Question

A 10.0 milliliter sample of NaOH (aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH (aq)?

The Answer is 2.0M but I am unsure how to do the math. Someone please help!!

Answer 1

$NaOH_{(aq)} + HCl_{(aq)} ------\ \textgreater \ NaCl_{(aq)} + H_{2} O_{(l)}$

moles of HCl =  (molarity) * (volume)
                     = $0.50 mol / dm^{3} * 0.040 dm^{3}$
                     = 0.02 mol 
                   
     mole ratio of NaOH : HCl according to the equation is  1 : 1
     ∴ mol of NaOH = mol of HCl
                              =  0.02mol

Since volume of NaOH is 0.010 $dm^{3}$

then molarity of NaOH = $\frac{moles}{volume}$
                                     = $\frac{0.02 mol}{0.010 dm^{3}}$
                                     = $2.0 mol / dm^{3} OR 2.0 M$

Note:  
1) $mol/ dm^{3}$ is the same as M
2) I converted the ml to l then to $dm ^{3}$                         

Answer 2

You need to know this equation
Molarity(of acid) x volume (of acid) = Molarity (of base) x Volume(of base)
which is normally just written as MxV=MxV
you are already given the molarity(M) and the volume of the acid so you plug them in 40x0.50=MxV
now you also have the volume of the base so plug that in to
40x0.50=Mx10
now your just missing the molarity of the base 
you now just have to solve for M
first, to make it simpler multiply the left side 
use your mind or a calculator to find 40x0.50=20
now  you have 20=Mx10
now you divide by 10 on both sides
20/10= 2
M=2
now you have that the molarity of the base is 2