In class, we discussed the titration of a 50.0 mL sample of 0.100 M HCL with a 0.100 M solution of NaOH.
How would this system change if we used a 0.100 M solution of Sr(OH)2 instead of NaOH? To answer this question, complete each of the following:
a) What volume of 0.100 M Sr(OH)2 solution would be needed to reach the equivalence point of this titration? In a short (2-3 sentence) paragraph, compare this volume to the volume of NaOH required, as discovered in class. Does your answer make sense? Why or why not?
b) What will be the pH of the combined HCl- Sr(OH)2 at this point (the equivalence point)? Once again, use a short (2-3 sentence) paragraph to compare your answer to that obtained in class for the titration of HCl with NaOH. Is this pH at the equivalence point the same in both titrations? If so why? If not, why not? In either case, do(es) the answer(s) obtained make sense?
. Ordinary commercially available vinegar is composed of acetic acid (HC2H3O2) in aqueous solution, and normally exhibits a pH value of approximately 2.50. Given the Ka value of acetic acid (1.80x10-5), determine the concentration of HC2H3O2 present in vinegar.
I would be interested in knowing what you discovered/discussed in class and what you don't understand in working this problem. The only difference, correct me if I'm wrong, is that Sr(OH)2 contains two OH^- per mole Sr(OH)2 while NaOH contains only one.
HCl + NaOH ==> NaCl + HOH
2HCl + Sr(OH)2 ==>SrCl2 + 2HOH
a) To determine the volume of 0.100 M Sr(OH)2 solution needed for the equivalence point, you would use the same process as for NaOH. The balanced chemical equation for the reaction between HCl and Sr(OH)2 is 2 HCl + Sr(OH)2 -> SrCl2 + 2 H2O. From the stoichiometry of the balanced equation, you would find that 1 mole of Sr(OH)2 reacts with 2 moles of HCl. Therefore, the volume of 0.100 M Sr(OH)2 solution needed would be twice the volume of 0.100 M NaOH solution used in class.
Comparing the volumes of NaOH and Sr(OH)2 solutions needed, it makes sense that the volume of 0.100 M Sr(OH)2 solution required would be twice the volume of NaOH because of the stoichiometry of the reaction. The balanced equation shows that two moles of HCl react with one mole of Sr(OH)2, hence requiring twice the volume of Sr(OH)2 as NaOH.
b) At the equivalence point, the pH of the combined HCl-Sr(OH)2 solution will be different from that obtained in class for the titration of HCl with NaOH. This is because the equivalence point corresponds to the complete neutralization of the acid with the base, resulting in the formation of water and a salt. The salt formed in the case of HCl and NaOH titration is NaCl, which is a neutral salt and does not affect the pH of the solution. On the other hand, the salt formed in the case of HCl and Sr(OH)2 titration is SrCl2, which is a salt of a strong base and a weak acid. This salt will hydrolyze in water and produce hydroxide ions, leading to a slightly basic pH at the equivalence point. Hence, the pH at the equivalence point will not be the same in both titrations.
The answers obtained make sense because they are based on the stoichiometry of the reaction and the properties of the resulting salts. The different salts formed in the two titrations have different effects on the pH, leading to the differences observed at the equivalence point.