A beaker is filled with 225.0 mL of a sodium hydroxide solution with an unknown concentration. A 0.0100 M solution of HCl is used in the titration.

Question

The equivalence point is reached when 16.4 mL of HCl have been added.

What is the initial concentration of NaOH in the beaker?

__________M

Answer 1

FIrst, understand that both sodium hydroxide (NaOH) and HCl are strong base and strong acid respectively. This means that when they dissociate into OH- ions and H+ ions they entirely contribute to the formation of OH- ions and H+ ions.

Since OH- ions and H+ ions basically cancel each other out by combining to form water this means that, according to the problem, when 16.4 mL of HCl is added to the NaOH solution that enough HCl has been added at that point to generate enough H+ ions to combine with all of the OH- ions in the NaOH solution. Thus in order to determine the concentration of NaOH in the initial solution we have to first determine the amount of mols of H+ ions added into the solution through HCl.

HCl --> H+ + Cl-

NaOH --> Na+ + OH-

1:1 mol ratio between the ionic compounds and their ion products.

M (HCl) = 0.0100 M

V (HCl) = 16.4 mL

M = mols of solute / V of solution (L)

1st: Convert V HCl to L

16.4 mL * (1 L/1000 mL) = 0.0064 L

2nd: Use the Molarity value you are given to calculate # of mols of HCl

M = mols/V --> mols = (M)(V) = (0.0100 M)(0.0064 L)

The L units cancel out leaving you with mols:

mols HCl = 0.000064 mols HCl

3rd: There is a 1:1 mol ratio between HCl and H+ ions, therefore the # of mols of HCl is the same as # of mols of H+ ions.

mols H+ = 0.000064 mols H+

4th: Since the addition of 16.4 mL of HCl is the equivalence point that means that enough mols of H+ were added to neutralize enough mols of OH- , thus the # mols of OH- is the same as # mols of H+ in this case.

mols OH- = 0.000064 mols OH-

5th: There is a 1:1 mol ratio between OH- and NaOH thus the # mols of NaOH is the same as the # mols of OH- that was just calculated.

mols NaOH = 0.000064 mols NaOH

6th: You know the mols of NaOH now as well as the initial Volume of the NaOH solution, refer to the molarity equation once again to calculate the Molarity of the NaOH solution.

M = mols solute / V (L) solution

V = 225.0 mL

V = 225.0 mL * (1 L/1000 mL) = 0.225 L

M = 0.000064 mols NaOH / 0.225 L NaOH

M = 0.00028 M NaOH

Admittedly, this seems like an extremely low concentration (M) however it makes sense due to the fact that you only had to titrate 16.4 mL of HCl into the large 225 mL solution of NaOH not to mention that the concentration of HCl was pretty low itself at 0.01 M