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?

Answer 1

#c=0.0432 mol dm^-3#

Finding the molar ratio in the reaction would be the first step. Strong acid-strong base reactions can generally be simplified by saying:

Base + Acid -> 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.

Applying the formula for concentration:

#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.

Thus:

#0.0054/(0.125)=c#

#c=0.0432 mol dm^-3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Naomi Aronson

The concentration of the HCl solution is 0.219 M.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.