Expressing Concentration of Solutions - Ninth Grade Science
In chemistry, stating that a solution is "concentrated" or "dilute" provides vague, qualitative information. Adding a pinch of table salt to a liter of water makes it slightly salty; adding five tablespoons makes it intensely salty. In quantitative science, medicine, and industrial manufacturing, precision is paramount: a pharmacist formulating intravenous saline, an agriculturalist spraying pesticide, or a chemical engineer synthesizing polymers must know the exact proportion of solute present in a solution. In physical chemistry, the measure of how much solute is dissolved in a given quantity of solution or solvent is called concentration. In this chapter, you will investigate how solubility works, distinguish between unsaturated, saturated, and supersaturated solutions, and master quantitative concentration calculations, including mass percentage, volume percentage, and mass-by-volume percentage.
Solubility and Saturation States
Solubility is defined as the maximum mass of solute in grams that can dissolve in exactly one hundred grams of a given solvent at a specific temperature to form a saturated solution.
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| Solution State | Molecular Saturation Definition |
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| Unsaturated Solution | Can dissolve MORE solute at that temperature; |
| | solute concentration is below solubility limit. |
| Saturated Solution | Contains MAXIMUM dissolved solute at that temp; |
| | dynamic equilibrium between dissolving & crystal.|
| Supersaturated | Contains MORE dissolved solute than normal max; |
| Solution | unstable state prepared by cooling a hot solution|
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Dynamic Equilibrium in a Saturated Solution
Dissolution
Solid Solute <===========> Dissolved Solute
Crystallization
Rate of Dissolving Solute = Rate of Recrystallizing Solute!
Unsaturated Solutions
An unsaturated solution is one in which the concentration of solute is less than its equilibrium solubility at that temperature. If you add additional solute crystals to an unsaturated solution, they dissolve completely upon stirring.
Saturated Solutions and Dynamic Equilibrium
A saturated solution contains the maximum possible quantity of dissolved solute that the solvent can support at a specific temperature. When a solution reaches saturation, a state of dynamic physical equilibrium is established: the rate at which solid solute particles dissolve into the liquid exactly equals the rate at which dissolved solute molecules collide and recrystallize back onto the solid surface. If you add more sugar crystals to a saturated sugar solution, the crystals remain at the bottom, unable to dissolve.
Supersaturated Solutions
A supersaturated solution is an unstable physical state in which the liquid holds more dissolved solute than its normal saturation limit at that temperature. Because the solubility of most solid solutes increases with temperature, a chemist can prepare a saturated solution at a high temperature (such as $80^\circ ext{C}$) and then cool it down slowly and undisturbed to room temperature ($20^\circ ext{C}$). If no dust or agitation triggers crystallization, the excess solute remains dissolved in an unstable state. However, if a tiny "seed crystal" of solute is dropped into the supersaturated liquid, the excess solute rapidly crystallizes out of solution, releasing latent heat of crystallization.
Factors Affecting Solubility
The solubility of a solute in a solvent depends on three primary physical factors:
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| Influencing Factor | Effect on Solid Solutes | Effect on Gases |
+-----------------------+-----------------------------+--------------------+
| Temperature Increase | Solubility INCREASES for | Solubility |
| | most solids (Endothermic) | DECREASES (Gas |
| | | escapes into air) |
| Pressure Increase | Negligible effect on solids | Solubility |
| | (Incompressible liquids) | INCREASES linearly |
| | | (Henry's Law) |
| Chemical Nature | "Like Dissolves Like" | Polar gases in H2O;|
| | (Polar in polar, non-polar) | non-polar in oils |
+-----------------------+-----------------------------+--------------------+
The Effect of Temperature
For most solid solutes (such as potassium nitrate, sugar, or copper sulfate), dissolution is an endothermic process ($\Delta H > 0$). According to Le Chatelier's Principle, raising the temperature drives the dissolution forward, increasing solubility. For example, 100 grams of water can dissolve roughly 36 grams of sodium chloride at $20^\circ ext{C}$, but can dissolve over 200 grams of sugar at that same temperature. Conversely, for gases dissolved in liquids (such as oxygen or carbon dioxide in water), dissolution is an exothermic process ($\Delta H < 0$). Raising the temperature decreases gas solubility. As water warms, dissolved oxygen escapes into the atmosphere, which explains why aquatic fish suffer respiratory distress in thermally polluted industrial waterways.
The Effect of Pressure: Henry's Law
Pressure has negligible effect on the solubility of solids and liquids because they are essentially incompressible. However, pressure profoundly dictates the solubility of gases in liquids. According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid:
S = k_H \, P
In carbonated beverage manufacturing, soda bottles are pressurized with carbon dioxide gas at several atmospheres of pressure, forcing immense quantities of $ ext{CO}_2$ to dissolve into the liquid. When you crack the bottle cap open, the internal pressure drops abruptly to atmospheric pressure (1 atm). In accordance with Henry's Law, the solubility of $ ext{CO}_2$ plunges, and excess carbon dioxide gas bubbles violently out of solution as effervescence.
Quantitative Methods of Expressing Concentration
In ninth-grade chemistry, concentration is expressed primarily through three percentage metrics that compare the quantity of solute to the total quantity of the solution.
