Precipitate-forming ions

The solubility of a precipitate can be improved by adding a ligand capable of forming a soluble complex with one of the precipitate s ions. For example, the solubility of Agl increases in the presence of NH3 due to the formation of the soluble Ag(NH3)2°" complex. As a final illustration of the systematic approach to solving equilibrium problems, let us find the solubility of Agl in 0.10 M NH3. 

Another important parameter that may affect a precipitate s solubility is the pH of the solution in which the precipitate forms. For example, hydroxide precipitates, such as Fe(OH)3, are more soluble at lower pH levels at which the concentration of OH is small. The effect of pH on solubility is not limited to hydroxide precipitates, but also affects precipitates containing basic or acidic ions. The solubility of Ca3(P04)2 is pH-dependent because phosphate is a weak base. The following four reactions, therefore, govern the solubility of Ca3(P04)2. 

Silver Thiosulfate. Silver thiosulfate [23149-52-2], Ag 2 y is an insoluble precipitate formed when a soluble thiosulfate reacts with an excess of silver nitrate. In order to minimize the formation of silver sulfide, the silver ion can be complexed by haUdes before the addition of the thiosulfate solution. In the presence of excess thiosulfate, the very soluble Ag2(S203) 3 and Ag2(S203) 3 complexes form. These soluble thiosulfate complexes, which are very stable, are the basis of photographic fixers. Silver thiosulfate complexes are oxidized to form silver sulfide, sulfate, and elemental sulfur (see Thiosulfates). 

If Q > K, the solution contains a higher concentration of ions than it can hold at equilibrium. A precipitate forms, decreasing the concentrations until the ion product becomes equal to Ksp and equilibrium is established. 

If Q < Ksp, the solution contains a lower concentration of ions than is required for equilibrium with the solid. The solution is unsaturated. No precipitate forms equilibrium is not established. 

Water from a well is found to contain 3.0 mg of calcium ion per liter. If 0.50 mg of sodium sulfate is added to one liter of the well water without changing its volume, will a precipitate form What should [S042-] be to just start precipitation ... 

Silver fluoride is soluble. Therefore, no precipitate forms as Ag ions are added to a solution of F ions. 

A 50 ml volume of 0.04 M Ca(N03)2 solution is added to 150 ml of 0.008 M (NH4)2S04 solution. Show that a trial value of the calcium sulfate ion product is 6 X 10-6. Will a precipitate form ... 

To a solution containing 0.1 Af of each of the ions Ag+, Cu+, Fe+ and Ca+4 is added 2 M NaBr solution, giving precipitate A. After filtration, a sulfide solution is added to the solution and a black precipitate forms, precipitate B. This precipitate is removed by filtration and 2 M sodium carbonate solution is added, giving precipitate C. What is the composition of each precipitate, A, B, and C ... 

The small amount of mercury(I) chloride in suspension has no appreciable effect upon the oxidising agent used in the subsequent titration, but if a heavy precipitate forms, or a grey or black precipitate is obtained, too much tin(II) solution has been used the results are inaccurate and the reduction must be repeated. Finely divided mercury reduces permanganate or dichromate ions and also slowly reduces Fe3+ ions in the presence of chloride ion. 

Cations forming insoluble chromates, such as those of silver, barium, mercury (I), mercury(II), and bismuth, do not interfere because the acidity is sufficiently high to prevent their precipitation. Bromide ion from the generation may be expected to form insoluble silver bromide, and so it is preferable to separate silver prior to the precipitation. Ammonium salts interfere, owing to competitive oxidation by bromate, and should be removed by treatment with sodium hydroxide. 

When the solution in Beaker 1 is mixed with the solution in Beaker 2, a precipitate forms. Using the following table, write the net ionic equation describing the formation of the precipitate, and then identify the spectator ions. 

You are given a solution and asked to analyze it for the cations Ag t, Ca2+, and Hg24. You add hydrochloric acid. Nothing appears to happen. You then add dilute sulfuric acid, and a white precipitate forms. You filter out the solid and add hydrogen sulfide to the solution that remains. A black precipitate forms. Which ions should you report as present in your solution ... 

It is sometimes possible to separate different cations from a solution by adding a soluble salt containing an anion with which they form insoluble salts. For example, seawater is a mixture of many different ions. It is possible to precipitate magnesium ions from seawater by adding hydroxide ions. However, other cations are also present in seawater. Their individual concentrations and the relative solubilities of their hydroxides determine which will precipitate first if a certain amount of hydroxide is added. Optimum separation of two compounds is achieved when Qsp exceeds the Ksp of one species but is less than the Ksp of the second species. Example 11.11 illustrates a strategy for predicting the order of precipitation. 

