Question #f8f90

1 Answer
Mar 3, 2017

Answer:

The concentration of the NaOH is 0.775 mol/L.

Explanation:

The equation for the reaction is

#"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + 2"H"_2"O"#.

1. Calculate the moles of #"H"_2"SO"_4#

#"Moles of H"_2"SO"_4 = "0.035 21" color(red)(cancel(color(black)("L H"_2"SO"_4))) ×( "0.550 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0. 019 37 mol H"_2"SO"_4#

2. Calculate the moles of #"NaOH"#

#"Moles of NaOH" = "0.019 37" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("2 mol NaOH")/(1 color(red)(cancel(color(black)("H"_2"SO"_4)))) = "0.038 73 mol NaOH"#

3. Calculate the molarity of the #"NaOH"#

#"Molarity" = "moles"/"litres" = "0.038 73 mol"/("0.050 00 L") = "0.775 mol/L"#