Hydroxides, solubility products

The OH concentration increases (decreases) by one order of magnitude for every unit increase (decrease) in pH. This means that the formation of a metal hydroxide (whether as a colloid or as a precipitate) in aqueous solution will be strongly dependent on temperature when the product of the free metal ions and OH ions is close to the hydroxide solubility product, although increase in Ksp with temperature may partially offset this effect. 

Chapter 3 in [72DER] includes a solubility study of Zr hydroxide in the presence of varying concentrations of carbonate, conducted at a constant ionic strength of 1 M in NH4NO3. The determinations of the hydroxide solubility product in the absence of carbonate and of the carbonic acid dissociation constants in the same medium, needed for the evaluation of the Zr-C03 data, are also reported in the same chapter. 

The fraction of dinuclear complex compared to that of free ferric ions is about 10%. It is the most important one after that of Fe +. (Note that the ferric hydroxide solubility product Ks = 2.96 x 10 is not reached. Actually, no precipitation occurs.)... 

Gnanaprakash G, Philip RB (2007) Effect of divalent metal hydroxide solubility product on the size of ferrite nanoparticles. Mater Lett 61 4545-4548... 

Aqueous ammonia can also behave as a weak base giving hydroxide ions in solution. However, addition of aqueous ammonia to a solution of a cation which normally forms an insoluble hydroxide may not always precipitate the latter, because (a) the ammonia may form a complex ammine with the cation and (b) because the concentration of hydroxide ions available in aqueous ammonia may be insufficient to exceed the solubility product of the cation hydroxide. Effects (a) and (b) may operate simultaneously. The hydroxyl ion concentration of aqueous ammonia can be further reduced by the addition of ammonium chloride hence this mixture can be used to precipitate the hydroxides of, for example, aluminium and chrom-ium(III) but not nickel(II) or cobalt(II). 

The sodium salt of methyl red may be prepared by dissolving the crude product in an equal weight of 35 per cent, sodium hydroxide which has been diluted to 350 ml., hitoring, and evaporating under diminished pressure (Fig. II, 37, I). The resulting sodium salt forms orange leaflets. This water-soluble product is very convenient for use as an indicator. Incidentally, the toluene extraction is avoided. 

The sodium salt of CS [9005-22-5] is prepared by reaction of cellulose with sulfuric acid in alcohol followed by sodium hydroxide neutrali2ation (20). This water-soluble product yields relatively stable, clear, and highly viscous solutions. Introduced as a thickener for aqueous systems and an emulsion stabilizer, it is now of no economic significance. 

Solubility Product — The solubility product constant commonly referred to as the solubility product provides a convenient method of predicting the solubility of a material in water at equilibrium. Copper hydroxide, for example, dissolves according to the following equilibrium ... 

The colloidal palladium solution is prepared as follows A solution of a palladium salt is added to a solution of an alkali salt of an acid of high molecular weight, the sodium salt of protalbinic acid being suitable. An excess of alkali dissolves the precipitate formed, and the solution contains tine palladium in the form of a hydrosol of its hydroxide. The solution is purified by dialysis, and the hydroxide reduced with hydrazine hydrate. On further dialysis and evaporation to dryness a water-soluble product is obtained, consisting of colloidal palladium and sodium protalbinate, the latter acting as a protective colloid. 

The data given in Tables 1.9 and 1.10 have been based on the assumption that metal cations are the sole species formed, but at higher pH values oxides, hydrated oxides or hydroxides may be formed, and the relevant half reactions will be of the form shown in equations 2(a) and 2(b) (Table 1.7). In these circumstances the a + will be governed by the solubility product of the solid compound and the pH of the solution. At higher pH values the solid compound may become unstable with respect to metal anions (equations 3(a) and 3(b), Table 1.7), and metals like aluminium, zinc, tin and lead, which form amphoteric oxides, corrode in alkaline solutions. It is evident, therefore, that the equilibrium between a metal and an aqueous solution is far more complex than that illustrated in Tables 1.9 and 1.10. Nevertheless, as will be discussed subsequently, a similar thermodynamic approach is possible. 

