# If it takes 54 mL of 0.100 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl?

Apr 8, 2018

$c = 0.0432 m o l {\mathrm{dm}}^{-} 3$

#### Explanation:

The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:

Acid+Base ->Salt+ Water

Hence:
$H C l \left(a q\right) + N a O H \left(a q\right) \to N a C l \left(a q\right) + {H}_{2} O \left(l\right)$

So our acid and base are in a 1:1 molar ratio in this case- so an equal amount of $N a O H$ must have reacted with $H C l$ for the solution to neutralize.

Using the concentration formula:

$c = \frac{n}{v}$

$c$=concentration in $m o l {\mathrm{dm}}^{-} 3$
$n$=number of moles of substance dissolved in solution volume $\left(v\right)$
$v$=volume of the solution in liters - ${\mathrm{dm}}^{3}$

We were given concentration and volume of $N a O H$, so we can find its number of moles:

$0.1 = \frac{n}{0.054}$

$n = 0.0054$ mol

Hence, this must be the number of moles of $H C l$ found in the 125-milliliter solution, as we established that they reacted in a 1:1 ratio.

So:

$\frac{0.0054}{0.125} = c$

$c = 0.0432 m o l {\mathrm{dm}}^{-} 3$