Concentration Notes

Concentrations

Definition of Concentration

Concentration refers to the amount of a solute contained in a given amount of solution.
$concentration = \frac{amount \ of \ solute}{amount \ of \ solution}$
The "amount" can be measured in:
- Mass
- Volume
- Moles

Types of Concentration

The type of concentration used depends on the type of solution.

Solid Solutions

The most convenient measure of the amount of a solid is mass.
We use mass/mass percentages (m/m%).
$m/m\% = \frac{m{solute}}{m{solution}} \times 100\%$
m/m% indicates the percentage of the mass of the solution that is the solute.

Example

What is the mass percentage of zinc in brass if 10.00 g of zinc are dissolved into 25.00 g of copper to make brass?
$m/m\% = \frac{10.00 \ g}{10.00 \ g + 25.00 \ g} \times 100\% = \frac{10.00 \ g}{35.00 \ g} \times 100\% = 28.57\%$
28.57% of the mass of this brass is due to the zinc atoms.

Practice

What mass of tin is required to make 1500.0 g of a bronze solution containing 12.0% tin in copper?
$m/m\% = \frac{m{tin}}{m{solution}} \times 100\%$
$12.0\% = \frac{m_{tin}}{1500.0 \ g} \times 100\%$
$m_{tin} = \frac{12.0\% \times 1500.0 \ g}{100\%} = 180 \ g$
180 g of tin is required to make 1500.0 g of bronze.

Liquid Solutions

Liquids are most conveniently measured using volume.
We use volume/volume percentages (v/v%) to discuss concentrations in liquid-liquid solutions.
Examples include alcohol (rubbing or beverages) or hydrogen peroxide.
The v/v% tells you the percentage of the volume that is due to the solute.
$v/v\% = \frac{v{solute}}{v{solution}} \times 100\%$

Example

What is the volume percentage of isopropanol in rubbing alcohol if 50.00 mL of isopropanol are dissolved into 20.00 mL of water?
$v/v\% = \frac{50.00 \ mL}{50.00 \ mL + 20.00 \ mL} \times 100\% = \frac{50.00 \ mL}{70.00 \ mL} \times 100\% = 71.43\%$
71.43% of the volume of this rubbing alcohol is due to the isopropanol.

Practice

What volume of ethanol must you add to 500.0 mL of water to make vodka, which is a 40.00% solution?
$v/v\% = \frac{v{ethanol}}{v{solution}} \times 100\%$
$40.00\% = \frac{v{ethanol}}{v{ethanol} + 500.0 \ mL} \times 100\%$
$0.40 = \frac{v{ethanol}}{v{ethanol} + 500.0 \ mL}$
$0.40 (v{ethanol} + 500.0 \ mL) = v{ethanol}$
$0.40 v{ethanol} + 200.0 \ mL = v{ethanol}$
$200.0 \ mL = 0.60 v_{ethanol}$
$v_{ethanol} = \frac{200.0 \ mL}{0.60} = 333.3 \ mL$

Solid-Liquid Solutions

Solutions of a solid solute in a liquid solution are often reported using mass/volume percentages (m/v%).
This is seen in hospitals on saline solutions (0.90% NaCl in water).
The m/v% is always listed in g/mL.
$m/v\% = \frac{m{solute} (g)}{v{solution} (mL)} \times 100\%$

Example

What mass of sodium chloride is required to make 300 mL of saline solution (0.90%)?
$0.90\% = \frac{m_{NaCl}}{300 \ mL} \times 100\%$
$m_{NaCl} = \frac{0.90\% \times 300 \ mL}{100\%} = 2.7 \ g$
2.7 g of sodium chloride must be dissolved into 300 mL of water to make 300 mL of saline solution.

Practice

What is the m/v% of sugar in a can of coke?
It's a 12% solution of sugar in water (m/v).

