# Question #f8f90

Mar 3, 2017

The concentration of the NaOH is 0.775 mol/L.

#### Explanation:

The equation for the reaction is

$\text{H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + 2"H"_2"O}$.

1. Calculate the moles of ${\text{H"_2"SO}}_{4}$

${\text{Moles of H"_2"SO"_4 = "0.035 21" color(red)(cancel(color(black)("L H"_2"SO"_4))) ×( "0.550 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0. 019 37 mol H"_2"SO}}_{4}$

2. Calculate the moles of $\text{NaOH}$

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

3. Calculate the molarity of the $\text{NaOH}$

$\text{Molarity" = "moles"/"litres" = "0.038 73 mol"/("0.050 00 L") = "0.775 mol/L}$