Solubility Product Ksp Calculator

Saturated Solutions & Precipitation Equilibrium Solver

Input Parameters

mol/L
Salt Dissolution Model Saturated
Solubility Product (Ksp)
1.7956e-10
Solution is saturated at this concentration.
Mathematical Working & Substitutions
Ksp = x^x × y^y × S^(x+y)

Understanding Solubility Product (Ksp)

What is the Solubility Product?

The Solubility Product Constant (Ksp) is a specialized chemical equilibrium constant that applies to sparingly soluble ionic compounds in water. It represents the degree to which an ionic solid dissolves into its constituent cations and anions in a saturated aqueous solution.

The Mathematical Expression

For a sparingly soluble salt that dissolves according to the equilibrium equation:

MxXy (s) ↔ x My+ (aq) + y Xx- (aq)

The solubility product expression is defined as:

Ksp = [My+]x × [Xx-]y

Where pure solids are excluded from the equilibrium expression, meaning the solid salt reactant is treated with an activity of 1.

Relating Solubility (S) to Ksp

If we define the molar solubility of the salt as S (in mol/L), then at equilibrium, the concentrations of the dissolved ions are:

Substituting these concentrations back into the equilibrium expression gives us the direct relationship:

Ksp = (x × S)x × (y × S)y = xx × yy × Sx+y

This allows us to convert between solubility product and molar solubility for any stoichiometry:

Precipitation Predictor & Reaction Quotient (Qsp)

The Ionic Product (Qsp) is calculated using the exact same formula as Ksp, but using the instantaneous concentrations of the mixed ions rather than their equilibrium values. By comparing Qsp with Ksp, we can predict whether a precipitate will form:

  1. Qsp < Ksp (Unsaturated): The solution can hold more ions. No precipitate forms, and any added solid will dissolve.
  2. Qsp = Ksp (Saturated): The solution is in dynamic equilibrium. The rates of dissolution and crystallization are equal.
  3. Qsp > Ksp (Supersaturated): The ion concentration exceeds the solubility limit. Precipitation occurs, and solid salt will settle out until Qsp returns to equal Ksp.

Solved Examples

Example 1: Calculating Ksp from Solubility

Problem: The molar solubility of Calcium Fluoride (CaF2) in water at 25°C is 2.0 × 10−4 mol/L. Calculate its solubility product constant Ksp.

Solution:

  1. Write the dissociation equation: CaF2(s) ↔ Ca2+(aq) + 2F−(aq). Here x = 1, y = 2.
  2. Write the relation: Ksp = xx × yy × Sx+y = 11 × 22 × S3 = 4S3.
  3. Substitute S = 2.0 × 10−4 mol/L:
  4. Ksp = 4 × (2.0 × 10−4)3 = 4 × (8.0 × 10−12) = 3.2 × 10−11.
Example 2: Calculating Solubility from Ksp

Problem: The solubility product constant Ksp of Silver Chloride (AgCl) is 1.8 × 10−10. Determine its molar solubility and mass solubility (Molar Mass of AgCl = 143.32 g/mol).

Solution:

  1. Dissociation: AgCl(s) ↔ Ag+(aq) + Cl−(aq). Stoichiometry is 1:1, so x = 1, y = 1.
  2. Relation: Ksp = S2 → S = √Ksp.
  3. Calculate molar solubility: S = √(1.8 × 10−10) = 1.34 × 10−5 mol/L.
  4. Calculate mass solubility: Smass = S × Molar Mass = 1.34 × 10−5 mol/L × 143.32 g/mol = 1.92 × 10−3 g/L.
Example 3: Predicting Precipitation on Mixing

Problem: If 50 mL of 2.0 × 10−5 M AgNO3 solution is mixed with 50 mL of 4.0 × 10−5 M NaCl solution, will silver chloride precipitate? (Ksp of AgCl = 1.8 × 10−10).

Solution:

  1. Mixing doubles the volume, so each ion concentration is halved:
  2. [Ag+] = (2.0 × 10−5 M) / 2 = 1.0 × 10−5 M.
  3. [Cl−] = (4.0 × 10−5 M) / 2 = 2.0 × 10−5 M.
  4. Calculate Ionic Product (Qsp): Qsp = [Ag+][Cl−] = (1.0 × 10−5) × (2.0 × 10−5) = 2.0 × 10−10.
  5. Compare Qsp and Ksp: Qsp (2.0 × 10−10) > Ksp (1.8 × 10−10).
  6. Since Qsp > Ksp, the solution is supersaturated, and a precipitate will form.

Frequently Asked Questions

What is the difference between solubility and solubility product?
Solubility (S) is the maximum quantity of solute that can dissolve in a given volume of solvent to form a saturated solution at a specific temperature. It is usually expressed in mol/L or g/L. The Solubility Product (Ksp) is the equilibrium constant for the dissolution process and is temperature-dependent, representing the product of dissolved ion concentrations raised to their stoichiometric powers.
How does temperature affect Ksp?
Like all chemical equilibrium constants, Ksp is highly dependent on temperature. For most salts, the dissolution process is endothermic (absorbs heat). Consequently, increasing the temperature shifts the equilibrium in the forward direction, which increases the solubility of the salt and raises the value of Ksp. For exothermic dissolution reactions, Ksp decreases as temperature increases.
What is the common ion effect and how does it affect solubility?
The Common Ion Effect occurs when an ionic compound is dissolved in a solution that already contains one of its constituent ions (a common ion). According to Le Chatelier's Principle, the presence of the common ion shifts the dissolution equilibrium backward (towards the solid reactant), thereby significantly decreasing the solubility of the sparingly soluble salt.
Why do pure solids and liquids not appear in the Ksp expression?
In thermodynamic equilibrium, we use chemical activities rather than concentrations. The activity of a pure solid or liquid is constant and defined as exactly 1.0. Because the concentration of solid reactants remains unchanged throughout the dissolution process, they do not impact the equilibrium constant ratio and are excluded from the final Ksp expression.