Solubility product, written as , helps students predict how much of a sparingly soluble ionic compound dissolves in water. This cheat sheet focuses on worked example patterns, including writing expressions, finding molar solubility, and deciding whether a precipitate forms. Students need these tools for equilibrium problems where solids, ions, and saturated solutions appear together.
The goal is to make each setup clear before doing the calculation.
The key idea is that a slightly soluble salt reaches equilibrium between its solid form and its dissolved ions. For a salt , the expression is , and the solid is not included. The ion product is compared with to predict precipitation: if , a precipitate forms.
Common ions reduce solubility because they shift the dissolution equilibrium toward the solid.
Key Facts
- For , the solubility product is .
- Pure solids and liquids are not included in expressions because their activities are treated as constant.
- For , if the molar solubility is , then .
- For , if the molar solubility is , then .
- The ion product has the same form as , so for , .
- If , the solution is unsaturated and more solid can dissolve.
- If , the solution is saturated and the system is at solubility equilibrium.
- If , the solution is supersaturated and precipitation is predicted until falls to .
Vocabulary
- Solubility product
- The equilibrium constant for the dissolving of a sparingly soluble ionic solid into its ions.
- Molar solubility
- The number of moles of a solute that dissolve per liter of solution, usually represented by in .
- Saturated solution
- A solution in which dissolved ions are in equilibrium with undissolved solid at a given temperature.
- Ion product
- The value calculated from current ion concentrations using the same exponent pattern as the expression.
- Common ion effect
- The decrease in solubility caused by adding an ion already present in the dissolution equilibrium.
- Precipitate
- An insoluble or slightly soluble solid that forms when ion concentrations exceed the solubility limit.
Common Mistakes to Avoid
- Including the solid in the expression is wrong because only uses dissolved ion concentrations, not the amount of solid present.
- Forgetting coefficients become exponents is wrong because gives , not .
- Using for every ion concentration without stoichiometry is wrong because produces and .
- Comparing and with different expressions is wrong because must be built using the same ion powers as .
- Ignoring a common ion is wrong because an initial concentration such as can greatly reduce the solubility of .
Practice Questions
- 1 Write the expression for .
- 2 The molar solubility of is . Calculate using .
- 3 For , . Find the molar solubility in pure water using .
- 4 A solution contains extra ions before dissolves. Explain why the molar solubility of is lower than in pure water.
Understanding Solubility Product Ksp Worked Examples
Worked problems become easier when the chemical equation is written before any numbers are used. The coefficients in that equation control every later step. A salt that releases one metal ion for every two negative ions does not give equal ion concentrations.
If its molar solubility is called s, the metal ion concentration is s while the negative ion concentration is two times s. Those different amounts must be substituted before powers are applied. Many errors come from squaring or cubing s too early, or from forgetting the coefficient on an ion.
Keep the concentration relationships on a separate line. Then use the given Ksp value to solve for the positive value of s, since a concentration cannot be negative.
Precipitation questions need one extra habit. Concentrations after two solutions are mixed are usually lower than the concentrations printed on the bottles. First find the moles of each relevant ion from concentration times volume.
Next divide each amount by the total mixed volume. Only then calculate the ion product. For example, combining a lead nitrate solution with a potassium iodide solution requires the final lead ion concentration and the final iodide ion concentration.
The iodide concentration has a stronger effect because its concentration is multiplied by itself in the ion product. This is why a small calculation error in that concentration can change the prediction.
The common ion effect is an equilibrium idea, but it is often handled with a practical approximation. Suppose a salt is placed in a solution that already contains one of its ions. The added ion is usually much more concentrated than the small amount produced by dissolving the solid.
In that case, treat its starting concentration as nearly unchanged while solving for the new solubility. Check this assumption afterward. If the calculated added amount is not very small compared with the starting concentration, the full equilibrium calculation is needed.
Acidity can matter too. Some negative ions react with hydrogen ions. Removing those ions from solution can allow more solid to dissolve, so solubility may depend on pH rather than on Ksp alone.
These ideas appear outside textbook exercises. Mineral scale forms in pipes when dissolved ions become concentrated enough to make an insoluble solid. Water treatment plants use precipitation to remove ions from water.
In a laboratory, washing a precipitate with the wrong solution can dissolve part of the product. When learning this topic, label each concentration by where it came from. Distinguish an initial concentration, a concentration after mixing, and an equilibrium concentration.
Include only dissolved ions in the calculation. Finally, check whether the size of the answer makes chemical sense. A very low Ksp should normally lead to a very small molar solubility in pure water, unless another reaction changes the ions present.