MOLARITY a concentration unit defined as moles of solute per liter of solution.
The symbol for molarity is M.

EXAMPLE 1
If 400.0 mL of a solution contains 5.00 x 10^-3 moles of AgNO₃, what is the molarity of this solution?

     5.00 x 10^-3 mol AgNO₃ / 0.4000 L = 0.0125 M AgNO₃
 

EXAMPLE 2
Calculate the molarity of an HCl solution which contains 18.23 g of HCl in 355.0 mL of solution.

Calculate the number of moles of HCl.

     (18.23 g HCl) (1 mol HCl / 36.46 g HCl) = 0.5000 mol HCl

Divide the number of moles of HCl by the total volume in liters.

     0.5000 mol HCl / 0.3550 L solution = 1.408 M HCl
 

EXAMPLE 3
How many moles of NaCl are contained in a 27.49 mL sample of  0.350 M NaCl?

Rearranging the equation of molarity (M = mol/L) to solve for moles gives:       mol = (M)(L)

     (0.350 mol/L)(0.02749 L) = 9.62 x 10^-3 mol NaCl
 

TITRATION an analytical procedure of determining the concentration of one substance in solution by reacting it with a solution of another substance whose concentration is known, called a standard solution.

Acid-Base titrations are commonly referred to as neutralizations.

        Acid     +  Base     ---->  Salt      +  Water

        HCl      +  NaOH  ---->  NaCl   +   H₂O

        H₂SO₄ +  2 KOH  --->  K₂SO₄ +  2 H₂O
 

EXAMPLE 4
What volume of 0.250 M HCl (hydrochloric acid) is required to neutralize a 10.0 mL sample of 0.750 M NaOH (sodium hydroxide) solution?

A.  Calculate the number of moles of NaOH in the 10.00 mL sample.

       (0.750 mol/L)(0.0100 L NaOH) = 7.50 x 10^-3 mol NaOH

B.  Calculate the number of moles of HCl required to neutralize the NaOH.

       1 HCl + 1 NaOH ----> 1 NaCl + 1 H2O

       (7.50 x 10^-3 mol NaOH)(1 mol HCl / 1 mol NaOH) = 7.50 x 10^-3 mol HCl

C.  Calculate the volume of HCl required.

      Rearranging the equation of molarity (M = mol/L) to solve for volume gives:   L = mol / M

     (7.50 x 10^-3 mol HCl) / (0.250 mol HCl / L) = 0.0300 L HCl = 30.0 mL HCl
 

EXAMPLE 5
In a titration, a 25.00 mL sample of sodium hydroxide solution was neutralized by 32.72 mL of hydrochloric acid. The molarity of the HCl is 0.129 M. Find the concentration of the NaOH solution.

A. Calculate the number of moles of HCl required in the titration.

     (0.03272 L HCl)(0.129 mol/L) = 4.22 x 10^-3 mol HCl

B. Calculate the number of moles of NaOH neutralized.

     HCl + NaOH ----> NaCl + H₂O

     (4.22 x 10^-3 mol HCl)(1 mol NaOH / 1 mol HCl) = 4.22 x 10^-3 mol NaOH

C. Calculate the molarity of the NaOH.

     M = mol / L

     4.22 x 10^-3 mol NaOH / 0.02500 L NaOH solution = 0.169 M NaOH
 

EXAMPLE  6
During a titration, a 20.00 mL portion of a 0.100 M H₂SO₄ solution was carefully measured into a flask.
The solution required 18.47 mL of NaOH to reach a phenolphthalein endpoint. What is the concentration of the NaOH?

A.   (0.100 mol H₂SO₄/L)(0.02000 L H₂SO₄) = 2.00 x 10^-3 mol H₂SO₄

B.   H₂SO₄ + 2 NaOH ---> Na₂SO₄ + 2 H₂O

       (2.00 x 10^-3 mol H₂SO₄)(2 mol NaOH / 1 mol H₂SO₄) = 4.00 x 10^-3 mol NaOH

C.    M = mol/L

       4.00 x 10^-3 mol NaOH / 0.01847 L NaOH solution = 0.217 M NaOH