Practice: Calculate the pH of the solution resulting from the mixing of 75.0 mL of 0.100 M NaC_{2}H_{3}O_{2} and 75.0 mL of 0.30 M HC_{2}H_{3}O_{2} 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 HClO_{3} and 50.0 mL of a 0.100 M NH_{3}. (K_{b} of NH_{3} 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 NaC_{2}H_{3}O_{2} and 75.0 mL of 0.30 M HC_{2}H_{3}O_{2} 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, HC_{3}H_{5}O_{3}. What is the pH of the buffer solution after the addition of 300.0 mL of 0.100 M hydrobromic acid, HBr? The K_{a} of HC_{3}H_{5}O_{3} 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 HClO_{3} and 50.0 mL of a 0.050 M NH_{3}. (K_{b} of NH_{3} is 1.75 x 10^{-5}).

Practice: Consider the titration of 100.0 mL of 0.100 M CH_{3}NH_{2} with 0.250 M HNO_{3} at the equivalence point. What would be the pH of the solution at the equivalence point? The K_{b} of CH_{3}NH_{2} 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 HClO_{3} and 50.0 mL of a 0.050 M NH_{3}. (K_{b} of NH_{3} is 1.75 x 10^{-5}).

Practice: A solution contains 100.0 mL of 0.550 M sodium nitrite, NaNO_{2}. Find the pH after the addition of 180.0 mL of 0.400 M HClO_{4}. The K_{a} of HNO_{2} is 4.6 × 10^{−4}.