1. Acids and Bases Calculating Excess

2. Mixing strong acids and bases During an experiment, a student pours 25.0 mL of 1.40 mol/L nitric acid into a beaker that contains 15.0 mL of 2.00 mol/L sodium hydroxide solution. Is the resulting solution acidic or basic? What is the concentration of the ion that causes the solution to be acidic and basic? What is the final pH of the solution?

3. Mixing strong acids and bases Step 1: Write the chemical equation for the reaction. HNO 3 (aq) + NaOH (aq)  NaNO 3 (aq) + H 2 O (l)

4. Mixing strong acids and bases Step 2: Calculate the amount of each reactant using M=n/V 25.0mL of 1.40mol/L nitric acid n=M*V n=1.40mol/L * 0.0250 L = 0.0350 mol 15.0mL of 2.00mol/L sodium hydroxide n=M*V n=2.00mol/L * 0.0150 L = 0.0300 mol

5. Mixing strong acids and bases Step 3: Determine the limiting reagent HNO 3 : NaOH 1:1 The reactants combine in a 1:1 ratio. The amount of NaOH (0.0300mol) is less than that of HNO 3 (0.0350mol) so the NaOH is the limiting reagent.

6. Mixing strong acids and bases Step 4: The reactant in excess is a strong acid or base. Thus, the excess amount results in the same amount of H 3 O + or OH - Amount of excess HNO 3 = 0.0350 mol – 0.0300mol = 0.0050 mol Therefore, the amount of H 3 O + is 5.0 x 10 -3 mol.

7. Mixing strong acids and bases Step 5: Calculate the concentration of the excess ion by using the amount in excess and the total volume of the solution. Calculate the final pH. Total volume of solution = 25.0mL + 15.0mL=40.0mL [H 3 O + ]= (5.0x10 -3 mol) / 0.0400 L = 0.12 mol/L The final pH= - log[H 3 O + ] = 0.92 Acidic solution, [H 3 O + ]= 0.12 mol/L, pH= 0.92

8. Problem #2 During an experiment, a student pours 20.0 mL of 1.25 mol/L sulfuric acid into a beaker that contains 15.0 mL of 2.00 mol/L potassium hydroxide solution. Is the resulting solution acidic or basic? What is the concentration of the ion that causes the solution to be acidic and basic? What is the final pH of the solution?

9. Practice P 586 #1-4