List the major species at points A, B, C, and D on the following titration curve of the titration of ammonia with HCl.
A = NH3, it has yet to be acidified.
B = NH3 and NH4 + in the buffering region.
C = NH4 +. At the equivalence point, all the NH3 has been protonated, and water molecules begin to take up acidic protons.
D = NH4 + and more acid in solution (H3O+).
Why is it acceptable to use an indicator whose pK a is not exactly the pH at the equivalence point?
As we can see in the following titration curve, even if the pK a of the indicator is several units away from the pH at the equivalence point, there is only a negligible change in volume of titrant added due to the steep slope of the titration curve near the equivalence point.
It takes 26.23 mL of a 1.008 M NaOH solution to neutralize a solution of 5 g of an unknown monoprotic acid in 150.2 mL of solution. What is the molecular weight of the unknown?
This is a standard stoichiometry problem for titration. Calculate the number of moles of base to know the number of moles of the unknown because it is a monoprotic acid. Once you know the number of moles of the unknown, divide the mass of the unknown by the number of moles to obtain the solution: the molecular weight of the unknown is 189.1 g/mol. Titration stoichiometry problems do not get much trickier than this.