Solubility, Solution Concentration Calculations, and Solubility Product Constant

Definition and Fundamentals of Solubility

• Definition of Solubility:

  • Solubility is a physical property of a substance that indicates the maximum amount of a solute that can dissolve in a specified quantity of a solvent at a specific temperature (and at a specific pressure in the case of gases) to form a stable, homogeneous solution.
  • Recorded under session dated $30/07/26$

• Determining Factors:

  • Solubility depends directly on the chemical nature of both the solute and the solvent.
  • It is affected by environmental and physical conditions, specifically temperature and pressure.

• Units of Expression:

  • Generally expressed in grams of solute per $10\,\text{g}$ of solvent (or per $100\,\text{g}$ of solvent depending on the context).
  • Alternatively expressed in moles per liter ($\text{mol/L}$).

Components and Types of Solutions

• Components of a Solution:

  • Solute (Soluto): The substance that is dissolved in the solution.
  • Solvent or Dissolvent (Solvente o Disolvente): The substance that dissolves the solute.

• Classification of Solutions by Solute Content:

  • Unsaturated Solution (Insaturada): A solution that contains less solute than the solvent is capable of dissolving at a given temperature and pressure.
  • Saturated Solution (Saturada): A solution that contains the exact maximum amount of solute that can be dissolved in the solvent at a given condition.
  • Supersaturated Solution (Sobresaturada): A solution that contains more dissolved solute than can normally be maintained in equilibrium under standard conditions.

Factors Influencing Solubility and Rate of Dissolution

• 1. Nature of Solute and Solvent:

  • Governed by the principle "Like dissolves like" ("Lo semejante disuelve a lo semejante").
  • Polar substances dissolve polar substances, while non-polar substances dissolve non-polar substances.

• 2. Temperature:

  • For Solids: For the majority of solid solutes, higher temperature yields higher solubility.
  • Example: Sugar disolves much faster in hot coffee than in cold water.
  • For Gases: Gas solubility exhibits the opposite relationship: higher temperature results in lower gas solubility.
  • Example: A hot carbonated beverage (gaseosa caliente) loses its gas rapidly.

• 3. Pressure:

  • Primarily affects the solubility of gaseous solutes.
  • Increasing the pressure increases the solubility of the gaseous solute due to the pressure maintained inside the container.
  • Example: Carbonated soft drinks contain dissolved carbon dioxide ($\text{CO}_2$). When the bottle is opened, the internal pressure drops, causing the gas to escape and form bubbles.

• 4. Agitation (Agitación):

  • Agitation does not increase total solubility.
  • It accelerates the speed or rate at which dissolution occurs.
  • Example: Stirring coffee causes sugar to dissolve more rapidly.

• 5. Particle Size (Tamaño de la partícula):

  • Smaller solute particle size increases the total contact surface area with the solvent.
  • Increased surface area results in a faster rate of dissolution.
  • Example: Powdered sugar (azúcar pulverizada) dissolves much faster than a solid sugar cube (terrón de azúcar).

Concentration Calculations: Mass-Mass Percentage (% m/m)

• Fundamental Formulas:

  • Mass-Mass Percentage Formula:     $\%\,\text{m/m} = \frac{\text{masa soluto}}{\text{masa solución}} \times 100$
  • Total Solution Mass Relationship:     $\text{masa solución} = \text{masa soluto} + \text{masa solvente}$
  • Solute Mass Derived Formula:     $\text{masa soluto} = \frac{\%\,\text{m/m} \times \text{masa solución}}{100}$

• Step-by-Step Exercise Solutions:

