Molarity



 

Learning Objective

  • Calculating solution concentrations using Molarity.

Key Points

    • Molarity (M) indicates the number of moles of solute per liter of solution (moles/Liter) and is one of the most common units used to measure the concentration of a solution.
    • Molarity can be used to calculate the volume of solvent or the amount of solute.
    • The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.

Terms

  • molarityThe concentration of a substance in solution, expressed as the number moles of solute per liter of solution.
  • SI unitThe modern form of the metric system used extensively in the sciences (abbreviated SI from French: Système International d’Unités).
  • dilutionThe process by which a solution is made less concentrated via addition of more solvent.
  • concentrationThe relative amount of solute in a solution.

In chemistry, concentration of a solution is often measured in molarity (M), which is the number of moles of solute per liter of solution. This molar concentration (ci) is calculated by dividing the moles of solute (ni ) by the total volume (V) of the :

[latex]c_i=\frac{n_i}{V}[/latex]

The SI unit for molar concentration is mol/m3. However, mol/L is a more common unit for molarity. A solution that contains 1 mole of solute per 1 liter of solution (1 mol/L) is called “one Molar” or 1 M. The unit mol/L can be converted to mol/m3 using the following equation:

1 mol/L = 1 mol/dm3 = 1 mol dm−3 = 1 M = 1000 mol/m3

Calculating Molarity

To calculate the molarity of a solution, the number of moles of solute must be divided by the total liters of solution produced. If the amount of solute is given in grams, we must first calculate the number of moles of solute using the solute’s molar mass, then calculate the molarity using the number of moles and total volume.

Calculating Molarity Given Moles and Volume

If there are 10.0 grams of NaCl (the solute) dissolved in water (the solvent) to produce 2.0 L of solution, what is the molarity of this solution?

First, we must convert the mass of NaCl in grams into moles. We do this by dividing by the molecular weight of NaCl (58.4 g/mole).

[latex]10.0 \text{ grams NaCl} \times \frac{\text{1 mole}}{58.4 \text {g/mole}} = 0.17 \text{ moles NaCl}[/latex]

Then, we divide the number of moles by the total solution volume to get concentration.

[latex]c_i=\frac{n_i}{V}[/latex]

[latex]c_i=\frac{0.17 \text{ moles NaCl}}{2 \text{ liters solution}}[/latex]

[latex]c_i = 0.1 \text{ M}[/latex]

The NaCl solution is a 0.1 M solution.

Calculating Moles Given Molarity

To calculate the number of moles in a solution given the molarity, we multiply the molarity by total volume of the solution in liters.

How many moles of potassium chloride (KCl) are in 4.0 L of a 0.65 M solution?

[latex]c_{i}=\frac{n_{i}}{V}[/latex]

[latex]0.65 \text{ M} = \frac{n_i}{4.0 \text{ L}}[/latex]

[latex]n_i = (0.65 \text{ M})(4.0 \text{ L}) = 2.6 \text{ moles KCl}[/latex]

There are 2.6 moles of KCl in a 0.65 M solution that occupies 4.0 L.

Calculating Volume Given Molarity and Moles

We can also calculate the volume required to meet a specific mass in grams given the molarity of the solution. This is useful with particular solutes that cannot be easily massed with a balance. For example, diborane (B2H6) is a useful reactant in organic synthesis, but is also highly toxic and flammable. Diborane is safer to use and transport if dissolved in tetrahydrofuran (THF).

How many milliliters of a 3.0 M solution of BH3-THF are required to receive 4.0 g of BH3?

First we must convert grams of BH3 to moles by dividing the mass by the molecular weight.

[latex]\frac{4.0 \text{ g }BH_3 }{13.84 \text{g/mole }BH_3} = 0.29 \text{ moles }BH_3[/latex]

Once we know we need to achieve 0.29 moles of BH3, we can use this and the given molarity (3.0 M) to calculate the volume needed to reach 4.0 g.

[latex]c_{i}=\frac{n_{i}}{V}[/latex]

[latex]3.0 \text{ M} = \frac{0.29 \text{moles BH}_3} {V}[/latex]

[latex]V = 0.1 L[/latex]

Now that we know that there are 4.0 g of BH3 present in 0.1 L, we know that we need 100 mL of solution to obtain 4.0 g of BH3.

Dilution

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. This relationship is represented by the equation c1V1 = c2V2 , where c1 and c2 are the initial and final concentrations, and V1 and V2 are the initial and final volumes of the solution.

Example 1

A scientist has a 5.0 M solution of hydrochloric acid (HCl) and his new experiment requires 150.0 mL of 2.0 M HCl. How much water and how much 5.0 M HCl should the scientist use to make 150.0 mL of 2.0 M HCl?

c1V1 = c2V2

c1 and V1 are the concentration and the volume of the starting solution, which is the 5.0 M HCl. c2 and V2 are the concentration and the volume of the desired solution, or 150.0 mL of the 2.0 M HCl solution. The volume does not need to be converted to liters yet because both sides of the equation use mL. Therefore:

[latex](5.0 \text{ M HCl})(V_1) = (2.0 \text{ M HCl})(150.0 \text{ mL})[/latex]

V1 = 60.0 mL of 5.0 M HCl

If 60.0 mL of 5.0 M HCl is used to make the desired solution, the amount of water needed to properly dilute the solution to the correct molarity and volume can be calculated:

150.0 mL – 60.0 mL = 90.0 mL

In order for the scientist to make 150.0 mL of 2.0 M HCl, he will need 60.0 mL of 5.0 M HCl and 90.0 mL of water.

Example 2

Water was added to 25 mL of a stock solution of 5.0 M HBr until the total volume of the solution was 2.5 L. What is the molarity of the new solution?

We are given the following: c1= 5.o M, V1= 0.025 L, V2= 2.50 L. We are asked to find c2, which is the molarity of the diluted solution.

(5.0 M)(0.025 L) = c2 (2.50 L)

[latex]c_2 = \frac{(5.0 M)(0.025 L) }{2.50 L} =0.05 M[/latex]

Notice that all of the units for volume have been converted to liters. We calculate that we will have a 0.05 M solution, which is consistent with our expectations considering we diluted 25 mL of a concentrated solution to 2500 mL.