How do I fine the molar cnoncentration?

Question

My problem ask? A 10.0ml sample of aqueous HCL requires 24.32 mL of 0.134 M NaOH to reach the endpoint. What is the molar concentration of HCL? The equations for the reaction is

NaOH + HCL -----> NaCl + H2O

Answer 1

Molarity = moles / liters
rearrange to get-
moles = molarity x liters
1 mole of HCl reacts with 1 mole of NaOH
moles of NaOH = 0.02432 L x 0.134 Molar = 0.003259 moles
Therefore, the moles of HCl also equal 0.003259
Molarity of HCl = 0.003259 mole / 0.0100 liters
Molarity of HCl = 0.3259

The formula can also be written in millimoles (moles/1000)
mL x Molarity = millimoles
(This makes it a little easier, since all titrations are done in "mL" and not Liters.)

Answer 2

Since there is a one-to-one correlation between the NaOH and the HCl, you can use M1V1 = M2V2.
M, HCl = M,NaOH (V, NaOH/V, HCl)
M, HCl = 0.134 M,NaOH (24.32mL/10.0mL)
Use your calculator.

Answer 3

Let the HCl sample be called HS. Let NaOH solution be called NaS

24.32mLNaS/10.0mLHS x 0.134molNaOH/1000mLNaS x 1molHCl/1molNaOH x 1000mLHS/1LHS = 0.326 mole HCl per L HS, which is molarity.

Answer 4

1) Find the number of moles of base NaOH you need: .02432 L * .134 M = .003 moles. The equation is mol/L = M
2) Find the ratio of moles of NaOH to HCl. It's 1: 1 here. So .003 mol NaOH * (1mol HCl/1mol NaOH) give you .003 moles of HCl
3) Use the equation mol/L = M for HCl, and do:
.003 moles of HCl/ .01 L = .3M