1. Acids and Bases Strengths of Acids and Bases Making Dilutions

2. Strengths of Acids and Bases Strong acids completely ionize (100%) in aqueous solutions HCl + H 2 O H 3 O + + Cl - (100 % ions ) Strong bases completely (100%) dissociate into ions in aqueous solutions. NaOH Na + (aq) + OH - (aq) (100 % ions)

3. NH 3, A Bronsted-Lowry Base When NH 3 reacts with water, most of the reactants remain dissolved as molecules, but a few NH 3 reacts with water to form NH 4 + and hydroxide ion. NH 3 + H 2 O NH 4 + (aq) + OH - (aq) acceptor donor+

4. Strong and Weak Acids and Bases Strong acids HCl, HNO 3, H 2 SO 4 Most other acids are weak. Strong bases NaOH, KOH, and Ca(OH) 2 Most other bases are weak.

5. Acid/Base Strength Check Identify each of the following as a 1) strong acid or base 2) weak acid 3) weak base A. ___ HCl (aq) B. ___ NH 3 (aq) C. ___ NaOH (aq) D. ___ H 2 CO 3 (aq)

6. Solution Identify each of the following as a 1) strong acid or base 2) weak acid 3) weak base A. _1__ HCl (aq) B. _3__ NH 3 (aq) C. _1__ NaOH (aq) D. _2__ H 2 CO 3 (aq)

7. Antacids Used to neutralize stomach acid (HCl) Many contain one or more weak bases Alka-Seltzer: NaHCO 3, citric acid, and aspirin Di-gel: CaCO 3 and Mg(OH) 2 Gelusil: Al(OH) 3 and Mg(OH) 2 Maalox: Al(OH) 3 and Mg(OH) 2 Mylanta: Al(OH) 3 and Mg(OH) 2

8. More Antacids Milk of Magnesia: Mg(OH) 2 Rolaids: AlNa(OH) 2 CO 3 aluminum sodium dihydroxy carbonate Tums: CaCO 3 Tempo: CaCO 3, Al(OH) 3, Mg(OH) 2

9. Dilutions Add water Volume increases. New concentration is less than initial

10. Concentration of A Diluted Solution A 25 mL sample of a 6.0 M KOH is diluted by adding 75 mL of water. Calculate the new concentration of the KOH solution. Moles KOH = 0.025 L x 6.0 moles = 0.15 moles 1 L New volume = 25 mL + 75 mL = 100. mL = 0.100 L New molarity = 0.15 moles = 1.5 M 0.100 L

11. Dilution Calculation A 125 mL sample of a 3.0 M HCl is diluted by adding 250 mL of water. The new concentration of the HCl solution is 1) 8.0 M 2) 1.5 M 3) 1.0 M

12. Solution moles HCl = 0.125 L x 3.0 moles = 0.375 mole 1 L new M = 0.375 mole =1.0 M HCl 0.375 L (new volume)

13. Calculation of Concentration A 50.0 mL sample of 15% sucrose solution is diluted by adding 200. mL of water. The new concentration of the sucrose solution is 1) 3.0 % 2) 7.5 % 3) 10. %

14. Solution A 50.0 mL sample of 15% sucrose solution is diluted by adding 200. mL of water. The new concentration of the sucrose solution is 1) 3.0 % g sucrose = 50.0 mL x 15 g sucrose = 7.5 g 100 mL new % = 7.5 g sucrose = 3.0 % sucrose 250 mL (new volume)