Molarity relates the number of moles of solute and the volume of solution, in liters.

\(\displaystyle M\;=\;\frac{moles\;of\;solute}{L\;of\;solution}\)

We can calculate a solution volume, in liters, if we know the number of moles of solute and the molar concentration of the solution.

\(\displaystyle V\;=\;M\;\times\;V\)

If the volume, in liters, and the molarity of a solution is known, the number of moles of solute can be calculated.

\(\displaystyle moles\;of\;solute\;=\;M\;\times\;V\)

For the stoichiometry problems we have worked on up to now, only mass has been used. In the lab we also use solutions of different concentrations for a chemical reaction. If given a molar concentration and the solution volume, the number of moles of solute is easy to calculate. Once the moles have been determined, we can then use the balanced chemical equation to solve stoichiometry problems that involve solutions.

Example 1. If 100.00 mL of 0.475 M NiCl₂ is reacted with 100.00 mL of 0.325 M Na₂S, how many grams of solid NiS are formed? Below is the balanced chemical equation.

NiCl₂ (aq) + Na₂S (aq) → NiS (s) + 2 NaCl (aq)

First, find the number of moles of NiCl₂ and Na₂S.

\(\displaystyle mol\;NiCl_2\;=\;0.10000\;L\times\frac{0.475\;mol}{L}\;=\;\mathbf{0.0475\;mol\;NiCl_2}\)
 
\(\displaystyle mol\;Na_2S\;=\;0.10000\;L\times\frac{0.325\;mol}{L}\;=\;\mathbf{0.0325\;mol\;Na_2S}\)

The limiting reactant is Na₂. There is a one to one mole ratio of Na₂ and NiS, therefore, 0.0325 moles of NiS is produced. The molar mass of NiS is 90.76 g/mol.

\(\displaystyle 0.0325\;mol\;NiS\times\frac{90.76\;g}{mol}\;=\;\mathbf{2.95\;g\;NiS\;is\;formed}\)
 
Example 2. How many mL of 0.485 M HCl are required to completely react with 0.124 moles of Pb(NO₃)₂? The unbalanced equation is below:

2 HCl (aq) + Pb(NO₃)₂ (aq) → PbCl₂ (s) + 2 HNO₃ (aq)

Calculate the number of moles of HCl required to react with 0.124 moles of Pb(NO₃)₂. Use the balanced chemical reaction to do this.

\(\displaystyle 0.124
/;mol;Pb(NO_3)_2\times\frac{2\;mol\;HCl}{1\;mol;Pb(NO_3)_2}\;=\;\mathbf{0.248\;mol\;HCl}\)
 
Finally, calculate the volume needed by using the molar concentration and the number of moles of HCl.

\(\displaystyle \frac{0.248\;mol\;HCl}{0.485\:M\;HCl}\;=\;\mathbf{0.511\;L\;=\;511\;mL\;HCl}\)
 
It would take 511 mL of 0.485 M HCl.

Worksheet: Volumetric Stoichiometry and Titration

Exercises

Exercise 1. What mass of NaCl is needed to precipitate Ag⁺ ions from 25.00 mL of 0.250 M AgNO₃ solution?

Check Answer/Solution to Exercise 1

Exercise 2. How many mL of 0.125 M HCl are required to react with 45.00 mL of 0.150 M Ba(OH)₂?

Check Answer/Solution to Exercise 2

Exercise 3. Acetic acid, CH₃COOH, is the main acid component of vinegar. If 3.56 mL of vinegar requires 38.25 mL of 0.125 M NaOH for reaction, how many grams of acetic acid are in 545 mL of vinegar?

Check Answer/Solution to Exercise 3

CH₃COOH (aq) + NaOH (aq) → NaCH₃COO (aq) + H₂O (l)

Exercise 4. Consider the following chemical equation:

AlCl₃ (aq) + Ba(OH)₂ (aq) → Al(OH)₃ (s) + BaCl₂ (aq)

How many grams of BaCl₂ are formed if 125.00 mL of 0.355 M AlCl₃ is reacted with 150.00 mL of 0.296 M Ba(OH)₂?

Check Answer/Solution to Exercise 4
 

 
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