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

0 Members and 1 Guest are viewing this topic. 19kevin87

• New Member
•  • Posts: 7
• Mole Snacks: +0/-0 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

• Retired Staff
• Sr. Member
• • Posts: 6648
• Mole Snacks: +479/-80
• Gender:  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

• New Member
•  • Posts: 7
• Mole Snacks: +0/-0 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 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info