Will a Precipitate Form? Ionic Product vs Ksp Calculations
"Will a precipitate form when these two solutions are mixed?" is one of the most repeated problem types in ionic equilibrium. The answer never needs guesswork. You compare one number, the ionic product Q, with another, the solubility product Ksp. The same comparison explains selective precipitation, where two ions are separated by adding a reagent slowly. These ideas are standard in IIT-JAM, GATE, CSIR-NET and CUET-PG physical and analytical chemistry.
The rule
For a salt MaXb: Ksp = [Mb+]a[Xa−]b (at equilibrium, saturated solution)
Q (ionic product) = same expression, but with the concentrations actually present right now
Q > Ksp: precipitate forms | Q = Ksp: just saturated | Q < Ksp: no precipitate
Q (ionic product) = same expression, but with the concentrations actually present right now
Q > Ksp: precipitate forms | Q = Ksp: just saturated | Q < Ksp: no precipitate
What to watch
- Use concentrations after mixing. When equal volumes are mixed, each concentration halves. This is the single biggest source of wrong answers.
- Powers come from the formula. For Ag2CrO4, Q = [Ag+]2[CrO42−].
- Ksp depends on temperature. The values below are the numbers a problem would hand you, taken at about 25 °C. In an exam use the Ksp printed in your question or data sheet.
- Ignore the activity effect unless the question mentions ionic strength. Concentrations stand in for activities.
Worked examples
Example 1: equal volumes, precipitate expected. 100 mL of 2.0 × 10−4 M AgNO3 is mixed with 100 mL of 2.0 × 10−4 M NaCl. Ksp(AgCl) = 1.8 × 10−10. Will AgCl precipitate?
Step 1: Total volume is 200 mL, so each concentration is halved: [Ag+] = [Cl−] = 1.0 × 10−4 M.
Step 2: Q = (1.0 × 10−4)(1.0 × 10−4) = 1.0 × 10−8.
Step 3: Compare: 1.0 × 10−8 > 1.8 × 10−10, a factor of about 56.
Answer: Yes, AgCl precipitates.
Step 1: Total volume is 200 mL, so each concentration is halved: [Ag+] = [Cl−] = 1.0 × 10−4 M.
Step 2: Q = (1.0 × 10−4)(1.0 × 10−4) = 1.0 × 10−8.
Step 3: Compare: 1.0 × 10−8 > 1.8 × 10−10, a factor of about 56.
Answer: Yes, AgCl precipitates.
Example 2: no precipitate. 50 mL of 1.0 × 10−5 M AgNO3 is mixed with 50 mL of 1.0 × 10−5 M NaCl.
Step 1: After mixing, both are 5.0 × 10−6 M.
Step 2: Q = (5.0 × 10−6)2 = 2.5 × 10−11.
Step 3: 2.5 × 10−11 < 1.8 × 10−10.
Answer: No precipitate. The solution is unsaturated.
Step 1: After mixing, both are 5.0 × 10−6 M.
Step 2: Q = (5.0 × 10−6)2 = 2.5 × 10−11.
Step 3: 2.5 × 10−11 < 1.8 × 10−10.
Answer: No precipitate. The solution is unsaturated.
Example 3: selective precipitation. A solution contains 0.010 M Cl− and 0.010 M CrO42−. Solid AgNO3 is added slowly (no volume change). Ksp(AgCl) = 1.8 × 10−10; Ksp(Ag2CrO4) = 1.1 × 10−12. Which precipitates first, and how much Cl− is left when the second begins?
Step 1: For AgCl to start: [Ag+] = Ksp / [Cl−] = 1.8 × 10−10 / 0.010 = 1.8 × 10−8 M.
Step 2: For Ag2CrO4 to start: [Ag+] = √(Ksp / [CrO42−]) = √(1.1 × 10−12 / 0.010) = √(1.1 × 10−10) = 1.05 × 10−5 M.
Step 3: AgCl needs far less Ag+, so AgCl precipitates first.
Step 4: When [Ag+] reaches 1.05 × 10−5 M and chromate starts to precipitate, [Cl−] = 1.8 × 10−10 / 1.05 × 10−5 = 1.7 × 10−5 M.
