I've been stuck on this question for awhile now. I'm having a lot of trouble figuring out how to start it.
If you had added exactly 20.00 mL do d.i. water before the titration, calculate the molar concentration of OAc- would have been at the end of the titration. (Ignore the volume of the HPhthn solution)
What were you titrating? HOAc?
What were you titrating with? NaOH or similar?
Take the volume of base you added, add 20 mL to that and that is the final volume. The OAc^- then will be (in molarity) mols HOAc you started with divided by the final volume in L.
To find the molar concentration of OAc- at the end of the titration, we need some additional information. Specifically, we need to know the initial concentration of the HPhthn solution and the volume of the HPhthn solution used in the titration.
Once we have that information, we can use the concept of molarity and the dilution equation to calculate the molar concentration of OAc-. The dilution equation is given by:
M1V1 = M2V2
Where:
M1 is the initial molar concentration of the solution (HPhthn)
V1 is the initial volume of the solution (HPhthn)
M2 is the final molar concentration of the solution (OAc-)
V2 is the final volume of the solution (HPhthn + d.i. water)
In this case, we are given that 20.00 mL of d.i. water is added before the titration. This means that the final volume (V2) is the sum of the initial volume of the HPhthn solution (V1) and the volume of d.i. water added (20.00 mL).
To solve the problem, we need to know the initial molar concentration (M1) and the initial volume (V1) of the HPhthn solution, as well as the volume of HPhthn solution used in the titration.
Once we have all the necessary information, we can rearrange the dilution equation and solve for M2, the molar concentration of OAc- at the end of the titration.