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%).
• $m/m\% = \frac{m<em>{solute}}{m</em>{solution}} \times 100\%$
• 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?
• $m/m\% = \frac{m<em>{tin}}{m</em>{solution}} \times 100\%$
• $12.0\% = \frac{m_{tin}}{1500.0 \ g} \times 100\%$
• $m_{tin} = \frac{12.0\% \times 1500.0 \ g}{100\%} = 180 \ g$
• 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.
• $v/v\% = \frac{v<em>{solute}}{v</em>{solution}} \times 100\%$

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?
• $v/v\% = \frac{v<em>{ethanol}}{v</em>{solution}} \times 100\%$
• $40.00\% = \frac{v<em>{ethanol}}{v</em>{ethanol} + 500.0 \ mL} \times 100\%$
• $0.40 = \frac{v<em>{ethanol}}{v</em>{ethanol} + 500.0 \ mL}$
• $0.40 (v<em>{ethanol} + 500.0 \ mL) = v</em>{ethanol}$
• $0.40 v<em>{ethanol} + 200.0 \ mL = v</em>{ethanol}$
• $200.0 \ mL = 0.60 v_{ethanol}$
• $v_{ethanol} = \frac{200.0 \ mL}{0.60} = 333.3 \ mL$

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.
• $m/v\% = \frac{m<em>{solute} (g)}{v</em>{solution} (mL)} \times 100\%$

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.
• $ppm = \frac{m<em>{solute}}{m</em>{solution}} \times 10^6$
• $ppb = \frac{m<em>{solute}}{m</em>{solution}} \times 10^9$

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.
• $392 \ ppm = \frac{m<em>{CO</em>2}}{575000 \ g} \times 10^6$
• $m<em>{CO</em>2} = \frac{392 \times 575000 \ g}{10^6} = 225 \ g$
• 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.
• $C<em>1V</em>1 = C<em>2V</em>2$
  • $C_1$ = 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?
• $C_1 = 1.0 \ mol/L$
• $V_1 = 3.0 \ mL$
• $C_2 = ?$
• $V_2 = 17.0 + 3.0 = 20.0 \ mL$
• $C<em>1V</em>1 = C<em>2V</em>2$
• $C<em>2 = \frac{C</em>1V<em>1}{V</em>2} = \frac{(1.0 \ mol/L)(3.0 \ mL)}{(20.0 \ mL)} = 0.15 \ mol/L$
• 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?
• $C_1 = 3.0 \ mol/L$
• $V_1 = 1.0 \ mL$
• $C_2 = ?$
• $V_2 = 50.0 \ mL$
• $C<em>1V</em>1 = C<em>2V</em>2$
• $C<em>2 = \frac{C</em>1V<em>1}{V</em>2} = \frac{(3.0 \ mol/L)(1.0 \ mL)}{(50.0 \ mL)} = 0.060 \ mol/L$
• 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?
• $C_1 = 0.50 \ mol/L$
• $V_1 = ?$
• $C_2 = 2.5 \times 10^{-4} \ mol/L$
• $V_2 = 25.0 \ mL$
• $C<em>1V</em>1 = C<em>2V</em>2$
• $V<em>1 = \frac{C</em>2V<em>2}{C</em>1} = \frac{(2.5 \times 10^{-4} \ mol/L)(25.0 \ mL)}{(0.50 \ mol/L)} = 0.012 \ mL$
• 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?