Predicting Precipitation: Qsp vs Ksp

To predict whether a precipitate forms when solutions are mixed, calculate the ion product Qsp from the concentrations after mixing and compare it with Ksp: if Qsp > Ksp a precipitate forms, if Qsp = Ksp the solution is just saturated, and if Qsp < Ksp no precipitate forms. In brief: mixing equal volumes of 1.0 × 10−3 M AgNO3 and 1.0 × 10−3 M NaCl gives Qsp = 2.5 × 10−7, which is far above Ksp (1.8 × 10−10) for AgCl, so AgCl precipitates.

The prediction of precipitation is an application of the reaction quotient to the dissolution equilibrium. It is used to design the separation of ions, to avoid scaling in pipes and to understand the formation of minerals. This article explains the rule, the importance of dilution on mixing, how to find the concentration at which a precipitate begins to form and how to separate two ions by fractional precipitation.

What is the rule?

For a salt AB with Ksp = [A+][B−], the ion product is Qsp = [A+][B−], calculated with the actual concentrations in the mixture, whether or not it is at equilibrium. For salts with other formulas, the exponents are the same as in Ksp.

Comparing the ion product with Ksp
Comparison State of the solution Result
Qsp < KspUnsaturatedNo precipitate; more solid would dissolve
Qsp = KspSaturatedEquilibrium; no net change
Qsp > KspSupersaturatedPrecipitate forms until Qsp = Ksp

How do you carry out the prediction?

  1. Write the formula of each possible precipitate and its Ksp expression. Use solubility rules to identify likely candidates.
  2. Calculate the concentration of each ion after mixing, taking into account the dilution: cfinal = cinitial × Vinitial / Vtotal.
  3. Calculate Qsp for each candidate.
  4. Compare with Ksp and decide.

Worked examples

Example 1: a precipitate forms

50.0 mL of 1.0 × 10−3 M AgNO3 is mixed with 50.0 mL of 1.0 × 10−3 M NaCl. Will AgCl precipitate?

  1. The total volume is 100.0 mL, so each concentration is halved: [Ag+] = [Cl−] = 5.0 × 10−4 M.
  2. Qsp = (5.0 × 10−4)(5.0 × 10−4) = 2.5 × 10−7.
  3. Since 2.5 × 10−7 > 1.8 × 10−10, AgCl precipitates.
[Ag+] = [Cl−] = 5.0 × 10−4 M Qsp = (5.0 × 10−4)(5.0 × 10−4) = 2.5 × 10−7 2.5 × 10−7 > 1.8 × 10−10 = Ksp

Example 2: no precipitate

100.0 mL of 2.0 × 10−4 M Pb(NO3)2 is mixed with 100.0 mL of 2.0 × 10−4 M KI. After mixing, [Pb2+] = 1.0 × 10−4 M and [I−] = 1.0 × 10−4 M. Qsp = [Pb2+][I−]2 = (1.0 × 10−4)(1.0 × 10−4)2 = 1.0 × 10−12. This is smaller than Ksp (9.8 × 10−9), so there is no precipitate.

Qsp = (1.0 × 10−4)(1.0 × 10−4)2 = 1.0 × 10−12 1.0 × 10−12 < 9.8 × 10−9 = Ksp

Example 3: the same salt at a higher concentration

If the same volumes of 2.0 × 10−2 M solutions are mixed, then [Pb2+] = [I−] = 1.0 × 10−2 M, and Qsp = (1.0 × 10−2)(1.0 × 10−2)2 = 1.0 × 10−6. This is above Ksp by a factor of about 100, so bright yellow PbI2 precipitates.

Qsp = (1.0 × 10−2)(1.0 × 10−2)2 = 1.0 × 10−6 1.0 × 10−6 > 9.8 × 10−9 = Ksp

Example 4: two possible precipitates

When a solution of Ba(OH)2 is added to a solution of MgSO4, two precipitates can form, BaSO4 (Ksp = 1.1 × 10−10) and Mg(OH)2 (Ksp = 5.6 × 10−12). Both have very small Ksp values, so both precipitate.

How do you find the concentration at which precipitation begins?

Set Qsp = Ksp and solve for the concentration of the added ion.

Example: magnesium hydroxide

At what pH does Mg(OH)2 begin to precipitate from a 0.0050 M solution of Mg2+?

  1. Ksp = [Mg2+][OH−]2 = 5.6 × 10−12.
  2. [OH−]2 = 5.6 × 10−12 / 0.0050 = 1.12 × 10−9.
  3. [OH−] = 3.35 × 10−5 M, so pOH = 4.47 and pH = 9.53.
[OH−]2 = 5.6 × 10−12 / 0.0050 = 1.12 × 10−9 [OH−] = 3.35 × 10−5 M pOH = 4.47 , pH = 14 − 4.47 = 9.53

Above a pH of 9.5, magnesium hydroxide precipitates. This is the basis for the production of magnesium from seawater, in which lime is added.

What is fractional precipitation?

When a reagent is added to a solution containing two ions that form precipitates with different Ksp values, the less soluble precipitate forms first. By stopping at the right time, the two ions can be separated.

Worked example: chloride and chromate with silver ions

A solution contains 0.010 M Cl− and 0.010 M CrO42−. Solid AgNO3 is added slowly (the volume change is ignored). Which precipitate forms first?

  1. For AgCl: [Ag+] = Ksp / [Cl−] = 1.8 × 10−10 / 0.010 = 1.8 × 10−8 M.
  2. For Ag2CrO4: [Ag+] = √(Ksp / [CrO42−]) = √(1.1 × 10−12 / 0.010) = √(1.1 × 10−10) = 1.05 × 10−5 M.
  3. AgCl needs a much lower silver ion concentration, so white AgCl precipitates first.
  4. The red Ag2CrO4 begins to form when [Ag+] reaches 1.05 × 10−5 M. At that point [Cl−] = 1.8 × 10−10 / 1.05 × 10−5 = 1.7 × 10−5 M, which is 0.17 % of the original chloride.
[Ag+] = 1.8 × 10−10 / 0.010 = 1.8 × 10−8 M [Ag+] = √(1.1 × 10−10) = 1.05 × 10−5 M 1.8 × 10−8 < 1.05 × 10−5

So 99.8 % of the chloride has been precipitated before any chromate comes down. This is the principle of the Mohr titration for chloride, in which chromate serves as an indicator: the first permanent red color of silver chromate shows that all the chloride has been used up.

Why do some precipitates form slowly?

The ion product can exceed Ksp without a precipitate appearing immediately. Solid formation needs a nucleus, and in a very clean solution a supersaturated state can persist. In practice, a slight excess of Qsp over Ksp may not give a visible precipitate for minutes or hours, and very small particles may form a colloid rather than a settling solid. The rule therefore tells you whether precipitation is thermodynamically possible, not how fast it takes place.

Where was this used first?

The use of selective precipitation for separating metal ions is the basis of the classical scheme of qualitative analysis, which was developed in the 19th century by Carl Remigius Fresenius and others. Karl Friedrich Mohr introduced the titration of chloride with silver nitrate and chromate as an indicator in 1856. The quantitative explanation in terms of solubility products came after Nernst’s work of 1889.

Which mistakes are common?