When a salt dissolves in water, it produces cations and anions. Lead(II) nitrate and potassium iodide are soluble salts. Lead(II) nitrate dissolves in water to generate Pb cations and NO3 anions. Potassium iodide dissolves in water to generate K and I ions. Mixing the solutions combines all four types of ions. A precipitate forms if any of the new combinations of the ions forms a salt that is insoluble in water. The new combinations when these two solutions mix are K combining with NO3 or Pb combining with I ... 

If we were to conduct a second solubility experiment in which solutions of KI and NaN03 were mixed, we would find that no precipitate forms. This demonstrates that K and NO3 ions do not form a solid precipitate, so the bright yellow precipitate must be lead(II) iodide, Pbl2. As the two salt solutions mix, cations and r anions combine to produce lead(II) iodide, which precipitates from the solution. On standing, the yellow precipitate settles, leaving a colorless solution that contains potassium cations and nitrate anions. The molecular blowups in Figure depict these solutions at the molecular level. 

A white precipitate forms when 2.00 X 10 mL of 0.200 M potassium phosphate solution is mixed with 3.00 X 10 mL of 0.250 M calcium chloride solution. Write the net ionic equation that describes this process. Calculate the mass of the precipitate that forms, and identify the ions remaining in solution. 

C04-0107. A white precipitate forms when aqueous calcium nitrate is mixed with aqueous ammonium sulfate, (a) Identify the precipitate and write the net ionic equation for the reaction, (b) What are the spectator ions ... 

In random structures, stoichiometry need not be exact and adventitious ions can be incorporated without causing disruption. Bonds are not highly directed, and neighbouring regions of precipitation, formed around different nuclei, can be accommodated within the structure. Continuous networks can be formed rapidly. Thus, random structures are conducive to cement formation and, in fact, most AB cements are essentially amorphous. Indeed, it often appears that the development of crystallinity is detrimental to cement formation. 

Hydrogen sulfide gas is used in analytical chemistry laboratories to detect certain metal ions in a solution. If the solution contains a Group IIA metal, such as calcium, a precipitate will form when an acidic solution containing hydrogen sulfide is added to it. The precipitate forms because one of the products of the chemical reaction is insoluble (which means it does not dissolve). 

Strong reducing agents such as sulfur dioxide, sodium bisulfite, sodium metabisulfite, and ferrous sulfate are used in the iron and steel finishing sites to reduce hexavalent chromium to the triva-lent form, which allows the metal to be removed from solution by chemical precipitation.21 23 Metal-containing wastewaters may also be treated by chemical precipitation or ion-exchange. 

Vinegar, or acetic acid, combines with calcium carbonate to dissolve the precipitate, form free calcium ions and water, and liberate carbon dioxide gas ... 

Test Ttibe 6 0.1 M Ni(N03)2 1 was able to identify this tube through observation since the Ni2+ ion is green. No other solution in the list should have had a green color. I was able to confirm this when I mixed samples of test tube 6 with test tube 3 (ammonia) producing Ni(NH3)62+, a known light blue solution. When I mixed samples from test tube 6 with test tube 4 which I believed to contain OH-, a green precipitate formed consistent with the color of Ni(OH)2. 

A student mixes 50.0 mL of 0.10 M Fe(N03)2 solution with 50.0 mL of 0.10 M KOH. A green precipitate forms, and the concentration of the hydroxide ion becomes very small. Which of the following correctly places the concentrations of the remaining ions in order of decreasing concentration ... 

Tarmins are important commercial products of complex and diverse chemistry. During the last years, the interest of biomaterials and specifically in tannins was growing. The ability of tannins to precipitate metal ions is due to their multiple adjacent phenolic hydroxyl groups, which can form stable complexes with many metal ions [8, 9]. 

In this section, you determined the solubility product constant, Kgp, based on solubility data. You obtained your own solubility data and used these data to calculate a value for Kgp. You determined the molar solubility of ionic solutions in pure water and in solutions of common ions, based on their Ksp values. In section 9.3, you will further explore the implications of Le Chatelier s principle. You will use a reaction quotient, Qsp, to predict whether a precipitate forms. As well, you will learn how selective precipitation can be used to identify ions in solution. 

The units are correct in the calculation of [Ag. It seems reasonable that a precipitate formed, since Kgp for silver chloride is very small compared with the concentration of chloride ions and silver ions. 

This diagram shows a scheme for identifying silver, Agh cadmium, Cd T and aluminium, AP+, ions. After each step, if a precipitate forms, it is separated by centrifugation. Then the solution is carefully poured into a new test tube and used in the next step. 

---