Precipitation Prediction Calculator
Enter the cation and anion concentrations, their stoichiometric exponents, and the solubility product Ksp to find out whether a precipitate forms when two solutions are mixed. The calculator compares the ion product Qsp with Ksp and predicts precipitation instantly.
Will a Precipitate Form When You Mix These Ions?
Precipitation Prediction
Where:
- \(Q_{sp}\) — ion product of the salt in the mixture (mol/L raised to the power \(m+n\))
- \([M^{+}]\) — molar concentration of the cation after mixing (mol/L)
- \([A^{n-}]\) — molar concentration of the anion after mixing (mol/L)
- \(m\) — number of cations in one formula unit of the salt
- \(n\) — number of anions in one formula unit of the salt
- \(K_{sp}\) — solubility product of the salt at 25 °C
The ion product has the same mathematical form as the solubility product, but it is built from the concentrations actually present in the mixture. Comparing \(Q_{sp}\) with \(K_{sp}\) tells you whether the solution is unsaturated, exactly saturated, or supersaturated.
Precipitation Prediction – Calculation Example
Example: equal volumes of 0.020 M AgNO₃ and 0.020 M NaCl are mixed, so after dilution each ion is at 0.010 mol/L. Does AgCl precipitate? Use \(m = n = 1\) and \(K_{sp} = 1.8 \times 10^{-5}\) for AgCl.
\[ Q_{sp} = [Ag^{+}]^{1}[Cl^{-}]^{1} = (0.010)(0.010) = 1.0 \times 10^{-4} \]
\[ \frac{Q_{sp}}{K_{sp}} = \frac{1.0 \times 10^{-4}}{1.8 \times 10^{-5}} \approx 5.56 \]
Result: \(Q_{sp} > K_{sp}\), so the solution is supersaturated and AgCl precipitates. Roughly 57.6% of both ions must leave the solution as solid AgCl before equilibrium is restored.
How to Use the Precipitation Prediction Calculator
- Enter the Cation Concentration in mol/L. This must be the concentration after mixing, not the concentration in the original solution.
- Set the Cation Exponent (m) — the number of cations in the salt's formula unit (e.g. 3 for Ca₃(PO₄)₂).
- Enter the Anion Concentration in mol/L, again using the post-mixing value.
- Set the Anion Exponent (n) — the number of anions in the formula unit (e.g. 2 for PbCl₂).
- Enter the Solubility Product (Ksp) of the salt at 25 °C. Scientific notation such as
1.8e-5is accepted. - Read the results: the Ion Product, the Saturation Ratio \(Q_{sp}/K_{sp}\), the Prediction, and the Calculation Details showing the full comparison.
What Is the Ion Product and Why Compare It with Ksp?
The solubility product \(K_{sp}\) is the equilibrium constant for a slightly soluble salt dissolving in water. For a salt \(M_mA_n\), it equals the product of the equilibrium ion concentrations, each raised to its stoichiometric power. At 25 °C, AgCl has \(K_{sp} = 1.8 \times 10^{-5}\), which corresponds to a molar solubility of about \(1.3 \times 10^{-5}\) mol/L in pure water.
When two soluble solutions are mixed, the ions may or may not exceed this equilibrium limit. The ion product \(Q_{sp}\) is computed from the concentrations actually present. It works like a reaction quotient: comparing it with \(K_{sp}\) reveals which way the dissolution equilibrium must shift.
The interpretation follows three simple rules:
- \(Q_{sp} < K_{sp}\) — unsaturated: no precipitate forms, and any added solid would keep dissolving.
- \(Q_{sp} = K_{sp}\) — exactly saturated: the system is at equilibrium, with no net change.
- \(Q_{sp} > K_{sp}\) — supersaturated: a precipitate forms, removing ions from solution until \(Q_{sp}\) falls back to \(K_{sp}\).
The saturation ratio \(Q_{sp}/K_{sp}\) adds a sense of scale. A ratio of 5.56 means the ion product is more than five times the equilibrium value, so precipitation is essentially certain and substantial. A ratio of 1.02 predicts precipitation too, but only a tiny amount of solid will form before equilibrium is reached.
Assumptions, Limitations and Practical Tips
Use post-mixing concentrations. The most common mistake is entering the concentrations of the original solutions. When you mix \(V_1\) and \(V_2\) of two solutions, each concentration is diluted by the total volume: \(C_{\text{mixed}} = C_{\text{initial}} \cdot V_1/(V_1 + V_2)\). In the example above, mixing equal volumes halves each 0.020 M solution to 0.010 M. Our Molarity Calculator can help you convert between concentration units first.
Ksp values are temperature-dependent. Tabulated values are normally quoted at 25 °C. At other temperatures the solubility product changes, and the prediction may shift — most salts become more soluble as temperature rises.
