Practice: Calculate the pH of the solution resulting from the mixing of 75.0 mL of 0.100 M NaC2H3O2 and 75.0 mL of 0.30 M HC2H3O2 with 0.0040 moles of HBr.
Titration where weak base is the titrate and strong acid is titrant.
Concept #1: Before the Equivalence Point
Before the Equivalence Point is where the moles of weak base is greater than moles of strong acid. Some conjugate acid is formed.
Example #1: Calculate the pH of the solution resulting from the titration between 25.0 mL of a 0.100 M HClO3 and 50.0 mL of a 0.100 M NH3. (Kb of NH3 is 1.75 x 10-5).
Practice: Calculate the pH of the solution resulting from the mixing of 75.0 mL of 0.100 M NaC2H3O2 and 75.0 mL of 0.30 M HC2H3O2 with 0.0040 moles of HBr.
Practice: In order to create a buffer 7.321 g of potassium lactate is mix with 550.0 mL of 0.328 M lactic acid, HC3H5O3. What is the pH of the buffer solution after the addition of 300.0 mL of 0.100 M hydrobromic acid, HBr? The Ka of HC3H5O3 is 1.4 × 10−4.
Concept #2: At the Equivalence Point
At the Equivalence Point is where the moles of weak base is equal to moles of strong acid. Only conjugate acid remains.
Example #2: Calculate the pH of the solution resulting from the titration between 25.0 mL of a 0.100 M HClO3 and 50.0 mL of a 0.050 M NH3. (Kb of NH3 is 1.75 x 10-5).
Practice: Consider the titration of 100.0 mL of 0.100 M CH3NH2 with 0.250 M HNO3 at the equivalence point. What would be the pH of the solution at the equivalence point? The Kb of CH3NH2 is 4.4 × 10−4.
Concept #3: After the Equivalence Point
After the Equivalence Point is where moles of weak base is less than the moles of strong acid. Excess strong acid remains.
Example #3: Calculate the pH of the solution resulting from the titration between 125.0 mL of a 0.100 M HClO3 and 50.0 mL of a 0.050 M NH3. (Kb of NH3 is 1.75 x 10-5).
Practice: A solution contains 100.0 mL of 0.550 M sodium nitrite, NaNO2. Find the pH after the addition of 180.0 mL of 0.400 M HClO4. The Ka of HNO2 is 4.6 × 10−4.
