# How many milliliters of 0.45M HCI will neutralize 25.0 mL of 1.00M KOH?

Jul 20, 2017

$56$ $\text{mL HCl soln}$

#### Explanation:

We're asked to find the volume (in $\text{mL}$) of a $0.45 M$ $\text{HCl soln}$ that will neutralize $25.0$ $\text{mL}$ of a $1.00 M$ $\text{KOH soln}$.

To do this, let's first write the chemical equation for the reaction:

$\text{HCl"(aq) + "KOH"(aq) rarr "KCl"(aq) + "H"_2"O} \left(l\right)$

What we can do is use the molarity equation to find the number of moles of $\text{KOH}$ that are present:

$\text{molarity" = "mol solute"/"L soln}$

"mol solute" = ("molarity")("L soln")

"mol KOH" = (1.00"mol"/(cancel("L")))overbrace((0.0250cancel("L")))^"converted to liters" = color(red)(0.0250 color(red)("mol KOH"

Now, we can use the coefficients of the chemical equation to find the relative number of moles of $\text{HCl}$:

color(red)(0.0250)cancel(color(red)("mol KOH"))((1color(white)(l)"mol HCl")/(1cancel("mol KOH"))) = color(red)(0.0250 color(red)("mol HCl"

Lastly, we can use the molarity equation again to find the volume of $\text{HCl}$ solution:

$\text{molarity" = "mol solute"/"L soln}$

$\text{L soln" = "mol solute"/"molarity}$

"L HCl soln" = (color(red)(0.0250)cancel(color(red)("mol HCl")))/(0.45(cancel("mol"))/"L") = color(blue)(0.056 color(blue)("L" = color(blue)(ul(56color(white)(l)"mL"