Solve Acid Base Titrations: 0.700-M KOH & 0.350-M H2SO4

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Discussion Overview

The discussion revolves around a homework problem involving the titration of sulfuric acid (H2SO4) with potassium hydroxide (KOH). Participants explore the stoichiometry of the reaction, the concepts of molarity and normality, and the interpretation of the problem statement.

Discussion Character

  • Homework-related
  • Debate/contested
  • Exploratory

Main Points Raised

  • One participant expresses confusion about the question, questioning whether it asks for the volume of KOH needed to neutralize H2SO4 and whether the reaction equation is necessary.
  • Another participant states that at the endpoint of a titration, the number of equivalents of base equals the number of equivalents of acid.
  • A participant outlines their calculations for the moles of H2SO4 and the corresponding volume of KOH needed, but expresses uncertainty about the relevance of the reaction equation provided.
  • One participant suggests there may be a typo in the question, proposing it should refer to lithium hydroxide (LiOH) instead of KOH, which they believe clarifies their confusion.
  • A participant reminds others that molarity is not the same as normality, indicating that the acid concentration should be expressed in normality for titration calculations.
  • Another participant suggests using an online resource for additional help with titration endpoints.
  • A participant confirms the suspicion of a typo, stating that the question should indeed refer to LiOH, which resolves their confusion about the problem.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the reaction equation and the correct interpretation of the problem statement. There is no consensus on whether the question contains a typo, although multiple participants suggest that it may refer to LiOH instead of KOH. The discussion remains unresolved regarding the application of normality versus molarity in this context.

Contextual Notes

Participants note the importance of understanding normality in relation to titration, but some indicate that they have not yet covered this concept in their studies. The discussion also highlights potential confusion stemming from the phrasing of the homework question.

ellaina
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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 probably figure this out but for some reason my brain can't reason out what I'm supposed to be aiming for here? Is it wanting to know how many mL of KOH are needed to neutralize the H2SO4? If so, I don't need the given equation do I? Maybe just a hint would send me in the right direction... Then again maybe I am just totally lost.

Thanks in advance for any and all help...
 
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At the endpoint, the number of equivalents of base is equal to the number of equivalents of acid.
 
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?
 
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!
 
Don't forget that molarity (M) is not the same as normality (N). The acid solution concentration needs to be expressed in terms of normality, not molarity.
 
Go to Google, type in "titratiom end points" and look at the top reference. Think that will help you. Good luck.

Steve
 
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 completely, I was so lost on why the equation had LiOH but the question was asking for KOH! I was afraid I had completely missed something important. I can do it now that I'm not trying to figure out how KOH came into play :)

Thanks for yalls help tho! Much appreciated :)
 

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