# Question b5150

##### 1 Answer
Feb 19, 2016

$\text{16.9 mL}$

#### Explanation:

You start by taking a look at the balanced chemical equation for this neutralization reaction. The stoichiometric coefficients of the reactants will tell you how many moles of each will be consumed by the reaction.

$\textcolor{b l u e}{2} {\text{KOH"_text((aq]) + "H"_2"SO"_text(4(aq]) -> "K"_2"SO"_text(4(aq]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

So, the $\textcolor{b l u e}{2} : 1$ mole ratio that exists between the two reactants tells you that a complete neutralization will always consume twice as many moles of potassium hydroxide than of sulfuric acid.

This means that if you know how many moles of potassium hydroxide you have in solution, you can use this mole ratio to find the number of moles of sulfuric acid needed for the complete neutralization.

Since you know the mass of potassium hydroxide used for your aqueous solution, you can use the compound's molar mass to determine how many moles you have present

0.503 color(red)(cancel(color(black)("g KOH"))) * "1 mole KOH"/(56.11color(red)(cancel(color(black)("g KOH")))) = "0.008965 moles KOH"

This means that your sulfuric acid solution must contain

0.008965color(red)(cancel(color(black)("moles KOH"))) * ("1 mole H"_2"SO"_4)/(color(blue)(2)color(red)(cancel(color(black)("moles KOH")))) = "0.004482 moles H"_2"SO"_4

As you know, a solution's molarity tells you how many moles of solute you get per liter of solution.

$\textcolor{b l u e}{c = \frac{n}{V}}$

You know the molarity of the sulfuric acid solution and how many moles of sulfuric acid it must contain, which means that you can determine the volume of solution

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

In your case, you will have

V_(H_2SO_4) = (0.004482color(red)(cancel(color(black)("moles"))))/(0.265color(red)(cancel(color(black)("moles")))"L"^(-1)) = "0.0169 L"#

Expressed in milliliters, the answer will be

${V}_{{H}_{2} S {O}_{4}} = \textcolor{g r e e n}{\text{16.9 mL}}$