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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!!

Verified Answer :

JoshEast

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

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 \(dm ^{3}\)                         



rivky

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

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