Now recall that solubility is a chemical property that deals with the ability of a solute to become dissolved in a solvent. Now connected to this idea of solubility is a new term, our solubility product constant, which uses the variable \( K_{sp} \). This is the equilibrium constant that deals with the solubility of ionic solids. So if we're looking at the solubility of ionic solids, we're dealing with \( K_{sp} \). We're going to say here it deals with their solubility, and solubility can also be referred to as concentration or molarity. Remember molarity uses capital M. The basic idea is the higher your \( K_{sp} \) value, then the more soluble your ionic solid. And the smaller your \( K_{sp} \) value, then the less soluble your ionic solid; the less of it will dissolve within a solvent. So just keep this in mind when looking at \( K_{sp} \).
- 1. Matter and Measurements4h 29m
- What is Chemistry?5m
- The Scientific Method9m
- Classification of Matter16m
- States of Matter8m
- Physical & Chemical Changes19m
- Chemical Properties8m
- Physical Properties5m
- Intensive vs. Extensive Properties13m
- Temperature (Simplified)9m
- Scientific Notation13m
- SI Units (Simplified)5m
- Metric Prefixes24m
- Significant Figures (Simplified)11m
- Significant Figures: Precision in Measurements7m
- Significant Figures: In Calculations19m
- Conversion Factors (Simplified)15m
- Dimensional Analysis22m
- Density12m
- Specific Gravity9m
- Density of Geometric Objects19m
- Density of Non-Geometric Objects9m
- 2. Atoms and the Periodic Table5h 23m
- The Atom (Simplified)9m
- Subatomic Particles (Simplified)12m
- Isotopes17m
- Ions (Simplified)22m
- Atomic Mass (Simplified)17m
- Atomic Mass (Conceptual)12m
- Periodic Table: Element Symbols6m
- Periodic Table: Classifications11m
- Periodic Table: Group Names8m
- Periodic Table: Representative Elements & Transition Metals7m
- Periodic Table: Elemental Forms (Simplified)6m
- Periodic Table: Phases (Simplified)8m
- Law of Definite Proportions9m
- Atomic Theory9m
- Rutherford Gold Foil Experiment9m
- Wavelength and Frequency (Simplified)5m
- Electromagnetic Spectrum (Simplified)11m
- Bohr Model (Simplified)9m
- Emission Spectrum (Simplified)3m
- Electronic Structure4m
- Electronic Structure: Shells5m
- Electronic Structure: Subshells4m
- Electronic Structure: Orbitals11m
- Electronic Structure: Electron Spin3m
- Electronic Structure: Number of Electrons4m
- The Electron Configuration (Simplified)22m
- Electron Arrangements5m
- The Electron Configuration: Condensed4m
- The Electron Configuration: Exceptions (Simplified)12m
- Ions and the Octet Rule9m
- Ions and the Octet Rule (Simplified)8m
- Valence Electrons of Elements (Simplified)5m
- Lewis Dot Symbols (Simplified)7m
- Periodic Trend: Metallic Character4m
- Periodic Trend: Atomic Radius (Simplified)7m
- 3. Ionic Compounds2h 18m
- Periodic Table: Main Group Element Charges12m
- Periodic Table: Transition Metal Charges6m
- Periodic Trend: Ionic Radius (Simplified)5m
- Periodic Trend: Ranking Ionic Radii8m
- Periodic Trend: Ionization Energy (Simplified)9m
- Periodic Trend: Electron Affinity (Simplified)8m
- Ionic Bonding6m
- Naming Monoatomic Cations6m
- Naming Monoatomic Anions5m
- Polyatomic Ions25m
- Naming Ionic Compounds11m
- Writing Formula Units of Ionic Compounds7m
- Naming Ionic Hydrates6m
- Naming Acids18m
- 4. Molecular Compounds2h 18m
- Covalent Bonds6m
- Naming Binary Molecular Compounds6m
- Molecular Models4m
- Bonding Preferences6m
- Lewis Dot Structures: Neutral Compounds (Simplified)8m
- Multiple Bonds4m
- Multiple Bonds (Simplified)6m
- Lewis Dot Structures: Multiple Bonds10m
- Lewis Dot Structures: Ions (Simplified)8m
- Lewis Dot Structures: Exceptions (Simplified)12m
- Resonance Structures (Simplified)5m
