A smaller Ksp usually suggests a less soluble salt. But that comparison is fair only when the salts break into the same number of ions.
Key idea
For salts with the same ion pattern (both 1:1, or both 1:2), the smaller $K_{sp}$ is the less soluble salt. For different patterns, compute each molar solubility $s$ and compare those instead.
Why it works
$K_{sp}$ is $s^2$ for a 1:1 salt but $4s^3$ for a 1:2 salt: different powers of $s$. Since $s$ is a small number, its cube is far smaller than its square. A 1:2 salt can therefore have a smaller $K_{sp}$ than a 1:1 salt and still dissolve more.
| Pattern | Example | $K_{sp}$ in terms of s | s |
|---|---|---|---|
| AB | AgCl | $s^2$ | $\sqrt{K_{sp}}$ |
| $\mathrm{AB_2}$ or $\mathrm{A_2B}$ | $\mathrm{PbI_2}$, $\mathrm{Ag_2CrO_4}$ | $4s^3$ | $\sqrt[3]{K_{sp}/4}$ |
| $\mathrm{AB_3}$ | $\mathrm{Fe(OH)_3}$ | $27s^4$ | $\sqrt[4]{K_{sp}/27}$ |
| $\mathrm{A_3B_2}$ | $\mathrm{Ca_3(PO_4)_2}$ | $108s^5$ | $\sqrt[5]{K_{sp}/108}$ |
Worked examples
Worked example: Smaller Ksp, more soluble
AgCl has $K_{sp} = 1.8 \times 10^{-10}$ and $\mathrm{Ag_2CrO_4}$ has $K_{sp} = 1.1 \times 10^{-12}$. Which is more soluble in water?
- AgCl: $s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}\ \text{M}$.
- $\mathrm{Ag_2CrO_4}$: $K_{sp} = (2s)^2s = 4s^3$, so $s = \sqrt[3]{1.1 \times 10^{-12}/4} = 6.5 \times 10^{-5}\ \text{M}$.
- $\mathrm{Ag_2CrO_4}$ dissolves about five times more, despite a $K_{sp}$ more than 100 times smaller.
Answer: $\mathrm{Ag_2CrO_4}$.
Worked example: Same pattern
Rank AgCl ($1.8 \times 10^{-10}$), AgBr ($5.0 \times 10^{-13}$) and AgI ($8.3 \times 10^{-17}$) from most to least soluble.
- All are 1:1, so the $K_{sp}$ values can be compared directly.
- Most soluble: AgCl, then AgBr, then AgI.
Answer: $\mathrm{AgCl > AgBr > AgI}$.
Worked example: A fluoride and a sulfate
$\mathrm{CaF_2}$ has $K_{sp} = 3.5 \times 10^{-11}$ and $\mathrm{BaSO_4}$ has $K_{sp} = 1.1 \times 10^{-10}$. Which dissolves more?
- $\mathrm{BaSO_4}$ is 1:1: $s = \sqrt{1.1 \times 10^{-10}} = 1.0 \times 10^{-5}\ \text{M}$.
- $\mathrm{CaF_2}$ is 1:2: $s = \sqrt[3]{3.5 \times 10^{-11}/4} = 2.1 \times 10^{-4}\ \text{M}$.
- $\mathrm{CaF_2}$ dissolves about 20 times more, though its $K_{sp}$ is smaller.
Answer: $\mathrm{CaF_2}$.
Common mistake
Ranking salts by $K_{sp}$ when they release different numbers of ions. Check the formulas first; if the patterns differ, convert each $K_{sp}$ to $s$ before comparing.
The safe habit: whenever two formulas differ in how many ions they release, convert both to molar solubility (or to grams per liter, if the question is about mass) before ranking.
The silver halide order is used in photography and in the Mohr titration for chloride, where red $\mathrm{Ag_2CrO_4}$ forms only after the white AgCl has finished precipitating.