Understanding Molarity

Molarity is a crucial concept in chemistry that measures the concentration of a solution. Defined as the number of moles of solute per liter of solution, it helps chemists quantify how much of a substance is dissolved in a solvent. This is particularly important in laboratory settings and industrial applications where precise concentrations are vital for chemical reactions and processes. Below are three diverse, practical examples of calculating molarity of a solution.

Example 1: Preparing a Sodium Chloride Solution

In a laboratory setting, a chemist needs to prepare a sodium chloride (NaCl) solution for an experiment. They want to create a 0.5 M solution using distilled water.

To calculate the amount of NaCl needed:

• Volume of solution required: 1 liter (L)
• Molarity desired: 0.5 M
• Molar mass of NaCl: 58.44 g/mol

Using the formula for molarity:

Molarity (M) = Moles of solute / Volume of solution (L)

Rearranging gives us:

Moles of solute = Molarity × Volume of solution

Moles of solute = 0.5 M × 1 L = 0.5 moles

Next, we convert moles to grams:

Mass (g) = Moles × Molar mass

Mass = 0.5 moles × 58.44 g/mol = 29.22 g

The chemist needs to weigh out 29.22 grams of sodium chloride and dissolve it in enough distilled water to make a final volume of 1 liter.

Notes

• If a different volume is required, the same calculation can be applied by adjusting the volume in the formula.
• Always ensure the solute is fully dissolved before measuring the final volume.

Example 2: Diluting a Concentrated Acid

A laboratory technician needs to dilute a concentrated hydrochloric acid (HCl) solution from 12 M to a 1 M solution for safety and ease of use in experiments.

The technician has:

• Initial concentration (C1): 12 M
• Final concentration (C2): 1 M
• Final volume required (V2): 2 L

To find the volume of the concentrated solution needed (V1), we can use the dilution equation:

C1 × V1 = C2 × V2

Rearranging this gives:

V1 = (C2 × V2) / C1

Substituting the values:

V1 = (1 M × 2 L) / 12 M = 0.1667 L = 166.7 mL

The technician measures out 166.7 mL of the 12 M HCl and adds distilled water until the total volume reaches 2 L.

Notes

• Always add acid to water, not the other way around, to avoid exothermic reactions that can cause splattering.
• This method can also be used for different concentrations and volumes by changing the values in the dilution equation.

Example 3: Determining Molarity from a Reaction

An environmental scientist is studying the concentration of lead (II) nitrate (Pb(NO3)2) in a water sample. They found that 0.1 moles of lead (II) nitrate are dissolved in 500 mL of water.

To find the molarity of the lead (II) nitrate solution:

• Moles of solute: 0.1 moles
• Volume of solution: 500 mL = 0.5 L

Using the molarity formula:

Molarity (M) = Moles of solute / Volume of solution (L)

Molarity = 0.1 moles / 0.5 L = 0.2 M

The concentration of lead (II) nitrate in the water sample is 0.2 M.

Notes

• This approach can be adapted for any solute and is useful in environmental monitoring where concentrations of pollutants need to be quantified.
• Ensure that the volume is converted to liters when performing calculations.