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| Three Core Concentration Formulas |
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| 1. Mass by Mass Percentage (w/w %): |
| Mass % = [ Mass of Solute / Total Mass of Solution ] * 100 |
| |
| 2. Volume by Volume Percentage (v/v %): |
| Volume % = [ Volume of Solute / Total Volume of Solution ] * 100 |
| |
| 3. Mass by Volume Percentage (w/v %): |
| Mass/Vol % = [ Mass of Solute in grams / Volume of Solution in mL ]*100|
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CRITICAL WARNING: The Denominator is ALWAYS the SOLUTION!
Solution = Solute + Solvent
If you dissolve 20 g of Salt in 80 g of Water:
Total Mass of Solution = 20 g + 80 g = 100 g
Mass Percentage = (20 / 100) * 100 = 20% (w/w)
Do NOT divide by Solvent alone: (20 / 80) is WRONG!
1. Mass Percentage (Mass by Mass % or w/w)
Mass percentage expresses the mass of solute in grams present in one hundred grams of the final solution:
\text{Mass \% of Solute} = \left( \frac{\text{Mass of Solute}}{\text{Total Mass of Solution}} \right) \times 100
Where:
- $ ext{Total Mass of Solution} = ext{Mass of Solute} + ext{Mass of Solvent}$
If an aqueous glucose solution is labeled $10\%$ mass-by-mass, it means that exactly 10 grams of glucose are dissolved in 90 grams of water, producing 100 grams of solution.
2. Volume Percentage (Volume by Volume % or v/v)
Volume percentage expresses the volume of liquid solute in milliliters present in one hundred milliliters of the final solution, commonly used when mixing liquids:
\text{Volume \% of Solute} = \left( \frac{\text{Volume of Solute}}{\text{Total Volume of Solution}} \right) \times 100
Commercial rubbing alcohol labeled $70\%$ isopropyl alcohol by volume contains 70 milliliters of pure isopropyl alcohol blended with water to yield a total volume of 100 milliliters.
3. Mass by Volume Percentage (w/v)
Mass by volume percentage expresses the mass of solute in grams dissolved in one hundred milliliters of solution, widely used in clinical medicine and pharmacology:
\text{Mass by Volume \%} = \left( \frac{\text{Mass of Solute (grams)}}{\text{Total Volume of Solution (milliliters)}} \right) \times 100
A standard medical saline drip labeled $0.9\%$ (w/v) $ ext{NaCl}$ contains $0.9 ext{ grams}$ of pure sodium chloride dissolved in every $100 ext{ mL}$ of solution, matching the osmotic pressure of human blood plasma.
Chapter Exercises
Exercise 1: A laboratory technician dissolves 35.0 grams of copper(II) sulfate crystals into 165.0 grams of distilled water. Calculate the total mass of the resulting solution, and calculate the concentration of copper(II) sulfate in the solution in terms of mass-by-mass percentage.
Exercise 2: An antiseptic mouthwash contains 24.0 milliliters of pure ethanol dissolved in sufficient water to produce a total volume of 300.0 milliliters of mouthwash solution. Calculate the volume-by-volume percentage concentration of ethanol in the mouthwash.
Exercise 3: A patient in a hospital requires an intravenous infusion of a 5.0% (w/v) glucose solution. If the nurse administers an intravenous infusion bottle containing 500.0 milliliters of this solution, calculate the total mass of pure glucose in grams delivered into the patient's bloodstream.
Solutions and Explanations
Solution 1: Identify the given masses: mass of solute (copper sulfate) is $m_{ ext{solute}} = 35.0 ext{ g}$, and mass of solvent (water) is $m_{ ext{solvent}} = 165.0 ext{ g}$. First, calculate the total mass of the solution: $m_{ ext{solution}} = m_{ ext{solute}} + m_{ ext{solvent}} = 35.0 ext{ g} + 165.0 ext{ g} = 200.0 ext{ g}$. Now apply the mass percentage formula: $ ext{Mass \%} = (m_{ ext{solute}} / m_{ ext{solution}}) imes 100 = (35.0 / 200.0) imes 100 = 0.175 imes 100 = 17.5\%$. The concentration of copper(II) sulfate in the solution is 17.5% by mass.
Solution 2: Identify the given volumes: volume of solute (ethanol) is $V_{ ext{solute}} = 24.0 ext{ mL}$, and total volume of the finished solution is $V_{ ext{solution}} = 300.0 ext{ mL}$. Applying the volume-by-volume percentage formula: $ ext{Volume \%} = (V_{ ext{solute}} / V_{ ext{solution}}) imes 100 = (24.0 / 300.0) imes 100 = 0.080 imes 100 = 8.0\%$. The mouthwash solution has an ethanol concentration of 8.0% by volume.
Solution 3: A concentration of 5.0% (w/v) means that there are 5.0 grams of dissolved glucose in every 100.0 milliliters of solution. We are given total solution volume $V = 500.0 ext{ mL}$. Using the mass-by-volume concentration formula: $ ext{Mass/Vol \%} = (m_{ ext{solute}} / V_{ ext{solution}}) imes 100$. Substitute the known values: $5.0 = (m_{ ext{solute}} / 500.0) imes 100$. Dividing both sides by 100 gives $0.050 = m_{ ext{solute}} / 500.0$. Multiplying both sides by 500.0 yields $m_{ ext{solute}} = 0.050 imes 500.0 = 25.0$ grams. The intravenous infusion delivers exactly 25.0 grams of pure glucose into the patient's bloodstream.