Study Notes on Measuring Solubility and Concentration in Chemistry

Instant Heat!

  • Presented by Jess McCartney and Edrolo, 2023

  • Focus on measuring solubility and concentration in VCE Chemistry

Overview of the Lesson

Study Design

  • Key topics discussed:

    • Units of concentration of solutions

    • Dilution

    • Saturated, unsaturated, and supersaturated solutions

  • Definitions include the concentration as a measure of the quantity of solute dissolved in a given mass or volume of solution (e.g., mol L⁻¹, g L⁻¹, %(m/v), %(v/v), ppm).

  • Includes unit conversions.

Solubility

Definition

  • Solubility: A measure of how much solute will dissolve in a given amount of solvent at a specified temperature.

Key Terms

  • Saturated Solution:

    • A solution in which no more solute can be dissolved at a particular temperature.

  • Unsaturated Solution:

    • A solution that contains less solute than needed to make it saturated and can dissolve more solute.

  • Supersaturated Solution:

    • An unstable solution that contains more dissolved solute than a saturated solution.

Deep Dive

  • Solvent (n.): A substance that dissolves solutes to form solutions.

  • Solute (n.): A substance that is dissolved in another substance.

  • Illustration of concepts: Solvent → Solution → Solute

Concentration

Definition

  • Concentration (n.): The quantity of substance dissolved in a quantity of solution.

Units of Concentration

Common Units

  • Mass of solute per litre of solution (g/L)

  • Moles of solute per litre of solution (mol/L)

  • Parts per million (ppm)

  • Percentage by mass (%m/m)

  • Percentage by volume (%v/v)

  • Percentage mass/volume (%m/v)

Concentration Calculations

Example Calculation

  • Concentration (g/L): Indicates the amount of solid solute (in grams) dissolved in solution, e.g., salt in seawater.

  • Ppm is used for very small quantities:

    • Equivalents: mg/L, µg/g

    • Formula: Concentration (g/L)=mass of solute (in g)volume of solution (in L)\text{Concentration (g/L)} = \frac{\text{mass of solute (in g)}}{\text{volume of solution (in L)}}

Percentage by Mass

  • Describes the mass of solute per 100 g of solution (%m/m).

  • Example:

    • Saline solution: 0.9%(w/w) indicates 0.9 g of sodium chloride per 100 g of saline solution.

    • Alcohol content: 15%(v/v) means 15.0 mL of alcohol for every 100 mL of wine.

Molarity

Definition

  • Molarity (n.): Number of moles of solute per litre of solution, denoted as M.

Calculation Methodology

  • If given the mass of solute:

    • Formula: n=mMrn = \frac{m}{M_r}

    • Concentration formula: c=nVc = \frac{n}{V}

  • Units:

    • Concentration in mol/L, Unit - M

    • Number of moles, Unit - mol

    • Volume in litres, Unit - L

Worked Examples

Example 1: Molar Concentration Calculation

  • Problem: Calculate the concentration, in mol/L, of a solution containing 16.8 mg of AgNO₃ dissolved in 150 mL.

  • Steps:

    1. Volume conversion: V(AgNO3)=1501000=0.150LV(AgNO_3) = \frac{150}{1000} = 0.150 L

    2. Mass conversion: m(AgNO3)=16.81000=0.0168gm(AgNO_3) = \frac{16.8}{1000} = 0.0168 g

    3. Molar mass calculation:

    • M<em>r(AgNO</em>3)=107.9+14.0+(3×16.0)=169.9g/molM<em>r(AgNO</em>3) = 107.9 + 14.0 + (3 \times 16.0) = 169.9 g/mol

    1. Calculate moles: n(AgNO<em>3)=mM</em>r=0.0168169.9=9.89×10−5moln(AgNO<em>3) = \frac{m}{M</em>r} = \frac{0.0168}{169.9} = 9.89 \times 10^{-5} mol

    2. Calculate molar concentration: C(AgNO3)=nV=9.89×10−50.150=6.59×10−4MC(AgNO_3) = \frac{n}{V} = \frac{9.89 \times 10^{-5}}{0.150} = 6.59 \times 10^{-4} M

Example 2: Dilution Calculation

Definition

  • Dilution (n.): Lowering the concentration of solute in a solution by adding more solvent.

Formula for Dilution

  • C<em>1V</em>1=C<em>2V</em>2C<em>1V</em>1 = C<em>2V</em>2

Worked Example for Dilution

  • Problem: Calculate the concentration of the solution formed when 10.0 mL of water is added to 5.0 mL of 1.2 M HCl.

  • Steps:

    1. Identify C₁ and V₁:

    • C<em>1=1.2M,V</em>1=5.0mLC<em>1 = 1.2M, V</em>1 = 5.0 mL

    1. Total volume: V2=10.0+5.0=15.0mLV_2 = 10.0 + 5.0 = 15.0 mL

    2. Find new concentration:

    • C<em>2=C</em>1V<em>1V</em>2=1.2×5.015.0=0.40MC<em>2 = \frac{C</em>1V<em>1}{V</em>2} = \frac{1.2 \times 5.0}{15.0} = 0.40 M

Multiple Choice Activities

Activity Example 1: Concentration of Ammonia

  • Calculate the amount in moles of ammonia (NH₃) in 25.0 mL of a 0.3277 M ammonia solution.

  • Choices:
    A. 81.9 mol
    B. 13.1 mol
    C. 0.00131 mol
    D. 8.19 × 10⁻³ mol
    E. I don’t know.

Activity Example 2: ppm Calculation

  • Concentration, in ppm, of a 0.00200 M solution of NaCl, remember ppm = mg/L.

  • Choices:
    A. 0.117 ppm
    B. 1.17 ppm
    C. 117 ppm
    D. 1117 ppm
    E. I don’t know.

Summary

  • Solubility: The amount of solute dissolved in the solvent.

  • Concentration: Quantity of solute divided by the quantity of solvent, represented in various units.

  • Molarity: Number of moles of solute per litre of solution.

  • Dilution: Process of decreasing the concentration of a solution by adding solvent.

Key Terms

  • Solvent

  • Solute

  • Concentration

  • Molarity

  • Dilution

Important Formulas

  • n=mMn = \frac{m}{M}

  • C<em>1V</em>1=C<em>2V</em>2C<em>{1}V</em>{1} = C<em>{2}V</em>{2}

  • c=nVc = \frac{n}{V}

  • Concentration in mol/L, Unit - M

  • Number of mole, Unit - mol

  • Volume in litres, Unit - L

Further Material

  • Video link: ClickHeat - Reusable Heat Pad [HD] on YouTube.

Disclaimer

  • Content has been prepared under the copyright of the Victorian Curriculum and Assessment Authority. Errors are acknowledged, and feedback is welcomed.