Parts per Million (ppm) or Billion (ppb)

Some solutions are very dilute, so their concentrations will be small using v/v% or m/m%.
We use parts per million (ppm) or parts per billion (ppb).
Similar to mass percentages.
ppm represents the number of grams of solute in 1 million grams of solution.
ppb is similar but for 1 billion grams of solution.
$ppm = \frac{m{solute}}{m{solution}} \times 10^6$
$ppb = \frac{m{solute}}{m{solution}} \times 10^9$

Example

The concentration of carbon dioxide in air is approximately 392 ppm. What mass of carbon dioxide is there in this classroom? The class contains approximately 575000 g of air.
$392 \ ppm = \frac{m{CO2}}{575000 \ g} \times 10^6$
$m{CO2} = \frac{392 \times 575000 \ g}{10^6} = 225 \ g$
There are roughly 225 g of carbon dioxide in this room.

Practice

The maximum safe concentration of arsenic in drinking water is 0.010 ppm. If you dissolved 0.0050 g of arsenic in a 20.0 L (20,000 g) water jug, would you be arrested for serving that water to your friend?
$ppm = \frac{0.0050 \ g}{20000 \ g} \times 10^6 = 0.25 \ ppm$
Since there is more arsenic dissolved than the safe concentration, you would be arrested for attempted murder if you served this to your friend!

Molar Concentration

For most applications in chemistry, it is convenient to have concentrations listed using moles of solute.
Molar concentration is listed in mol/L (or M).
Used for any solution, independent of the states of the solute/solvent.
$Molar \ concentration = \frac{moles \ of \ solute}{volume \ of \ solution (L)}$

Example

What is the molar concentration of hydrochloric acid if 0.547 g of hydrogen chloride is dissolved into 30.0 mL of water?
Convert grams of HCl to moles:
- Molar mass of HCl = 1.008 + 35.45 = 36.458 g/mol
- $moles \ of \ HCl = \frac{0.547 \ g}{36.458 \ g/mol} = 0.015 \ mol$
Convert mL to L:
- $30.0 \ mL = 30.0 \ mL \times \frac{1 \ L}{1000 \ mL} = 0.030 \ L$
$Molar \ concentration = \frac{0.015 \ mol}{0.030 \ L} = 0.500 \ M$
The concentration of hydrochloric acid is 0.500 M.

Practice

What is the molar concentration of potassium hydroxide if 0.0421 g of solid potassium hydroxide is dissolved into 25.0 mL of water?
Convert grams of KOH to moles:
- Molar mass of KOH = 39.0983 + 15.999 + 1.008 = 56.1053 g/mol
- $moles \ of \ KOH = \frac{0.0421 \ g}{56.1053 \ g/mol} = 0.00075 \ mol$
Convert mL to L:
- $25.0 \ mL = 25.0 \ mL \times \frac{1 \ L}{1000 \ mL} = 0.025 \ L$
$Molar \ concentration = \frac{0.00075 \ mol}{0.025 \ L} = 0.0300 \ M$
The concentration of potassium hydroxide is 0.0300 M.

Dilution

Useful solutions often have low concentrations.
Low enough that it is impractical to weigh out the mass of solute required.
Example: insulin injections.
Insulin for diabetes contains a $6.20 \times 10^{-4}$ mol/L solution of insulin in water.
Each 1.00 mL dose contains only 0.0036 g of insulin.
This is too small an amount to measure out and dissolve into 1.00 mL of water.
To get around this issue, we can dilute solutions by adding more solvent.
Adding solvent causes the concentration to decrease.
There is the same amount of solute in a larger volume of solution.

Stock Solutions

We can make a stock solution and use it to make diluted samples.
A stock solution has a relatively high concentration.
We take out an aliquot (a small sample) of the stock solution and place it in a new container.
We then add more solvent to it.
This will create a diluted solution.
Ex: if we take 1 mL of solution and add it to 9 mL of water, we make a diluted solution.

