Titrimetric Analyses - pH

1. Titrimetric Analyses - pH Diprotic Weak Acid: H2A + −OH  14.00 pH 7.00 0.00 0.0000 1/2Ve1 Ve1 3/2Ve1 Ve2 Volume of Standardized Strong Base Added (L) Determining pH at various points: A. No Base Added – This is a weak acid problem, and it can be treated as a monoprotic acid at this point only iff one has compared Ka1/Ka2 and found the value to be ≤ 5%. If this is true, then one sets this up as follows: B. Base Added Up to Ve1 – This is a buffer problem involving the acid H2A and its conjugate base HA−. The pH is calculated using the Henderson-Hasselbach equation:

2. B. continued There are some key points here between 0 L base added and Ve1 that are readily calculated: At ½ Ve1, At ¼ Ve1, At 1/3 Ve1, At 2/3 Ve1, At ¾ Ve1, C. Volume Of Base Added Is Ve1 – Here, we have the amphiprotic species HA− in aqueous solution, and it can undergo base hydrolysis with water to yield either H2A or acid hydrolysis to yield A2 −. The pH is merely the average of the two pKa values, D. Base Added Beyond Ve1 but Before Ve1 – Now there is the buffer system based on the A2 − and HA− conjugate base/acid pair. The pH is dictated by the Henderson-Hasselbach equation:

3. D. continued There are some key points here between 0 L base added and Ve1 that are readily calculated: At 3/2 Ve1, At 5/4 Ve1, At 4/3 Ve1, At 5/3 Ve1, At7/4 Ve1, E. Volume Of Base Added Is Ve2 – Here, we have the diprotic species A2− in aqueous solution, and it can undergo base hydrolysis with water to first yield HA− and the latter can react further with water to yield H2A . This situation is one wherein A2− can be treated as a monoprotic base at this point only iff one has compared Kb1/Kb2 and found the value to be ≤ 5%. If this is true, then one sets this up as follows, with the caveat that the formal conc. of A2− must be calculated based on the new volume in which it is contained:

4. F. Volume Of Base Added Is Ve2 + VXS – Here, we have the diprotic species A2− in aqueous solution as in E above. Usually, we do not have to worry about it being a stronger base than hydroxide, but one must be cautious when calculating the pH; if the pH value is calculated to be lower than in E above, then it is clear that one must use systematic treatment of equilibria to solve for pH. Typically one first just calculates the pH by determining how many moles excess hydroxide is added and then calculates the new conc. of hydroxide as follows: Of course, you should be able to do the same concept map and calculations for a diprotic base, B. This is something you should be able to do for this course and the examination on this material.