1. Strong Acid-Base Titrations Chapter 17

2. Neutralization Reactions Review Generally, when solutions of an acid and a base are combined, the products are a salt and water. HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l)

3. Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq )  Na + ( aq ) + Cl - ( aq ) + H 2 O ( l ) Net Ionic Equation: H + ( aq ) + OH - ( aq )  H 2 O ( l ) The K for this reaction is 10 14

4. Neutralization reactions When a strong base is added to a solution of weak acid, the following neutralization reaction occurs: Conj. Acid + OH -  conj. Base + H 2 O When a strong acid is added to a solution of a weak base, a neutralization reaction occurs: Conj base + H 3 O +  conj acid + H 2 O

5. Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

6. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

7. The initial pH (initial acid) is determined by the initial concentration of the strong acid. Major substances present: H + and Cl - Titration of a Strong Acid with a Strong Base

8. From the start of the titration to near the equivalence point, the pH goes up slowly. As strong base is added, the pH increases slowly because the base is neutralized by the acid. Major substances present: H +, Cl -, Na + Titration of a Strong Acid with a Strong Base

9. The pH of the solution before the equivalence point increases quickly because it is determined by the concentration of the aced that has not yet been neutralized.

10. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base based on stoichiometry of reaction, and the solution contains only water and the salt from the cation of the base and the anion of the acid. The pH at equivalence point for a strong acid and strong base is always 7 exactly.

11. Titration of a Strong Acid with a Strong Base The pH after the equivalence point is determined by the concentration of the excess NaOH. As more base is added, the increase in pH again levels off

12. Practice Problem Calculate the pH when 49.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. Step 1: figure out how many moles of H + are present in solution 0.0500 L soln x 0.100 mol H + /1 L soln = 5.00 x 10 -3 mol H + Step 2: figure out how many moles of OH - are added 0.0490 L soln x 0.100 mol OH - /1 L soln = 4.90 x 10 -3 mol OH -

13. Calculate the pH when 49.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. Step 3: Because the equivalence point has not been reached, there are more moles of H+ present. ICE Table H+H+ OH - H 2 O(l) Initial5.00 x 10 -3 mol 0- Change4.90 x 10 -3 mol - Equilibrium0.10 x 10 -3 mol 0-

14. H+H+ OH - H 2 O(l) Initial (before addition) 5.00 x 10 -3 mol 0- Change (addition) 4.90 x 10 -3 mol - Equilibrium (after addition) 0.10 x 10 -3 mol 0- Step 4: Determine the total volume of the mixed solution 0.050 L + 0.049 L = 0.0990 L Step 5: Determine concentration of H+ and then pH [H+] = 0.10 x 10 -3 mol/0.0990 L = 1.0 x 10 -3 M pH = -log(1.0x10 -3 )