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 =
`(moles)/(volume)`
=
`(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