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Titration of H2SO4 and NaOH

  • Thread starter Kitty808
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Homework Statement


Ok, this is a lab related question. Given a diluted solution of approximately 1.25M H2SO4, determine the amount of solution needed to use for titration with the base you prepared and standardized. Do not use a volume of battery acid that will require more than 20mL of base to reach the equivalence point. Then determine the concentration of H2SO4 in the diluted solution you were given.


Homework Equations


Balanced chemical equation: H2SO4 + 2NaOH --> 2H2O + Na2SO4
[NaOH] = 0.0939M

The Attempt at a Solution


I determined the amount of diluted H2SO4 to be 18.9 mL (I used 18 mL)

20mL NaOH(1L/1000mL)(0.0939M NaOH)(1 mol H2SO4/2 mol NaOH) = 0.0945M H2SO4
(1.25M)(V1)=(0.0945M)(250mL); V1 = 18.9mL
It's at this point where I get stuck. I'm supposed to calculate the concentration of H2SO4 in the diluted acid solution then Find the original [H2SO4] in the undiluted battery acid (assume diluted 1:5). I know I'm supposed to use a titration calculation to find the concentration of the diluted acid but I'm not sure how to set it up and which numbers to use to end with mol/L H2SO4. I'm also not sure how to find the initial concentration. Any ideas?
 

Answers and Replies

  • #2
symbolipoint
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You showed no steps for dilution of the sample. Your basic calculation arrangement gave you a prediction (V1) for how much sample of battery acid to take to use your 20 ml. volume of titrant. Now, do you want to make an analytical dilution of your sample? What size volumetric flask and what aloquat do you want to use?

edited: next posting uses improved understanding of what you asked.
 
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  • #3
symbolipoint
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I reviewed your question again; your DILUTED sample is about 1.25M, so when you find the concentration of your DILUTED sample, then you use the knowledge of how it was diluted (you gave a ratio of 1:5 , so use it.). What is the meaning of this dilution?
 

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