A simple calculation based on the solubility product of ferrous hydroxide and assuming an interfacial pH of 9 (due to the alkalisation of the cathodic surface by reaction ) shows that, according to the Nernst equation, at -0-85 V (vs. CU/CUSO4) the ferrous ion concentration then present is sufficient to permit deposition hydroxide ion. It appears that the ferrous hydroxide formed may be protective and that the practical protection potential ( —0-85 V), as opposed to the theoretical protection potential (E, = -0-93 V), is governed by the thermodynamics of precipitation and not those of dissolution. 

Although the hydroxides of the alkaline earth elements become more soluble in water as we go down the column, the opposite trend is observed in the solubilities of the sulfates and carbonates. For example, Table 21-VII shows the solubility products of the alkaline earth sulfates. 

Example 3. The solubility product of magnesium hydroxide is 3.4 x 10-11 mol3 L 3. Calculate its solubility in grams per L. 

The relative molecular mass of magnesium hydroxide is 58.3. Each mole of magnesium hydroxide, when dissolved, yields 1 mole of magnesium ions and 2 moles of hydroxyl ions. If the solubility is xmolL-1, [Mg2 + ] = x and [OH-] = 2x. Substituting these values in the solubility product expression ... 

The extent of hydrolysis of (MY)(n 4)+ depends upon the characteristics of the metal ion, and is largely controlled by the solubility product of the metallic hydroxide and, of course, the stability constant of the complex. Thus iron(III) is precipitated as hydroxide (Ksal = 1 x 10 36) in basic solution, but nickel(II), for which the relevant solubility product is 6.5 x 10 l8, remains complexed. Clearly the use of excess EDTA will tend to reduce the effect of hydrolysis in basic solutions. It follows that for each metal ion there exists an optimum pH which will give rise to a maximum value for the apparent stability constant. 

Solutions which prevent the hydrolysis of salts of weak acids and bases. If the precipitate is a salt of weak acid and is slightly soluble it may exhibit a tendency to hydrolyse, and the soluble product of hydrolysis will be a base the wash liquid must therefore be basic. Thus Mg(NH4)P04 may hydrolyse appreciably to give the hydrogenphosphate ion HPO and hydroxide ion, and should accordingly be washed with dilute aqueous ammonia. If salts of weak bases, such as hydrated iron(III), chromium(III), or aluminium ion, are to be separated from a precipitate, e.g. silica, by washing with water, the salts may be hydrolysed and their insoluble basic salts or hydroxides may be produced together with an acid ... 

B. Precipitation and separation of hydroxides at controlled hydrogen ion concentration or pH. The underlying theory is very similar to that just given for sulphides. Precipitation will depend largely upon the solubility product of the metallic hydroxide and the hydroxide ion concentration, or since pH + pOH = pKw (Section 2.16), upon the hydrogen ion concentration of the solution. 

Values for the solubility products of metallic hydroxides are, however, not very precise, so that it is not always possible to make exact theoretical calculations. The approximate pH values at which various hydroxides begin to precipitate from dilute solution are collected in Table 11.2. 

Seif-Test 12.10B Use the tables in Appendix 2B to calculate the solubility product of cadmium hydroxide, Cd(OH)2. 

The compound Cr(OH), is very insoluble in water therefore, electrochemical methods must be used to determine its fCsp. Given that the reduction of Cr(OH)3(s) to Cr(s) and hydroxide ions has a standard potential of —1.34 V, calculate the solubility product for Cr(OH)3. 

C18-0073. For the following salts, write a balanced equation showing the solubility equilibrium and write the solubility product expression for each (a) silver chloride (b) barium sulfate (c) iron(H) hydroxide and (d) calcium phosphate. 

---