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Topic: How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4  (Read 6950 times)

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19kevin87

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How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4
« on: June 14, 2010, 11:45:03 PM »
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!

AWK

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Re: How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4
« Reply #1 on: June 15, 2010, 01:51:11 AM »
Densities of both solutions are needed
AWK

19kevin87

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Re: How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4
« Reply #2 on: June 15, 2010, 04:30:39 AM »
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

Borek

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Re: How much Vol of 48.5% of NaOH needed to neutralize 21mL of 4.7% H2SO4
« Reply #3 on: June 15, 2010, 05:42:18 AM »
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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