Molar concentration?

Question

If 14.8 mL of 0.100 M NaOH sol are needed to react with 25.0 mL of an unknown HCl sol, what is the molar concentration of the HCI solution?

Answer 1

There is a key word missing from your question..."[needed to react] completely". We can pour any arbitrary amount of NaOH and HCl together and they will react, but to solve this problem we need to assume they completely neutralize eachother or we need to be told a final pH value of the solution.
But we can assume, for simplicity sake, that the number of moles of NaOH origially present equals the number of moles of HCl so they can completel neutralize eachother in the following chemical reaction:
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)

Molar concentration = Moles of solute / Liter of solution

moles of solute = Molar concentration * Liter of solution
moles of NaOH = .100 Molar * .0148 Liters
moles of NaOH = .00148 moles
Which should be equal to the moles of HCl
moles of HCl = .00148 moles

If this number of moles of HCl is dissolved in 25 mL,
Molar concentration = .00148 moles / .025 Liters
Molar HCl concentration = .0592 Molar

Answer 2

Basically you need to know the equation
Moles=ConcentrationxVolume/1000

1) Work out the Moles of NaOH as you have the C and V.
Moles=0.100x14.8/1000=0.00148 moles

2) Secondly one should know that if there is no number infront of NaOH or HCl assume the ratio is 1:1

3) Therefore Moles of NaOH =Moles of HCl = 0.00148

4) We can know find the Concentration of HCl
Concentration=Moles/Volumex1000
C=0.00148/25x1000 = 0.0592M

Hope thats helpful.