculate the number of grams (same compound).

21. Molarity Calculation with Gram/M Conversion KOH H 3 PO 4 K 3 PO 4 + → + H2OH2OH2OH2O Potassium hydroxide Hydrogen phosphate Potassium phosphate Hydrogen hydroxide (Water) 3 113 KOH moles (KOH) H 3 PO 4 moles (H 3 PO 4 ) K 3 PO 4 moles (K 3 PO 4 ) H 2 O moles (H 2 O) KOH grams (KOH) H 3 PO 4 grams (H 3 PO 4 ) K 3 PO 4 grams (K 3 PO 4 ) H 2 O grams (H 2 O) KOH m.mass (KOH) H 3 PO 4 m.mass (H 3 PO 4 ) K 3 PO 4 m.mass (K 3 PO 4 ) H 2 O m.mass (H 2 O) KOH Molarity (KOH) H 3 PO 4 Molarity (H 3 PO 4 ) K 3 PO 4 Molarity (K 3 PO 4 ) H 2 O Molarity (H 2 O) KOH vol,ml (KOH) H 3 PO 4 vol,ml (H 3 PO 4 ) K 3 PO 4 vol,ml (K 3 PO 4 ) H 2 O vol,ml (H 2 O)

22. Titration of Acids and Bases – Calculations, M and grams You can use M and volume to calculate moles (same compound). You can use moles to calculate the number of grams (same compound). You can use the M and volume of one compound to calculate the M of another compound if you have its volume.

23. Titration of Acids and Bases - Calculations If you know the Molarity of the acid, you can calculate the M of the base. If you know the Molarity of the base, you can calculate the M of the acid. Known volume Known Molarity Known volume Calculated Molarity Known Molarity

24. Molarity Calculation with Gram/Mole Conversion KOH H 3 PO 4 K 3 PO 4 + → + H2OH2OH2OH2O Potassium hydroxide Hydrogen phosphate Potassium phosphate Hydrogen hydroxide (Water) 3 113 KOH moles (KOH) H 3 PO 4 moles (H 3 PO 4 ) K 3 PO 4 moles (K 3 PO 4 ) H 2 O moles (H 2 O) KOH grams (KOH) H 3 PO 4 grams (H 3 PO 4 ) K 3 PO 4 grams (K 3 PO 4 ) H 2 O grams (H 2 O) KOH m.mass (KOH) H 3 PO 4 m.mass (H 3 PO 4 ) K 3 PO 4 m.mass (K 3 PO 4 ) H 2 O m.mass (H 2 O) KOH Molarity (KOH) H 3 PO 4 Molarity (H 3 PO 4 ) K 3 PO 4 Molarity (K 3 PO 4 ) H 2 O Molarity (H 2 O) KOH vol,ml (KOH) H 3 PO 4 vol,ml (H 3 PO 4 ) K 3 PO 4 vol,ml (K 3 PO 4 ) H 2 O vol,ml (H 2 O)

25. Practice 9.A. Calculate the M of a KOH solution if 25 ml of 6M H 3 PO 4 was used to titrate 50 ml of KOH. Look at the Molarity/Gram to mole Conversion Guide Slide (#24). First box in what you have (start). Next box in what you need to do (finish). Can you see we need three steps? 1.Convert the M &V given (start step) to moles H 3 PO 4. 2.Convert moles H 3 PO 4 to moles K 3 PO 4. 3.Convert moles K 3 PO 4 to M using the V given (finish step).

26. Molarity Calculation with Gram/Mole Conversion KOH H 3 PO 4 K 3 PO 4 + → + H2OH2OH2OH2O Potassium hydroxide Hydrogen phosphate Potassium phosphate Hydrogen hydroxide (Water) 3 113 KOH moles (KOH) H 3 PO 4 moles (H 3 PO 4 ) K 3 PO 4 moles (K 3 PO 4 ) H 2 O moles (H 2 O) KOH grams (KOH) H 3 PO 4 grams (H 3 PO 4 ) K 3 PO 4 grams (K 3 PO 4 ) H 2 O grams (H 2 O) KOH m.mass (KOH) H 3 PO 4 m.mass (H 3 PO 4 ) K 3 PO 4 m.mass (K 3 PO 4 ) H 2 O m.mass (H 2 O) KOH Molarity (KOH) H 3 PO 4 Molarity (H 3 PO 4 ) K 3 PO 4 Molarity (K 3 PO 4 ) H 2 O Molarity (H 2 O) KOH vol,ml (KOH) H 3 PO 4 vol,ml (H 3 PO 4 ) K 3 PO 4 vol,ml (K 3 PO 4 ) H 2 O vol,ml (H 2 O)

27. Practice 9.A. Step 1. moles (compound in solution) M (compound in solution), moles/l = ---------------------------------, volume (compound in solution), l H3PO4 moles ( H3PO4 in solution) H3PO4 6M ( H3PO4 in solution), moles/l = --------------------------------- H3PO4 0.025 l ( H3PO4 in solution) H3PO4H3PO4 H3PO4 moles ( H3PO4 in solution) = 6M ( H3PO4 in solution) * 0.025 l ( H3PO4 in solution) Remember that M=moles/l H3PO4H3PO4 moles ( H3PO4 in solution) = 0.150 moles ( H3PO4 in solution)

28. Practice 9.A; Step 1 to Step 2 Look at the Molarity/Gram to mole Conversion Guide Slide (#24). You should be able to see what we do next. Can you see we need three steps? 1.Convert the M &V given (start step) to moles H3PO4 2.Convert moles H3PO4 to moles K3PO4. 3.Convert moles K3PO4 to M using the V given (finish step). We have converted the given (start step 1) to moles H 3 PO 4. Can you see we have to next convert moles H 3 PO 4 to moles KOH (step 2).

29. Practice 9.A; Step 2 [From ] (Convert it) [word ] * [From] = [ problem] [Equality/ Equation] Equation] 3 1 (H3PO4) 0.150 moles (H3PO4) * # moles (Comp A) * # # (H3PO4) [ moles (H3PO4) ] (KOH) [ moles (KOH) ] [ moles (Comp B) ] [ moles (Comp A) ] (KOH) ——— = ? moles (KOH) ——————— = ? moles (Comp B) (KOH) = 0.450 moles (KOH) KOH H 3 PO 4 K 3 PO 4 + → + H2OH2OH2OH2O 3 113

30. Practice 9.A; Step 2 to Step 3 Look at the Molarity/Gram to mole Conversion Guide Slide (#24). You should be able to see what we do next. Can you see we need three steps? 1.Convert the M &V given (start step) to moles H3PO4 2.Convert moles H3PO4 to moles K3PO4. 3.Convert moles K3PO4 to M using the V given (finish step). We have converted the given (start step 1) to moles H 3 PO 4. We have to next convert moles H 3 PO 4 to moles KOH (step 2). Our last step is to convert moles KOH to M using the V given (finish step).

31. Practice 9.A.; Step 3. moles (compound in solution) M (compound in solution), moles/l = ---------------------------------, volume (compound in solution), l KOH 0.450 moles ( KOH in solution) KOH M ( KOH in solution), moles/l = --------------------------------- KOH 0.050 l ( KOH in solution) KOHKOH M ( KOH in solution), moles/l = 0.09 M ( KOH in solution)

32. Association: Make up two questions to show what you have learned.

33. Achievements Molar Mass Mole Mole to Mole Conversion Mole to Gram Conversion.

34. Homework Do the association problems. Do the Practice Problems. Try the odd numbered problems at the end of each section. The answers are at the end of the chapter. If you have problems, open it up for discussion.