Step 5: Fraction left = 1.7 × 10−5 / 0.010 = 0.17%.
Answer: AgCl first. About 99.8% of the chloride is removed before Ag2CrO4 begins. This is the idea behind the Mohr titration, where chromate acts as the indicator.
Note: compare the Ag+ needed, not the Ksp values. The salt with the smaller Ksp is not always first, because the formulas differ (Ag2CrO4 has the smaller Ksp here, yet precipitates second).
Step 1: For AgCl to start: [Ag+] = Ksp / [Cl−] = 1.8 × 10−10 / 0.010 = 1.8 × 10−8 M.
Step 2: For Ag2CrO4 to start: [Ag+] = √(Ksp / [CrO42−]) = √(1.1 × 10−12 / 0.010) = √(1.1 × 10−10) = 1.05 × 10−5 M.
Step 3: AgCl needs far less Ag+, so AgCl precipitates first.
Step 4: When [Ag+] reaches 1.05 × 10−5 M and chromate starts to precipitate, [Cl−] = 1.8 × 10−10 / 1.05 × 10−5 = 1.7 × 10−5 M.
Step 5: Fraction left = 1.7 × 10−5 / 0.010 = 0.17%.
Answer: AgCl first. About 99.8% of the chloride is removed before Ag2CrO4 begins. This is the idea behind the Mohr titration, where chromate acts as the indicator.
Note: compare the Ag+ needed, not the Ksp values. The salt with the smaller Ksp is not always first, because the formulas differ (Ag2CrO4 has the smaller Ksp here, yet precipitates second).
Example 4: pH needed to start precipitating a hydroxide. A solution is 0.010 M in Mg2+. Ksp[Mg(OH)2] = 5.6 × 10−12. At what pH does Mg(OH)2 begin to precipitate (25 °C)?
Step 1: Ksp = [Mg2+][OH−]2, so [OH−]2 = 5.6 × 10−12 / 0.010 = 5.6 × 10−10.
Step 2: [OH−] = 2.37 × 10−5 M.
Step 3: pOH = −log(2.37 × 10−5) = 4.63. pH = 14 − 4.63 = 9.37.
Answer: Precipitation starts at pH ≈ 9.4. Below this pH, the solution stays clear. This is why a buffer of NH4Cl and NH3 can keep Mg2+ in solution while a more insoluble hydroxide is precipitated.
Step 1: Ksp = [Mg2+][OH−]2, so [OH−]2 = 5.6 × 10−12 / 0.010 = 5.6 × 10−10.
Step 2: [OH−] = 2.37 × 10−5 M.
Step 3: pOH = −log(2.37 × 10−5) = 4.63. pH = 14 − 4.63 = 9.37.
Answer: Precipitation starts at pH ≈ 9.4. Below this pH, the solution stays clear. This is why a buffer of NH4Cl and NH3 can keep Mg2+ in solution while a more insoluble hydroxide is precipitated.
Decision table
| Question wording | What to compute |
|---|---|
| Will a precipitate form on mixing? | Concentrations after mixing, then Q against Ksp |
| Minimum concentration to start precipitation | Set Q = Ksp and solve for the unknown |
| Which ion precipitates first? | Required precipitant concentration for each, smaller one first |
| Is separation complete? | Concentration left of the first ion when the second begins |
| Hydroxide or sulfide precipitation | Find [OH−] or [S2−], then convert to pH |
- Forgetting dilution on mixing. Using 2.0 × 10−4 M directly in Example 1 gives Q = 4.0 × 10−8, which is wrong even though the conclusion happens to be the same.
- Judging solubility from Ksp alone. Comparing Ksp values to decide which salt is more soluble works only for salts with the same formula type. Calculate the real solubilities for mixed types.
- Dropping the exponent. For Mg(OH)2 the OH− term is squared.
- Treating Q = Ksp as precipitate formed. Exactly at Q = Ksp the solution is just saturated, with no solid.
- Ignoring the common ion. A solution already containing one of the ions needs less of the other to precipitate.
Exam patterns and permitted data differ between JAM, GATE, NET and CUET-PG. Check the latest official notification and syllabus for your exam.
Use the Ksp tool to check solubility and the ionic-product comparison after you have worked the problem by hand.
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