Tools for Accurate Dilutions

To make solutions with accurate concentrations we need to use tools that measure volume accurately.
Volumetric flasks:
- Measure 1 volume very accurately (4-5 digits).
- Used as a container for the new solution.
Pipettes:
- Measure aliquot volumes to 3-4 digits.
- We pipette an aliquot into the volumetric flask and then fill to the line with solvent.

Dilution Calculations

Calculating the new concentration of a diluted sample is fairly easy.
The moles in the new dilute solution are the same as in the concentrated solution.
$C1V1 = C2V2$
- $C_1$ = initial concentration
- $V_1$ = initial volume
- $C_2$ = final concentration
- $V_2$ = final volume
This equation can be used to determine the concentrations and volumes involved with a dilution.

Practice 1

If 3.0 mL of a 1.0 M solution of HCl(aq) is added to 17.0 mL of water, what is the new concentration?
$C_1 = 1.0 \ mol/L$
$V_1 = 3.0 \ mL$
$C_2 = ?$
$V_2 = 17.0 + 3.0 = 20.0 \ mL$
$C1V1 = C2V2$
$C2 = \frac{C1V1}{V2} = \frac{(1.0 \ mol/L)(3.0 \ mL)}{(20.0 \ mL)} = 0.15 \ mol/L$
Therefore, the new solution will have a hydrochloric acid concentration of 0.15 mol/L.

Practice 2

A 1.0 mL aliquot is removed from a 500.0 mL solution of 3.0 M HCl(aq) and is added to a 50.0 mL volumetric flask. Water is added to the flask until the mark is reached. What is the new concentration?
$C_1 = 3.0 \ mol/L$
$V_1 = 1.0 \ mL$
$C_2 = ?$
$V_2 = 50.0 \ mL$
$C1V1 = C2V2$
$C2 = \frac{C1V1}{V2} = \frac{(3.0 \ mol/L)(1.0 \ mL)}{(50.0 \ mL)} = 0.060 \ mol/L$
Therefore, the new solution will have a hydrochloric acid concentration of 0.060 mol/L.

Practice 3

What volume aliquot should be removed from a 0.50 M HCl(aq) solution in order to create 25.0 mL of a $2.5x10^{-4}$ M solution?
$C_1 = 0.50 \ mol/L$
$V_1 = ?$
$C_2 = 2.5 \times 10^{-4} \ mol/L$
$V_2 = 25.0 \ mL$
$C1V1 = C2V2$
$V1 = \frac{C2V2}{C1} = \frac{(2.5 \times 10^{-4} \ mol/L)(25.0 \ mL)}{(0.50 \ mol/L)} = 0.012 \ mL$
You would need 0.012 mL of the original solution to create the diluted sample.

Spectrometry

UV/Vis spectroscopy can be used to measure the concentration of certain compounds in solution.
Because compounds absorb different wavelengths of light to differing extents.
Ex:
- Red dye absorbs 500 nm strongly.
- Observing this wavelength over time can show you changes in ethanol concentration.

Beer’s Law

The amount of light absorbed by a sample depends on two things:
- The identity of the compound(s) in the sample.
- The number of particles that the light encounters as it passes through the sample.
You can alter the #particles the light encounters via:
- The length of the sample cell.
- The concentration of the sample.
$A = \varepsilon l c$
- A = Absorbance
- $\varepsilon$ = molar absorptivity (L mol⁻¹ cm⁻¹)
- l = path length (cm)
- c = concentration (mol/L)
Molar absorptivity is a measure of how much light a substance absorbs for each 1 mol/L and 1 cm path length.

Practice

If the sample that creates the blue line on the right had a concentration of 0.10 mol/L and the sample cell was 1 cm long, what is the molar absorptivity at:
- 500 nm
- 700 nm

Concentration Lab

For this experiment, we are going to determine the concentration of copper (II) sulfate in an unknown sample using spectroscopy.
With your group members, figure out a method for determining this concentration.
What are you going to need to know?
How are you going to experimentally determine the things you need to know?