- Valence Shell Electron Pair Repulsion Theory (Simplified)4m
- Electron Geometry (Simplified)8m
- Molecular Geometry (Simplified)11m
- Bond Angles (Simplified)11m
- Dipole Moment (Simplified)15m
- Molecular Polarity (Simplified)7m
- 5. Classification & Balancing of Chemical Reactions3h 17m
- Chemical Reaction: Chemical Change5m
- Law of Conservation of Mass5m
- Balancing Chemical Equations (Simplified)13m
- Solubility Rules16m
- Molecular Equations18m
- Types of Chemical Reactions12m
- Complete Ionic Equations18m
- Calculate Oxidation Numbers15m
- Redox Reactions17m
- Spontaneous Redox Reactions8m
- Balancing Redox Reactions: Acidic Solutions17m
- Balancing Redox Reactions: Basic Solutions17m
- Balancing Redox Reactions (Simplified)13m
- Galvanic Cell (Simplified)16m
- 6. Chemical Reactions & Quantities2h 35m
- 7. Energy, Rate and Equilibrium3h 46m
- Nature of Energy6m
- First Law of Thermodynamics7m
- Endothermic & Exothermic Reactions7m
- Bond Energy14m
- Thermochemical Equations12m
- Heat Capacity19m
- Thermal Equilibrium (Simplified)8m
- Hess's Law23m
- Rate of Reaction11m
- Energy Diagrams12m
- Chemical Equilibrium7m
- The Equilibrium Constant14m
- Le Chatelier's Principle23m
- Solubility Product Constant (Ksp)17m
- Spontaneous Reaction10m
- Entropy (Simplified)9m
- Gibbs Free Energy (Simplified)18m
- 8. Gases, Liquids and Solids3h 25m
- Pressure Units6m
- Kinetic Molecular Theory14m
- The Ideal Gas Law18m
- The Ideal Gas Law Derivations13m
- The Ideal Gas Law Applications6m
- Chemistry Gas Laws16m
- Chemistry Gas Laws: Combined Gas Law12m
- Standard Temperature and Pressure14m
- Dalton's Law: Partial Pressure (Simplified)13m
- Gas Stoichiometry18m
- Intermolecular Forces (Simplified)19m
- Intermolecular Forces and Physical Properties11m
- Atomic, Ionic and Molecular Solids10m
- Heating and Cooling Curves30m
- 9. Solutions4h 10m
- Solutions6m
- Solubility and Intermolecular Forces18m
- Solutions: Mass Percent6m
- Percent Concentrations10m
- Molarity18m
- Osmolarity15m
- Parts per Million (ppm)13m
- Solubility: Temperature Effect8m
- Intro to Henry's Law4m
- Henry's Law Calculations12m
- Dilutions12m
- Solution Stoichiometry14m
- Electrolytes (Simplified)13m
- Equivalents11m
- Molality15m
- The Colligative Properties15m
- Boiling Point Elevation16m
- Freezing Point Depression9m
- Osmosis16m
- Osmotic Pressure9m
- 10. Acids and Bases3h 29m
- Acid-Base Introduction11m
- Arrhenius Acid and Base6m
- Bronsted Lowry Acid and Base18m
- Acid and Base Strength17m
- Ka and Kb12m
- The pH Scale19m
- Auto-Ionization9m
- pH of Strong Acids and Bases9m
- Acid-Base Equivalents14m
- Acid-Base Reactions7m
- Gas Evolution Equations (Simplified)6m
- Ionic Salts (Simplified)23m
- Buffers25m
- Henderson-Hasselbalch Equation16m
- Strong Acid Strong Base Titrations (Simplified)10m
- 11. Nuclear Chemistry56m
- BONUS: Lab Techniques and Procedures1h 38m
- BONUS: Mathematical Operations and Functions47m
- 12. Introduction to Organic Chemistry1h 34m
- 13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
- 14. Compounds with Oxygen or Sulfur1h 6m
- 15. Aldehydes and Ketones1h 1m
- 16. Carboxylic Acids and Their Derivatives1h 11m
- 17. Amines38m
- 18. Amino Acids and Proteins1h 51m
- 19. Enzymes1h 37m
- 20. Carbohydrates1h 46m
- Intro to Carbohydrates4m
- Classification of Carbohydrates4m
- Fischer Projections4m
- Enantiomers vs Diastereomers8m
- D vs L Enantiomers8m
- Cyclic Hemiacetals8m
- Intro to Haworth Projections4m
- Cyclic Structures of Monosaccharides11m
- Mutarotation4m
- Reduction of Monosaccharides10m
- Oxidation of Monosaccharides7m
- Glycosidic Linkage14m
- Disaccharides7m
- Polysaccharides7m
- 21. The Generation of Biochemical Energy2h 8m
- 22. Carbohydrate Metabolism2h 22m
- 23. Lipids2h 26m
- Intro to Lipids6m
- Fatty Acids25m
