Question

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`

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`