# What volume of 0.100 M "HCl" solution is needed to neutralize 50.0 mL of 0.350 M "KOH"?

Jul 11, 2016

$\text{175 mL}$

#### Explanation:

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

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

This means that a complete neutralization requires equal numbers of moles of hydrochloric acid, i.e. of hydronium cations, ${\text{H"_3"O}}^{+}$, and of potassium hydroxide, i.e. of hydroxide anions, ${\text{OH}}^{-}$.

As you know,m molarity is defined as the number of moles of solute present in $\text{1 L}$ of solution.

Now, notice that the potassium hydroxide solution, which has a molarity of $\text{0.350 M}$, is

$\left(0.350 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(0.100color(red)(cancel(color(black)("M}}}}\right) = \textcolor{red}{3.5}$

times more concentrated than the hydrochloric acid solution, which has a molarity of $\text{0.100 M}$. In other words, for the same volume of both solutions, the potassium hydroxide solution contains $\textcolor{red}{3.5}$ times more moles of solute than the hydrochloric acid solution.

This means that in order to have equal numbers of moles of both solutes, you need to have a volume of hydrochloric acid solution that is $\textcolor{red}{3.5}$ times bigger than the volume of the sodium hydroixde solution.

Since the sodium hydroixde solution has a volume of $\text{50.0 mL}$, it follows that the volume of hydrochloric acid needed will be

V_"HCl" = color(red)(3.5) xx "50.0 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("175 mL")color(white)(a/a)|)))-> to three sig figs