Question #efddd

1 Answer
Apr 22, 2015

You need 21.8 mL of your sulfuric acid solution to neutralize that much sodium hydroxide.

Start with the balanced chemical equation for this neutralization reaction

#H_2SO_(4(aq)) + color(red)(2)NaOH_((aq)) -> Na_2SO_(4(Aq)) + 2H_2O_((l))#

Notice that you have a #1:color(red)(2)# mole ratio between sulfuric acid and sodium hydroxide. This means that you need 2 moles of the lattter for every 1 mole of the former in order for complete neutralization to take place.

If you have less than 1 mole of sulfuric acid for every 2 moles of sodium hydroxide, you won't get complete neutralization, i.e. you'll have base remaining in solution.

Use sodium hydroxide's molar mass to determine how many moles of #NaOH# must be neutralized

#0.220cancel("g") * "1 mole NaOH"/(39.997cancel("g")) = "0.00550 moles NaOH"#

This means that you'd need

#0.00550cancel("moles NaOH") * ("1 mole "H_2SO_4)/(color(red)(2)cancel("moles NaOH")) = "0.00275 moles" # #H_2SO_4#

Since you're dealing with a sulfuric acid solution, you can use its molarity to determine what volume would contain this many moles

#C = n/V => V = n/C#

#V = (0.00275cancel("moles"))/(0.126cancel("moles")/"L") = "0.02183 L"#

Expressed in mL and rounded to three sig figs, the answer will be

#V = color(green)("21.8 mL")#