Concentrations approximate activities. In dilute solutions molar concentrations and thermodynamic activities are nearly equal, but in ionic media above roughly 0.1 mol/L the effective \(K_{sp}\) appears larger because of ionic interactions. For such solutions, check the Ionic Strength Calculator to see whether activity corrections matter for your case.
Side reactions are ignored. The calculator assumes both ions remain free in solution. Complex formation (e.g. Ag⁺ binding to NH₃), acid–base reactions of the anion (e.g. CO₃²⁻ partially converting to HCO₃⁻), or redox processes can change the free ion concentrations and alter the outcome. For acid–base effects on weak-acid anions, the pH Calculator is a useful companion.
Kinetics are not predicted. The calculator tells you whether precipitation is thermodynamically favored, not how fast it happens. Some supersaturated solutions (especially with sparingly soluble carbonates and sulfates) can persist for minutes or longer before visible crystals appear.
To explore the reverse question — how much of a salt dissolves before the solution saturates — the same \(K_{sp}\) comparison applies, and the Equilibrium Constant Calculator covers the broader family of equilibrium problems.
Frequently Asked Questions
How do I calculate Qsp by hand when mixing two solutions?
First apply the dilution: each concentration becomes \(C_{\text{initial}} \cdot V/(V_{\text{total}})\). Then raise the cation concentration to the power \(m\) and the anion concentration to the power \(n\), and multiply the results: \(Q_{sp} = [M^{+}]^{m}[A^{n-}]^{n}\). Finally compare this number with the tabulated \(K_{sp}\) — above it means a precipitate, below it means all ions stay dissolved.
Does the ion product Qsp have units?
Formally yes: \(Q_{sp}\) carries units of \(\text{mol}^{m+n}/\text{L}^{m+n}\), so a 1:1 salt like AgCl gives mol²/L² while Ca₃(PO₄)₂ gives mol⁵/L⁵. In practice chemists often treat \(K_{sp}\) and \(Q_{sp}\) as dimensionless, so tabulated values are pure numbers. The key point for your calculation: \(Q_{sp}\) and \(K_{sp}\) must be built the same way, with concentrations in mol/L and the same exponents, so the comparison is valid.
How do I convert Ksp to molar solubility (and back)?
For a salt \(M_mA_n\) dissolving to a molar solubility \(s\), the equilibrium gives \(K_{sp} = (ms)^m(ns)^n\). Solve for \(s\): \( s = \left(\frac{K_{sp}}{m^m \, n^n}\right)^{1/(m+n)} \). For a 1:1 salt this simplifies to \(s = \sqrt{K_{sp}}\). Going the other way, plug \(s\) back into \(K_{sp} = (ms)^m(ns)^n\). Remember to include the stoichiometric multipliers — forgetting the factor of 2 for PbCl₂ is a classic error.
Can I enter concentrations in ppm or mg/L instead of mol/L?
No — convert to mol/L first. Divide the mg/L value by the molar mass of the ion (in g/mol), then divide by 1000 to get mol/L. For example, 35.5 mg/L of Cl⁻ is \(35.5/35.45/1000 \approx 1.0 \times 10^{-3}\) mol/L. Using ppm directly will give a wildly wrong \(Q_{sp}\), because the solubility product is defined strictly in terms of molar concentrations.
What is the most common mistake when predicting precipitation?
Forgetting dilution on mixing. If you combine equal volumes of two solutions, every concentration halves, and this directly changes \(Q_{sp}\). The second most common error is skipping the exponents: for Ca₃(PO₄)₂ you need \([\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2\), not a simple product. A third is using a \(K_{sp}\) value for the wrong salt or the wrong temperature — always verify the salt formula and the 25 °C reference.
What happens if Qsp is only slightly above Ksp?
A precipitate still forms, but only a very small amount. The system removes ions until \(Q_{sp}\) falls back to exactly \(K_{sp}\), so when the saturation ratio is close to 1 (say 1.02), just a tiny fraction of the ions leaves the solution. The precipitate may be barely visible. Conversely, a large ratio like 50 or 100 means substantial precipitation. Kinetics also matter: some supersaturated solutions persist for minutes before the first crystals appear.
How do I predict which of two salts precipitates first?
For each candidate salt, calculate the minimum anion concentration needed to start precipitation: \( [A^{n-}]_{\text{crit}} = \left(\frac{K_{sp}}{[M^{+}]^m}\right)^{1/n} \). The salt with the smaller critical concentration precipitates first as you add the precipitating reagent. This is the basis of selective precipitation, used to separate ions such as separating Ag⁺ from Pb²⁺ by slowly adding chloride, since their \(K_{sp}\) values differ by orders of magnitude.