Ksp from Solubility Calculator

Enter Calculation

Molar Solubility, s (mol/L)

Cation subscript (n in A_nB_m)

Anion subscript (m in A_nB_m)

Results

Solubility Product (K_sp)

8.787999999999997E-12

for A₁B₂

Cation term (n·s)^n	1.3E-4
Anion term (m·s)^m	6.759999999999998E-8

What Is the Ksp from Solubility Calculator?

This calculator finds the solubility product constant (K_sp) of a sparingly soluble ionic salt from its molar solubility, s. It works for any general salt of the form A_nB_m that dissolves to give n cations and m anions per formula unit. It is a universal chemistry tool and applies anywhere.

How to Use It

Enter the molar solubility s in mol/L (for example 1.3×10⁻⁴ written as 1.3e-4), then enter the cation subscript n and the anion subscript m from the formula A_nB_m. The calculator returns K_sp along with the individual cation and anion terms.

The Formula Explained

When A_nB_m dissolves: A_nB_m → n A^m+ + m Bⁿ⁻. If the molar solubility is s, then [A] = n·s and [B] = m·s. The solubility product is the product of ion concentrations each raised to its stoichiometric coefficient:

$$K_{sp} = \left(\text{n} \cdot \text{s}\right)^{\text{n}} \cdot \left(\text{m} \cdot \text{s}\right)^{\text{m}}$$

Each formula unit yields n cations and m anions, giving ion concentrations n·s and m·s.

A salt AnBm dissolves into n cations and m anions in solution.

Worked Example

For CaF₂ (a 1:2 salt, n=1, m=2) with s = 1.3×10⁻⁴ mol/L: cation term = (1·s)¹ = 1.3×10⁻⁴, anion term = (2·s)² = (2.6×10⁻⁴)² = 6.76×10⁻⁸. So

$$K_{sp} = 1.3\times10^{-4} \times 6.76\times10^{-8} \approx 8.79\times10^{-12}$$

matching the textbook value.

FAQ

What units does Ksp have? K_sp is conventionally treated as dimensionless, though concentrations are in mol/L.

What are n and m for AgCl? AgCl is a 1:1 salt, so n = 1 and m = 1, giving \(K_{sp} = s^2\).

What about Ag₂CO₃? It is a 2:1 salt (n = 2, m = 1): \(K_{sp} = (2s)^2(s) = 4s^3\).