- Physical Properties of Fatty Acids6m
- Waxes4m
- Triacylglycerols12m
- Triacylglycerol Reactions: Hydrogenation8m
- Triacylglycerol Reactions: Hydrolysis13m
- Triacylglycerol Reactions: Oxidation7m
- Glycerophospholipids15m
- Sphingomyelins13m
- Steroids15m
- Cell Membranes7m
- Membrane Transport10m
- 24. Lipid Metabolism1h 45m
- 25. Protein and Amino Acid Metabolism1h 37m
- 26. Nucleic Acids and Protein Synthesis2h 54m
- Intro to Nucleic Acids4m
- Nitrogenous Bases16m
- Nucleoside and Nucleotide Formation9m
- Naming Nucleosides and Nucleotides13m
- Phosphodiester Bond Formation7m
- Primary Structure of Nucleic Acids11m
- Base Pairing10m
- DNA Double Helix6m
- Intro to DNA Replication20m
- Steps of DNA Replication11m
- Types of RNA10m
- Overview of Protein Synthesis4m
- Transcription: mRNA Synthesis9m
- Processing of pre-mRNA5m
- The Genetic Code6m
- Introduction to Translation7m
- Translation: Protein Synthesis18m
Solubility Product Constant (Ksp) - Online Tutor, Practice Problems & Exam Prep
Solubility, a key chemical property, indicates how well a solute dissolves in a solvent, represented by the solubility product constant (Ksp). A higher Ksp value signifies greater solubility of ionic solids. The dissolution process involves breaking ionic solids into their respective ions, leading to an equilibrium expression that excludes solids. Ksp can be calculated from known solubility values, while solubility can also be derived from known Ksp values, emphasizing the relationship between concentration and ionic dissociation.
Solubility Product Constant (Ksp) is associated with any ionic compound, which measures how soluble the compound will be in a solvent.
Ksp
Solubility Product Constant (Ksp) Concept 1
Video transcript
Solubility Product Constant (Ksp) Example 1
Video transcript
Here it asks, which substance is the most soluble? So here we have silver chloride, magnesium carbonate, calcium sulfate, and copper(II) sulfide, each of them with their given Ksp value. Remember, we want the most soluble of all of these choices. Remember, we're going to say the greater your Ksp value, then the more soluble you are as an ionic solid. So if we look here, we'd say that the answer has to be option C, since it's 7.1×10-5. It has the smallest negative, so it's the largest value there. It would be the most soluble out of all four choices. So here, option C would be the correct answer.
Solubility Product Constant (Ksp) Concept 2
Video transcript
Now, placing an ionic solid within a solvent involves two competing processes. We have the dissolution reaction, so basically where it's being broken up, and we also have its reverse. Okay? So dissolution is looking at the forward direction of the reaction where ionic solid dissolves into its ions. Here we have a b3a2 solid. So if we were to break this up into its ions, we'd say we have two a's. Remember this 3 didn’t come from b, it came from a. So, this is a 3+ ions in solution are aqueous, plus three b, the 2 that a possesses didn’t come from a, came from b, 2 minus aqueous. And remember, we know that the first ion is positive and the second one is negative because that’s the order in which we write ionic compounds. Now we're going to say that from this equilibrium equation, its equilibrium expression can be determined. The equilibrium expression is just the ratio of concentrations of products over reactants. And we're going to say that recall the equilibrium expression ignores solids and liquids for its ratios. Right? So, it's important that we know how to break up our ionic solids into their respective ions in order to successfully write your equilibrium expression later on. So keep this in mind in terms of the parameters in which it operates. It's products over reactants and it ignores solids and liquids.
