[Vader]

I'm struggling with this question guys, would appreciate some help if possible.

A 25.00-mL sample of a hydrochloric acid solution requires 24.16 mL of 0.106 mol L−1 sodium hydroxide solution for complete neutralisation.

What is the concentration of the original hydrochloric acid solution?

I can work out the equation and balance it but I am not sure how to answer the concentration question???

[Alpha-Omega]

This is an acid base neutrlization/double displacement reaction.

NaOH (aq) + HCl (aq) --> NaCl (aq) + H₂O (l)

Species always react mole to mole.  So if you look at tha balanced reaction equation you can see 1 mole of NaOH Reacts with 1 mole of HCl to form 1 mole of NaCl and 1 mole of water.

Again, this is a neutralization/double displacement reaction.  You are titrating the HCl  (initial volume 25.00 mL) with the NaOH. The final volume of NaOH is 24.16 mL and the molarity of the NaOH is given as 0.106M.

So start by writing down what you have:

V_HCl = 25.00 mL
V_NaOH = 24.16 mL
M_NaOH = 0.106M

You need to find the M_HCl

Now, they react 1:1 and mole:mole.  So just use the eqn: 

M_HCl x V_HCl = M_NaOH x V_NaOH

Substitute the knows you have and solve for M_HCl

Try that and see if it works.  BTW, FYI:  This is the procedure used for standardizing an acid (determining its concentration) using a base of known concentration.  Sometimes it helps to connect the problem to something it is used for so you remember the WHY!!!!!