# What volume of 1.6 M potassium hydroxide would be needed in order to neutralize 100.0 mL of 2.50 M hydrochloric acid?

Jun 7, 2016

$\text{156 mL}$

#### Explanation:

The first thing to do here is write a balanced chemical equation that describes this neutralization reaction

${\text{KOH"_ ((aq)) + "HCl"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Potassium hydroxide, $\text{KOH}$, and hydrochloric acid, $\text{HCl}$, react in a $1 : 1$ mole ratio to produce aqueous potassium chloride, $\text{KCl}$, and water.

This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.

Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction

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

You will have

n_"HCl" = "2.50 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{\text{HCl" = "0.250 moles HCl}}$

This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles

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

You will have

${V}_{\text{KOH" = (0.250color(red)(cancel(color(black)("moles"))))/(1.6color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.1563 mL}}$

I'll leave the answer rounded to three sig figs and express it in milliliters

"volume of KOH solution" = color(green)(|bar(ul(color(white)(a/a)color(black)("156 mL")color(white)(a/a)|)))