Solubility Product Constant (Ksp) Example 2
Video transcript
Here we need to provide the equilibrium expression for calcium nitrate. So step 1 says we need to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Remember, in solution, our ions are in the aqueous form. Alright. So we have calcium nitrate solid. It breaks up into calcium, which is in group 2A, so it's \( Ca^{2+} \) (that's where this 2 came from). Not only is that where the 2 came from, but that also means that we have 2 nitrate ions, \( 2\,NO_3^{-} \), remember the charge of nitrate ion, aqueous. Now that we've done this, we're going to say step 2, using \( K_{sp} \), write the equilibrium expression based on the equilibrium equation. Since the reactant is a solid, set it equal to 1 within the equilibrium expression. Remember, it's products over reactants. So we'd have calcium in brackets, so \( [Ca^{2+}] \times [NO_3^-]^2 \) (and remember this 2 here, whatever the coefficient is, becomes the power). As the reactant is a solid, we just ignore it and replace it with 1. But realize here that we have this expression over 1, which just translates into those concentrations times each other. And nitrates would be squared. Okay. So that would just be our equilibrium expression for this calcium nitrate solid. So these are the steps you need to take in breaking up your ionic solid into its respective ions, and then correctly giving the equilibrium expression for it.
Solubility Product Constant (Ksp) Concept 3
Video transcript
Here we can say if the solubility or molar concentrations of ions within an ionic solid are known, the Ksp can be calculated. If we take a look here, we calculate the Ksp value for silver phosphate, which is Ag₃PO₄, which has a solubility of 1.8×10-18 at 25 degrees Celsius. Here we write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Alright. So here we have our silver phosphate solid. It is going to break up into its ions. It is going to break up into 3 silver ions, remember ions are aqueous in solution, plus our phosphate ion aqueous. Next, we write the equilibrium expression based on the equilibrium equation. So remember, the equilibrium expression is Ksp = products over reactants. But your reactant is a solid, so we are going to ignore it. So it is just going to be Ag+, remember the coefficient here is going to become the power, so cubed times PO43-. Now some new stuff. Step 3, we are going to make concentrations of the ions equal to their coefficients multiplied by the x variable. Alright. So what do I mean by that? We are going to say this is equal to the coefficient in front of the silver ion in our equation as a 3, so this equals 3x and it is going to be cubed. And there is a coefficient of 1 here that is invisible, so that would be times 1x or just x.
Next, we are going to substitute in the given solubility value for the x variable and solve for Ksp. Alright. So before we do that, we have to work out this algebraic expression here. So 3x cubed means that 3 is going to be cubed and x is going to be cubed. 3 times 3 times 3 equals 27 x cubed times x. So 27 x cubed times x is 27 x to the 4th. So that is what my Ksp is equal to. So now we are going to do what step 4 said. It is that substitute in the given solubility value, we are told the solubility is this 1.8×10-18. So Ksp equals 27 times 1.8×10-18 to the 4th. Order of operations. We do what is in the parentheses first. So 1.8×10-18 to the 4th. When you punch that into your calculator, it is going to give you back 1.04976×10-71. Then that multi gets multiplied by 27. When we do that, we get as our answer 2.8×10-70 as the Ksp for our silver phosphate here. Here our answer has 2 significant figures because the solubility given to us in the beginning also has 2 significant figures.
