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 pKa is not exactly the pH at the equivalence point?
As we can see in the following titration curve, even if the pKa 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.
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