October 17, 2019, 02:11:05 PM
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Topic: How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4  (Read 4759 times)

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Offline 19kevin87

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I've calculated and my answer is 1.66 mL.
But I'm not sure bout the ans plus my working is too complicated, can you show me how u get the ans?
Correct me if I'm wrong.
Thanks! :)
my version:
1) to change % to concentration
   For H2SO4: 4.74*10/MWH2SO4 = 0.48 mol L-1
   For NaOH: 48.5*10/MWNaOH = 12.13 mol L-1
2) 0.48mol of 21mL H2SO4 is eq to 0.01008 mol
3) since acid to alkali ratio is 1:2, therefore moles of NaOH needed is 0.02016 mol
4) therefore volume needed for 48.5% NaOH is
   0.02016*1000/12.13 = 1.66 mL

Thanks again!

Offline AWK

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Densities of both solutions are needed
AWK

Offline 19kevin87

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Densities of both solutions are needed

Ya, from the formula, the densities are needed.. since both sulphuric acid and caustic soda have density around 1gcm-3, i treat it as 1

Offline Borek

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since both sulphuric acid and caustic soda have density around 1gcm-3, i treat it as 1

That's true only for diluted solutions, one of the solutions in question have density around 1.5 g/mL.
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