In this final section, we look at how solubility equilibria and complex-ion formation can be used to detect the presence of particular metal ions in solution. Before the development of modern analytical instrumentation, it was necessary to analyze mixtures of metals in a sample by what were called wet chemical methods. For example, an ore sample that might contain several metallic elements was dissolved in a concentrated acid solution that was then tested in a systematic way for the presence of various metal ions.

Qualitative analysis determines only the presence or absence of a particular metal ion, whereas quantitative analysis determines how much of a given substance is present. Even though wet methods of qualitative analysis have become less important in the chemical industry, they are frequently used in general chemistry laboratory programs to illustrate equilibria, to teach the properties of common metal ions in solution, and to develop laboratory skills. Typically, such analyses proceed in three stages: (1) The ions are separated into broad groups on the basis of solubility properties. (2) The ions in each group are separated by selectively dissolving members in the group. (3) The ions are identified by means of specific tests.

A scheme in general use divides the common cations into five groups (FIGURE 17.23). The order in which reagents are added is important in this scheme. The most selective separations—those that involve the smallest number of ions—are carried out first. The reactions used must proceed so far toward completion that any concentration of cations remaining in the solution is too small to interfere with subsequent tests.

Let's look at each of these five groups of cations, briefly examining the logic used in this qualitative analysis scheme.

Group 1. Insoluble chlorides: Of the common metal ions, only Ag+, Hg22+, and Pb2+ form insoluble chlorides. When HCl is added to a mixture of cations, therefore, only AgCl, Hg2Cl2, and PbCl2 precipitate, leaving the other cations in solution. The absence of a precipitate indicates that the starting solution contains no Ag+, Hg22+, or Pb2+.


If a solution contained a mixture of Cu2+ and Zn2+ ions, would this separation scheme work? After which step would the first precipitate be observed?

FIGURE 17.23 Qualitative analysis. A flowchart showing a common scheme for identifying cations.

Group 2. Acid-insoluble sulfides: After any insoluble chlorides have been removed, the remaining solution, now acidic, is treated with H2S. Only the most insoluble metal sulfides—CuS, Bi2S3, CdS, PbS, HgS, As2S3, Sb2S3, and SnS2—precipitate. (Note the very small values of Kspfor some of these sulfides in Appendix D.) Those metal ions whose sulfides are somewhat more soluble—for example, ZnS or NiS—remain in solution.

Group 3. Base-insoluble sulfides and hydroxides: After the solution is filtered to remove any acid-insoluble sulfides, it is made slightly basic, and (NH4)2S is added. In basic solutions the concentration of S2– is higher than in acidic solutions. Thus, the ion products for many of the more soluble sulfides are made to exceed their Ksp values and precipitation occurs. The metal ions precipitated at this stage are Al3+, Cr3+, Fe3+, Zn2+, Ni2+, Co2+, and Mn2+ (The Al3+, Fe3+, and Cr3+ ions do not form insoluble sulfides; instead they precipitate as insoluble hydroxides, asFigure 17.23 shows.)

Group 4. Insoluble phosphates: At this point the solution contains only metal ions from groups 1A and 2A of the periodic table. Adding (NH4)2HPO4 to a basic solution precipitates the group 2A elements Mg2+, Ca2+, Sr2+, and Ba2+ because these metals form insoluble phosphates.

Group 5. The alkali metal ions and NH4+: The ions that remain after removing the insoluble phosphates are tested for individually. A flame test can be used to determine the presence of K+, for example, because the flame turns a characteristic violet color if K+ is present.


If a precipitate forms when HCl is added to an aqueous solution, what conclusions can you draw about the contents of the solution?


A sample of 1.25 L of HCl gas at 21 °C and 0.950 atm is bubbled through 0.500 L of 0.150 M NH3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L.


The number of moles of HCl gas is calculated from the ideal-gas law,

The number of moles of NH3 in the solution is given by the product of the volume of the solution and its concentration,

The acid HCl and base NH3 react, transferring a proton from HCl to NH3, producing NH4+ and Cl ions,

To determine the pH of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction,

Thus, the reaction produces a solution containing a mixture of NH3, NH4+, and Cl. The NH3 is a weak base (Kb = 1.8 × 10–5), NH4+ is its conjugate acid, and Cl is neither acidic nor basic. Consequently, the pH depends on [NH3] and [NH4+],

We can calculate the pH using either Kb for NH3 or Ka for NH4+. Using the Kb expression, we have

Hence, pOH = –log(9.4 × 10–6) = 5.03 and pH = 14.00 – pOH = 14.00 – 5.03 = 8.97.