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?

1 Answer
Apr 8, 2018

Answer:

#c=0.0432 mol 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:
#HCl(aq)+NaOH(aq) -> NaCl(aq) + H_2O(l)#

So our acid and base are in a 1:1 molar ratio in this case- so an equal amount of #NaOH# must have reacted with #HCl# for the solution to neutralize.

Using the concentration formula:

#c=(n)/(v)#

#c#=concentration in #mol dm^-3#
#n#=number of moles of substance dissolved in solution volume #(v)#
#v#=volume of the solution in liters - #dm^3#

We were given concentration and volume of #NaOH#, so we can find its number of moles:

#0.1=(n)/0.054#

#n=0.0054# mol

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

So:

#0.0054/(0.125)=c#

#c=0.0432 mol dm^-3#