The Water Equilibrium

Up to this point most attention has been given to the direction in which an equilibrium is favored. This does not mean we can ignore the unfavored direction. Consider, for example, the reaction NHjCaq) + HOH( ) NH4 (aq) -I- OH (aq). Although this reaction proceeds only slightly in the forward direction, many of the properties of household ammonia—its cleaning power, in particular—depend on the presence of OH ions. 

Redox-Before-Acid-Base Option if you have already studied Chapter 19, you know about the electron-transfer character of oxidation-reduction reactions. There are several similarities between those reactions and the proton-transfer reactions of Chapter 17. These are identified in the next section. If you have not yet studied Chapter 19, you may omit Section 17.7, which is repeated at the appropriate place in Chapter 19. 

7 Acid -Base Reactions and Redox Reactions Compared 

11 Compare and contrast acid-base reactions with redox reactions. 

At this point it may be useful to pause briefly and point out how acid-base reactions resemble redox reactions. 

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We shall be interested in determining the effect of electrolytes of low molecular weight on the osmotic properties of these polymer solutions. To further simplify the discussion, we shall not attempt to formulate the relationships of this section in general terms for electrolytes of different charge types-2 l, 2 2, 3 1, 3 2, and so on-but shall consider the added electrolyte to be of the 1 1 type. We also assume that these electrolytes have no effect on the state of charge of the polymer itself that is, for a polymer such as, say, poly (vinyl pyridine) in aqueous HCl or NaOH, the state of charge would depend on the pH through the water equilibrium and the reaction... 

The water equilibrium illustrates the amphiprotic nature of H2 O. In this reaction, one water molecule acts as a proton donor (acid), and another acts as a proton acceptor (base). 

The water equilibrium describes an inverse relationship between [H3 0+] eq nd [OH-]gq. When an acid dissolves in water, the hydronium ion concentration increases, so the hydroxide ion concentration must decrease to maintain the product of the concentrations at 1.0 X 10. Similarly, the hydroxide ion concentration increases when a base dissolves in water, so the hydronium ion concentration must decrease. 

The water equilibrium always exists in aqueous solution ... 

The strong acid dissociates completely, so the water equilibrium is the only one that applies to this problem . eq = - w = [H3 [OH jg = 1.0 X... 

In any solution of an acid, the total hydronium and hydroxide ion concentrations include the 10" M contribution from the water reaction. This example illustrates, however, that the change in hydronium ion concentration due specifically to the water equilibrium is negligibly small in an aqueous solution of a strong acid. This is true for any strong acid whose concentration is greater than 10 M. Consequently, the hydronium ion concentration equals... 

For any aqueous strong base, the hydroxide ion concentration can be calculated directly from the overall solution molarity. As is the case for aqueous strong acids, the hydronium and hydroxide ion concentrations are linked through the water equilibrium, as shown by Example. ... 

For this example, we summarize the first four steps of the method The problem asks for the concentration of ions. Sodium hydroxide is a strong base that dissolves in water to generate Na cations and OH- anions quantitatively. The concentration of hydroxide ion equals the concentration of the base. The water equilibrium links the concentrations of OH" and H3 O" ", so an equilibrium calculation is required to determine the concentration of hydronium ion. What remains is to organize the data, carry out the calculations, and check for reasonableness. 

Just as for solutions of strong acids, the water equilibrium contributes negligibly to the total hydroxide ion concentration in any solution of strong base whose concentration is greater than 10 M. Consequently, the... 

To convert from pH to ion concentrations, first apply Equation to calculate [H3 O ]. pH = - log [H3 O J Then make use of the water equilibrium to calculate [OH ]. 

Use of these definitions and the properties of logarithms leads to a statement of the water equilibrium constant in... 

To find the pH of a solution, first compute [H3 O ] or [OH ] and then apply Equation or. Sodium hydroxide is a strong base, and the water equilibrium provides the link between hydroxide and hydronium ion concentrations. 

Combining these two equilibria leads to cancellation of HF and F, so the sum of the two is the water equilibrium. How are the equilibrium constants for these three equilibria related ... 

The water equilibrium seldom is the dominant one when a solution contains any solute with acid-base properties, so it is reasonable that the pH is determined by. ... 

The water equilibrium always exists in aqueous solution. In general, we can focus our initial attention on the equilibria involving other major species (NH3 in this example). Nevertheless, the water equilibrium does exert its effect on the concentrations of OH and H3 O. In this example, the concentration of hydroxide anion is established by the ammonia equilibrium, but the concentration of hydronium cations must be found by applying the water equilibrium. We use this feature in several of our examples in this chapter. 