  • Exercise 1: A solution is prepared with $25\,\text{g}$ of $\text{NaCl}$ in $175\,\text{g}$ of $\text{H}_2\text{O}$. Calculate the mass-mass percentage.
  • Solute mass = $25\,\text{g}$ of $\text{NaCl}$
  • Solvent mass = $175\,\text{g}$ of $\text{H}_2\text{O}$
  • Total solution mass = $25\,\text{g} + 175\,\text{g} = 200\,\text{g}$
  • Calculation:       $\%\,\text{m/m} = \frac{25\,\text{g}}{200\,\text{g}} \times 100 = 12.5\%$
  • Exercise 2: A solution contains $18\,\text{g}$ of sugar in $200\,\text{g}$ of solution. What is its mass percentage?
  • Solute mass = $18\,\text{g}$
  • Total solution mass = $200\,\text{g}$
  • Calculation:       $\%\,\text{m/m} = \frac{18\,\text{g}}{200\,\text{g}} \times 100 = 9\%$
  • Exercise 3: How many grams of $\text{NaOH}$ are needed to prepare $500\,\text{g}$ of a $12\%$ solution?
  • Target solution mass = $500\,\text{g}$
  • Concentration = $12\%$
  • Calculation:       $\text{masa soluto} = \frac{500\,\text{g} \times 12}{100} = \frac{6000}{100} = 60\,\text{g NaOH}$
  • Exercise 4: A solution contains $45\,\text{g}$ of $\text{KNO}_3$ and $255\,\text{g}$ of water ($\text{H}_2\text{O}$). Calculate the mass-mass percentage.
  • Solute mass = $45\,\text{g}$ of $\text{KNO}_3$
  • Solvent mass = $255\,\text{g}$ of $\text{H}_2\text{O}$
  • Total solution mass = $45\,\text{g} + 255\,\text{g} = 300\,\text{g}$
  • Calculation:       $\%\,\text{m/m} = \frac{45\,\text{g KNO}_3}{300\,\text{g solución}} \times 100 = 15\%$

Solubility Limit Scaling Calculations

• Exercise 5 (Potassium Nitrate Solubility):

  • Given condition: The solubility of $\text{KNO}_3$ is $32\,\text{g}$ per every $100\,\text{g}$ of water at $20^\circ\text{C}$.
  • Problem: How many grams can dissolve in $250\,\text{g}$ of water?
  • Set-up:     $\frac{32\,\text{g KNO}_3}{100\,\text{g H}_2\text{O}} = \frac{x\,\text{g KNO}_3}{250\,\text{g H}_2\text{O}}$
  • Calculation:     $x = \frac{32\,\text{g KNO}_3 \times 250\,\text{g H}_2\text{O}}{100\,\text{g H}_2\text{O}} = \frac{8000\,\text{g KNO}_3}{100} = 80\,\text{g KNO}_3$

• Exercise 6 (Sodium Chloride Solubility):

  • Given condition: The solubility of sodium chloride ($\text{NaCl}$) is $36\,\text{g}$ per every $100\,\text{g}$ of water.
  • Problem: How much can dissolve in $500\,\text{g}$ of water?
  • Set-up:     $\frac{36\,\text{g NaCl}}{100\,\text{g H}_2\text{O}} = \frac{x\,\text{g NaCl}}{500\,\text{g H}_2\text{O}}$
  • Calculation:     $x = \frac{36\,\text{g NaCl} \times 500\,\text{g H}_2\text{O}}{100\,\text{g H}_2\text{O}} = \frac{18000\,\text{g NaCl}}{100} = 180\,\text{g NaCl}$

Solubility Product Constant ($K_{sp}$) and Precipitation

• Classification of Salt Mixtures in Solvents:

  • Recorded under topic dated $06/08/26$
  • Daily observed salt mixtures fall into three distinct categories:
  • Very soluble (Muy solubles)
  • Slightly soluble / Poorly soluble (Poco solubles)
  • Insoluble (Insolubles)

• Definition and Meaning of $K_{sp}$:

  • The solubility product constant ($K_{sp}$) is an equilibrium constant that indicates the degree of solubility of a slightly soluble salt.
  • Inverse relationship: The smaller the value of $K_{sp}$, the less soluble the substance is.
  • Direct interpretation: A very large $K_{sp}$ value indicates that the salt is more soluble; a small $K_{sp}$ value indicates that the salt is poorly soluble (poco soluble).

• Chemical Equilibrium Equations and Constants:

  • Barium Sulfate Dissolution Equilibrium:$\text{BaSO}_{4(s)} \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$
  • Given specific ion concentration values:       $[\text{Ba}^{2+}] = 3 \times 10^{-10}$$[\text{SO}_4^{2-}] = 5 \times 10^{-9}$
  • Silver Chloride Dissolution Equilibrium:$\text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)}$

• Comparative $K_{sp}$ Values:

  • Silver Chloride ($\text{AgCl}$): $K_{sp} = 1.8 \times 10^{-10}$
  • Barium Sulfate ($\text{BaSO}_4$): $K_{sp} = 1.1 \times 10^{-10}$
  • Calcium Carbonate ($\text{CaCO}_3$): $K_{sp} = 4.8 \times 10^{-9}$

• Precipitate Formation (Formación de Precipitados):

  • When two separate solutions are mixed together, two or three distinct chemical phenomena or outcomes can occur regarding solid precipitation.