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.

Question

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?

Answer 1

. 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.

Answer 2

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