Question

Titration question (attached file). Can someone tell me, in as much detail as possible, how I should approach this exercise?

Answer 1

idk if this would help but your bascially copying something on the top graph to the bottom.

Answer 2

yes with the difference that you include the pH range of the bromophenol blue indicator

Answer 3

Well, I don't believe that the pH range of the indicator is the only difference? For example, the initial pH of the acid should be different I guess (but how do I know where? Is it a value that I just have to learn by heart, or?). Also the half-equivalence point should be different, since you have a weak acid with strong base in this case compared to the first graph (but again, how do I figure this out?).

Answer 4

OK dear:
you are in the right way, the propanoic acid is a weak acid with a pKa 4.88, and the potassium hydroxide is a strong base. Then the pHrange of the indicator is 3.0-4.6.
The pH of the acid at 0mL titration can be calculated with the formula
\[[H ^{+}]= \sqrt{Ka \times Ca}\]
where
pKa= -log Ka = 4.88
Ka= 10^(-pka)
Ca= initial concentration of the acid = 0.1M

The after that you can calculate all the points until the equivalent point with the Henderson Hasselbach equation for a buffer. At the 50% titration (12.5mL) the pH will be equal to the pKa of the acid.

then the pH at the equivalence point is going to be a basic salt (potasssium propanoate) an you have to calculate the [OH-] and with the Kb and the concentration of the salt and then calculate the pH

Answer 5

do you need more details?

Answer 6

yes you are right I misread the question