Practice: In order to create a buffer 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? Ka = 4.9 × 10−10.
Titration where weak acid is the titrate and strong base is titrant.
Concept #1: Before the Equivalence Point
Before the Equivalence Point is where the moles of weak acid is greater than moles of strong base. Some conjugate base is formed.
Example #1: Consider the titration of 75.0 mL of 0.0300 M HC3H3O3 (Ka = 4.1 X 10-3) with 12.0 mL of 0.0450 M KOH. Calculate the pH.
Practice: In order to create a buffer 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? Ka = 4.9 × 10−10.
Concept #2: At the Equivalence Point
At the Equivalence Point is where the moles of weak acid is equal to moles of strong base. Only conjugate base remains.
Example #2: Consider the titration of 75.0 mL of 0.0300 M HC3H3O3 (Ka = 4.1 X 10-3) with 50.0 mL of 0.0450 M KOH. Calculate the pH.
Practice: Consider the titration of 75.0 mL of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The Ka of HNO2 is 4.6 × 10−4.
Concept #3: After the Equivalence Point
After the Equivalence Point is where moles of weak acid is less than the moles of strong base. Excess strong base remains.
Example #3: Consider the titration of 75.0 mL of 0.0300 M HC3H3O3 (Ka = 4.1 X 10-3) with 75.0 mL of 0.0450 M KOH. Calculate the pH.
Practice: Calculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH.
