Key Terms & Equations Needed for Chemistry 3.1

In this internal assessment you will need to carry out a practical investigation using volumetric analysis (titration).

There are a number of terms and equations that you will use and understand

Concentration

Concentration the is either the mass or amount of a substance in a particular volume of solution.

There are four units of concentration that you are likely to come across in high school Chemistry.

1. Concentration in moles per litre, mol L^-1

2. Concentration in grams per litre, g L^-1

3. Percentage concentration, % concentration

4. Concentration in parts per million or milligrams per litre, mg L^-1 or ppm

Concentration, c, in mol L^-1

• The concentration, c, of a solution is measured in moles per litre (mol L^-1 ), and can be calculated using the following equation

• c = n/V which can be re-arranged to n = cV and V = n/c

• c = concentration in moles per litre, mol L^-1

• n = amount of dissolved solute in moles, mol.

• V = volume of solution in litres, L

Determining the concentration of a standard solution from a mass and volume

Some quantitative problems involving concentration may involve the calculation of the amount of substance, n (in mol) from mass, m (in g) and molar mass, M (in g mol^-1 ) first.

This will mean that the n = m/M will need to be used, before the concentration (in mol L^-1) can be calculated using c = n/V

Example calculation

Anhydrous (without water) sodium carbonate can be used to make a standard solution.

3.00 g of anhydrous sodium carbonate, Na₂CO₃ is dissolved in distilled water and the volume of the solution is made up to 250 mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution in mol L^-1.

M(Na₂CO₃) = (2 x 23.0) + 12.0 + (3 x 16.0) = 106 g mol^-1

V(Na₂CO₃) = 250 mL = 0.250 L

Step 1: Calculate the number of moles of Na₂CO₃ using n = m/M.

n(Na₂CO₃) = m/M = 3.00/106 = 0.028301886 mol = 0.0283 mol (3 s.f.)

Step 2: Calculate the concentration of the solution using the unrounded number of moles from step 1.

c(Na₂CO₃) = n/V = 0.028301886/0.250 = 0.113207547 mol L^-1

c(Na₂CO₃) = 0.113 mol L^-1 (3 s.f.)

IMPORTANT: You will need to do one of these calculations in your assessment.

Concentration by mass

Concentration us also commonly expressed in terms of the mass of the solute in a litre of water.

The units for mass concentration are grams per litre, g L^-1

Converting between mol L^-1 and g L^-1 and vice versa

To convert from mol L^-1 to g L^-1 just multiply the concentration in mol L^-1 by the molar mass, M in g mol ^-1

Example 1

Convert 0.113 mol L^-1 sodium carbonate, Na₂CO₃ into g L^-1

c(Na₂CO₃) = 0.113 mol L^-1 & M(Na₂CO₃) = 106 g mol^-1

mass concentration = 0.113 x 106 = 11.978 g L^-1

Example 2

Convert 0.123 g L^-1 sodium carbonate, Na₂CO₃ , into mol L^-1

mass concentration(Na₂CO₃) = 0.123 g L^-1 & M(Na₂CO₃) = 106 g mol^-1

c(Na₂CO₃) = 0.123/106 = 1.16 x 10^-3 mol L^-1 (3 s.f.)

IMPORTANT: You will need to do one of these calculations in your assessment.

Percentage concentration

The percentage concentration of a solution is often expressed as mass percent.

In Chemistry you might have to make a percentage concentration of a solution. This is quite easy to do. To make a 2 % solution, you would dissolve 2 g of the solute in 100 mL of distilled water. To make 200 mL of a 5 % solution, you would need to use 5 g per 100 mL of solution. Since 200 mL is needed you would need to have 2 x 5 g = 10 g of solute dissolved and made up to 200 mL with distilled water.

Concentration in parts per million, ppm

As a senior Chemistry student you might come across concentration in parts per million, or ppm.

This unit is often used when dealing with very small concentrations, such as oxygen concentration in water.

Parts per million, ppm, is the same as milligrams per litre, mg L^-1

To calculate concentration in ppm, just multiply the mass concentration in g L^-1 by 1000.

Convert 5.67 g L^-1 of dissolved oxygen to ppm (mg L^-1 )

5.67 x 10^-3 g L^-1 = 5.67 x 10^-3 x 1000 = 5.67 ppm

Convert 0.00789 g L^-1 of dissolved oxygen to ppm (mg L^-1 )

0.00789 g L^-1 = 0.00789 x 1000 = 7.89 ppm