After completing our analysis of the effects of the dominant equilibrium, we may need to consider the effects of other equilibria. The calculation of [H3 O ] in a solution of weak base illustrates circumstances where this secondary consideration is necessary. Here, the dominant equilibrium does not include the species, H3 O, whose concentration we wish to know. In such cases, we must turn to an equilibrium expression that has the species of interest as a product. The reactants should be species that are involved in the dominant equilibrium, because the concentrations of these species are determined by the dominant equilibrium. We can use these concentrations as the initial concentrations for our calculations based on secondary equilibria. Look again at Example for another application of this idea. In that example, the dominant equilibrium is the reaction between hypochlorite anions and water molecules H2 0 l) + OCr(c2 q) HOCl((2 q) + OH ((2 q) Working with this equilibrium, we can determine the concentrations of OCl, HOCl, and OH. To find the concentration of hydronium ions, however, we must invoke a second equilibrium, the water equilibrium 2 H2 0(/) H3 O (a q) + OH (a q)... 

Because 1 is several orders of magnitude larger than 2 or, we identify 1 as dominant. Notice, however, that the water equilibrium generates some hydroxide ions in the solution, so this equilibrium must be used to find the concentration of hydroxide ions. The third reaction involves a minor species, HCO3, as a reactant, so it cannot be the dominant equilibrium. However, just as the water equilibrium generates some hydroxide ions, the hydrogen carbonate equilibrium generates some carbonate anions, whose concentration must be determined. 

Because sulfurous acid is diprotic, a second proton transfer equilibrium has an effect on the ion concentrations, and the water equilibrium also plays a secondaiy role ... 

In solution, sodium ethanoate is alkaline and so contains more OH" ions than H ions. This means that some of the H+ ions present in the water equilibrium must have been removed. 

The material within the pellet would be then more reducing than the sediment in which it lies, and would have a higher local Eh potential. Thus, the initial impetus to the process is a A Eh between sediment and pellet. The concentrations of the elements which must be present in the sea water solution to promote glauconite formation are largely governed by the type of sediment in contact with the water. Equilibrium is thus established punctually between pellet and sea water effecting a transfer of material between the two media. On a larger but still somewhat local scale the dissolution of detrital silicates in sea water provides the basic "reservoir" of material in solution and hence determines the activities of various elements in the solution. 

Immersion in pure water must lead to the same result as in a saturated atmosphere. If water contains solutes, its vapor pressure decreases. The corresponding equilibrium concentration, linked to the water activity, is proportional to the water vapor pressure. In other words, the water equilibrium concentration is a decreasing function of the solute concentration salt water is less active than pure water. 

In this problem, we have made the assumption that the contribution of water to [OH-] (equal to [H+], or 2.4 x 10-11 M) is negligible compared with that of NH3. Kw is used to calculate [H+], since water is the only supplier of H+. In general, [H+] for acidic solutions can be calculated without regard to the water equilibrium then Kw is used to calculate [OH-]. Conversely, [OH-] for basic solutions can usually be calculated without regard to the water equilibrium then, Kw is used to calculate [H+]. 

In comparing the results of this problem with Problem 17.26, note that the percent hydrolysis of NH]j is greater in the presence of a hydrolyzing anion (like acetate). The reason is that the removal of some of the products of the two hydrolyses, H+ and OH-, by the water equilibrium reaction (H2O = H4" + OH-) allows both hydrolyses to proceed to an increasing extent. 

But is autoionization an important source of H+ ions In pure water at 2S°C, [H+] is 10-7 M. In 1.0 M HCl the water will produce even less than 10-7 M H+, since by Le Chatelier s principle the H+ from the dissociated HCl will drive the position of the water equilibrium to the left. Thus the amount of H+ contributed by water is negligible compared with the 1.0 M H+ from the dissociation of HCl. Therefore we can say that [H+] in the solution is 1.0 M and that... 

Water is the most indispensable factor of life. By means of carefully coordinated regulatory mechanisms, the water equilibrium and hence the reservoir of body water is held constant. It is important to keep water intake and output in balance to maintain iso-volaemia. (s. fig. 16.1)... 

Electrolytes are subject to dissociation into negatively charged anions and positive cations. The vital electrolyte balance guarantees the respective uptake and discharge and ensures the correct presence and distribution. This regulatory process is closely linked to the water equilibrium. The intracellular and extracellular spaces differ in their electrolyte content, (s. tab. 16.1)... 

Recall that an equilibrium-constant expression relates the concentrations of species involved in an equilibrium. The relationship for the water equilibrium is simply [H30" ][0H ] = K q. This equilibrium constant, called the self-ionization constant of water, is so important that it has a special symbol, Its value can be found from the known concentrations of the hydronium and hydroxide ions in pure water, as follows ... 

When nitric acid is added to water, large numbers of H3O+ ions are produced. The large increase in [H3O+] shifts the water equilibrium far to the left (LeChatelier s Principle), and the [OR-] decreases. 

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