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titrations...like everyone else :S (1 Viewer)

royaaran

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what volume of 0.48M sulfuric acid is needed to neutralise 26.0ml of 0.51M sodium hydroxide solution?

what would be the molarity of your sodium hydroxide solution if you used the same volume of sulfuric acid at the same concentration, instead of the HCL acid? explain.

what volume of sulfuric acid would have been added to neutralise your hydroxide solution?

omg.....as if we are gonna need this in the future. sorry for asking so many questions but i have NO IDEA how to do them :(
 

Riviet

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H2SO4 + 2NaOH -> Na2SO4 + 2H2O

n(NaOH) = cV
= 0.51 x 0.026 L
= 0.01326 moles

n(sulfuric acid) = 1/2 x n(NaOH), using the above equation.

.'. n(sulfuric acid) = 1/2 x 0.01326
= 0.00663

V=n/c
=0.0063/0.48
=0.0138125L
=13.8ml of sulfuric acid.

Hope that helps.
 

sarah5629

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i wish it had helped me as much, i have an assessment on friday in which i have to do a titration, unfortunately i think my teacher missed the part of his teaching degree where he actually learnt to teach and something tells me our assessment is going to be a little bit harder than just watching the drippy thing until the stuff at the bottom turns pink.....:jedi:
 

Lizcat

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read your text book.. it helps.. and also.. try to learn the names of the apparatus.
 

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