Recent content by ellaina
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Solve Acid Base Titrations: 0.700-M KOH & 0.350-M H2SO4
Well I don't know about the normality vs molarity thing, we haven't covered that yet... As far as this question goes, I was able to confirm this evening that there is a typo in the question and it should be asking how many mL of LiOH are needed (KOH is a typo). That clears things up for me...- ellaina
- Post #7
- Forum: Biology and Chemistry Homework Help
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Solve Acid Base Titrations: 0.700-M KOH & 0.350-M H2SO4
Ok, I am starting to think that there is a typo in this question and that it should actually be asking "how many mL of *LiOH* are needed..." not "KOh"... Then it would make sense to me!- ellaina
- Post #4
- Forum: Biology and Chemistry Homework Help
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Solve Acid Base Titrations: 0.700-M KOH & 0.350-M H2SO4
Ok so here's what I've done: 1) .235L H2O4 solution * .350M = .0823 mol H2SO4 2).0823 mol KOH/.700M = 117.5mL KOH solution Here's where I'm confused, and maybe I've missed a core concept somewhere...but what am I supposed to do with the given reaction equation?- ellaina
- Post #3
- Forum: Biology and Chemistry Homework Help
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Solve Acid Base Titrations: 0.700-M KOH & 0.350-M H2SO4
Homework Statement Given H2SO4 + 2LiOH --> Li2SO4 + 2H2O, how many mL of a 0.700-M solution of KOH are needed to react with 235 mL of a 0.350-M H2SO4 solution? The Attempt at a Solution Really, I have none. My problem is that I don't understand what the question is asking. I could...- ellaina
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- Acid Acid base Base
- Replies: 7
- Forum: Biology and Chemistry Homework Help