Solubility Product Constant (Ksp) Concept 4
Video transcript
Conversely, the solubility of an ionic solid can be determined when its Ksp value is already known. So here it says the Ksp value for strontium fluoride, SrF2, is 7.9 × 10 - 10 at 25 ° C . Calculate its solubility in molarity. Right. So step 1, we're going to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. The part doesn't change. We have strontium fluoride solid which breaks up into strontium ion plus two fluoride ions. Next, we write the equilibrium expression based on the equilibrium equation. So that's Ksp equals our strontium ion times our fluoride ion. And remember the coefficient becomes the power. Next, we solve for the solubility variable x based on the given Ksp value. Alright. So remember, to do that we're going to say, we have our Ksp here, which is 7.9 × 10 - 10 , and that equals x , which stands in for the strontium ion, times, remember the coefficient is going to play a part here, it's going to be 2 x , and it's still squared. So we're going to have 7.9 × 10 - 10 equals x times, 2 squared is 2 times 2, so that's 4, x 2 . We're solving this like a math problem. So it's going to be 7.9 × 10 - 10 equals 4 x 3 . We're going to divide both sides now by 4 in order to isolate x 3 , so when we do that we're going to get here x 3 equals 1.975 × 10 - 10 . Here we need to take the cube root of both sides in order to isolate my x . So we're going to say here, when we take the cube root of both sides, x equals at the end 5.8 × 10 - 4 molar as my final concentration for this strontium fluoride solid.
Determine the equilibrium expression of the barium nitride solid.
Manganese (V) hydroxide has a measured solubility of 3.4×10–5 M at 25ºC. Calculate its Ksp value.
The Ksp value for strontium fluoride, SrF2, is 7.9×10–10 at 25ºC. Calculate its solubility in g/L.
Do you want more practice?
Here’s what students ask on this topic:
What is the solubility product constant (Ksp) and how is it used?
The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of ionic solids in a solvent. It represents the product of the molar concentrations of the constituent ions, each raised to the power of their coefficients in the balanced equation. Ksp is used to predict the solubility of a compound; a higher Ksp value indicates greater solubility, while a lower Ksp value indicates lesser solubility. It helps in calculating the solubility of a compound from its Ksp value and vice versa.
How do you calculate Ksp from solubility?
To calculate Ksp from solubility, follow these steps: 1) Write the balanced dissolution equation for the ionic solid. 2) Express the molar concentrations of the ions in terms of the solubility (s). 3) Write the Ksp expression using these concentrations. 4) Substitute the solubility value into the Ksp expression and solve. For example, for Ag₃PO₄ with solubility s, the equation is Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq). The Ksp expression is Ksp = [Ag⁺]³[PO₄³⁻] = (3s)³(s) = 27s⁴.
How do you determine the solubility of an ionic solid from its Ksp value?
To determine the solubility of an ionic solid from its Ksp value, follow these steps: 1) Write the balanced dissolution equation. 2) Write the Ksp expression. 3) Let the solubility be x and express the ion concentrations in terms of x. 4) Substitute the Ksp value and solve for x. For example, for SrF₂ with Ksp = 7.9 × 10⁻¹⁰, the equation is SrF₂(s) ⇌ Sr²⁺(aq) + 2F⁻(aq). The Ksp expression is Ksp = [Sr²⁺][F⁻]² = x(2x)² = 4x³. Solving 7.9 × 10⁻¹⁰ = 4x³ gives x = 5.8 × 10⁻⁴ M.
What is the relationship between Ksp and the solubility of ionic compounds?
The relationship between Ksp and the solubility of ionic compounds is direct: a higher Ksp value indicates a more soluble compound, while a lower Ksp value indicates a less soluble compound. Ksp quantifies the extent to which an ionic solid dissolves in a solvent to form its constituent ions. The solubility can be calculated from Ksp by setting up the equilibrium expression and solving for the ion concentrations. Conversely, Ksp can be determined from the known solubility of the compound.
How do you write the equilibrium expression for a dissolution reaction?
To write the equilibrium expression for a dissolution reaction, follow these steps: 1) Write the balanced chemical equation for the dissolution of the ionic solid. 2) Identify the ions produced and their stoichiometric coefficients. 3) Write the expression as the product of the molar concentrations of the ions, each raised to the power of their coefficients. For example, for the dissolution of CaF₂: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), the equilibrium expression is Ksp = [Ca²⁺][F⁻]².