# Question b77c6

##### 1 Answer
Apr 13, 2016

$\text{100 mL}$

#### Explanation:

Your starting point here will be the balanced chemical equation for this neutralization reaction. Sulfuric acid, ${\text{H"_2"SO}}_{4}$, will react with sodium hydroxide, $\text{NaOH}$, to form aqueous sodium sulfate, ${\text{Na"_2"SO}}_{4}$, and water

${\text{H"_ 2"SO"_ (4(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

Notice that every mole of sulfuric acid that takes part in the reaction consumes $\textcolor{red}{2}$ moles of sodium hydroxide. This means that if you figure out how many moles of sodium hydroxide you have present in your sample, you can use this mole ratio to find the number of moles of acid needed.

Use the molarity and volume of the sodium hydroxide solution to determine how many moles of strong base you have present

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

n_(NaOH) = "0.13 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(80 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

$= \text{ 0.0104 moles NaOH}$

This means that you must supply

0.0104 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_2"SO"_4)/(color(red)(2)color(red)(cancel(color(black)("moles NaOH")))) = "0.00520 moles H"_2"SO"_4

Now all you have to do is use the known molarity of the sulfuric acid solution to determine how many milligrams would contain that many moles of acid

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies V_"solution" = n_"solute}} / c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

V_(H_2SO_4) = (0.00520 color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.104 L"#

Expressed in millliliters and rounded to one significant figure, the answer will be

${V}_{{H}_{2} S {O}_{4}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{100 mL} \textcolor{w h i t e}{\frac{a}{a}} |}}}$