The autoprotolysis of water

As indicated earlier water may function both as an acid and as a base. Such a property is often referred to as an amfolyte and water is an amfolyte. This means that water may react with it self in the following reaction  

The equilibrium lies nevertheless far to the left. The equilibrium constant for the specific equilibrium is referred to as K having the following values. 

This means that there are only relatively few water molecules that are transferred to HjO and OH ions. This equilibrium sets itself in pure water and in aqueous solutions. In pure water [HsO ] = [OH ] as the ions are produced in an 1 1 relation. At a temperature of 25 C in pure water  

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The autoprotolysis of water contributes significantly to the pH when the acid is so dilute or so weak that the calculation predicts an H30+ concentration close to 10 mol-E l. In such cases, we must use the procedures described in Sections 10.18 and 10.19. We can ignore the contribution of the autoprotolysis of water in an acidic solution only when the calculated H30+ concentration is substantially (about 10 times) higher than 10-7 mol-L-1, corresponding to a pH of 6 or less. 

The blue square in the grid in the final box represents the percentage of the acid molecules that are deprotonated. We see that x is less than 5% of 0.10, and the approximation is valid. Because the pH < 6, the assumption that the autoprotolysis of water can be ignored is valid. 

We calculate the pH of solutions of weak bases in the same way as we calculate the pH of solutions of weak acids—by using an equilibrium table. The protonation equilibrium is given in Eq. 9. To calculate the pH of the solution, we first calculate the concentration of OH ions at equilibrium, express that concentration as pOH, and then calculate the pH at 25°C from the relation pH + pOH = 14.00. For very weak or very dilute bases, the autoprotolysis of water must be taken into consideration. 

The calculation of x can often be simplified, as explained in Toolbox 10.1. We ignore contributions from the autoprotolysis of water to the hydroxide ion concentration if the concentration of hydroxide ions is greater than 10 h mol-L-. Step 5 Determine the pOH of the solution and then calculate the pH from the pOH by using Eq. 6b. 

The approximation that x is less than 5% of 0.15 is valid (by a large margin). Moreover, the H30+ concentration (9.2 X 10 6 mol-F ) is much larger than that generated by the autoprotolysis of water (1.0 X 10 mol-L ), and so ignoring the latter contribution is valid. 

We assume that the polyprotic acid is the solute species present in largest amount. We also assume that only the first deprotonation contributes significantly to [H3Ot] and that the autoprotolysis of water does not contribute significantly to H30+] or [OH-]. 

Suppose we were asked to estimate the pH of 1.0 X 10 x m HCl(aq). If we used the techniques of Example 10.3 to calculate the pH from the concentration of the acid itself, we would find pH = 8.00. That value, though, is absurd, because it lies on the basic side of neutrality, whereas HC1 is an acid The error stems from there being two sources of hydronium ions, whereas we have considered only one. At very low acid concentrations, the supply of hydronium ions from the autoprotolysis of water is close to the supply provided by the very low concentration of HC1, and both supplies must be taken into account. The following two sections explain how to take autoprotolysis into account, first for strong acids and bases and then for weak ones. 

In very dilute solutions of strong acids and bases, the pH is significantly affected by the autoprotolysis of water. The pH is determined by solving three simultaneous equations the charge-balance equation, the material-balance equation, and the expression for Kw. 

Autoprotolysis also contributes to the pH of very dilute solutions of weak acids. In fact, some acids, such as hypoiodous acid, HIO, are so weak and undergo so little deprotonation that the autoprotolysis of water almost always contributes significantly to the pH. To find the pH of these solutions, we must take into account the autoprotolysis of water. 

EXAMPLE 10.15 Sample exercise Estimating the pH of a dilute aqueous solution of a weak acid when the autoprotolysis of water must be considered. 

In aqueous solutions of very weak acids, the autoprotolysis of water must be taken into account if the hydronium ion concentration is less than 10 6 mol-L The expressions for Kw and Ka are combined with the equations for charge balance and material balance to find the pH. 

Find the pH of an aqueous solution of strong acid or base that is so dilute that the autoprotolysis of water significantly affects the pH (Example 10.14). 

M phenol(aq), ignoring the effect of the autoprotolysis of water, (b) Repeat the calculations, taking into account the autoprotolysis of water. 

STRATEGY Decide whether the salt present at the stoichiometric point provides an ion that acts as a weak base or as a weak acid. If the former, expect pH > 7 if the latter, expect pH < 7. To calculate the pH at the stoichiometric point, proceed as in Example 10.10 or 10.11, noting that the amount of salt at the stoichiometric point is equal to the initial amount of acid and the volume is the total volume of the combined analyte and titrant solutions. The Kb of a weak base is related to the / a of its conjugate acid by Ka X Kb = fCw Ka is listed in Table 10.1. Assume that the autoprotolysis of water has no significant effect on the pH, and then check that assumption. 

Step 5 Use an equilibrium table to find the H.O concentration in a weak acid or the OH concentration in a weak base. Alternatively, if the concentrations of conjugate acid and base calculated in step 4 are both large relative to the concentration of hydronium ions, use them in the expression for /<, or the Henderson—Hasselbalch equation to determine the pH. In each case, if the pH is less than 6 or greater than 8, assume that the autoprotolysis of water does not significantly affect the pH. If necessary, convert between Ka and Kh by using Kw = KA X Kb. 

The ammonia is released and the protons remain in the zeolite, which then can be used as acidic catalysts. Applying this method, all extra-framework cations can be replaced by protons. Protonated zeolites with a low Si/Al ratio are not very stable. Their framework structure decomposes even upon moderate thermal treatment [8-10], A framework stabilization of Zeolite X or Y can be achieved by introducing rare earth (RE) cations in the Sodalite cages of these zeolites. Acidic sites are obtained by exchanging the zeolites with RE cations and subsequent heat treatment. During the heating, protons are formed due to the autoprotolysis of water molecules in the presence of the RE cations as follows ... 

In aqueous solutions, whenever protonation equilibria are involved, the autoprotolysis of water needs to be incorporated into the model. Thus, for an acetic acid titration the model comprises two equilibria... 

The autoprotolysis of water can be included into the algorithm in a very compact way. As already indicated in Table 1, hydroxo species are given the notation H i. According to equation (3.22) we can write ... 

STRATEGY Expect pH > 7, because amines are bases. Assume that the presence of base dominates the pH and therefore that the autoprotolysis of water need not be considered. Calculate the molarity of OH ions by using the equilibrium table as explained in Toolbox 10.1, but use Kh instead of fCa. Calculate [OH-], convert it to pOH, and then convert that pOH to pH by using the relation pH + pOH = 14.00. 

In most solutions of weak acids, the concentration of hydronium ions is so high that [H30+] > 1CT6 (that is, pH < 6). Under these conditions, Xw/[H30+] < 10-8 and we can ignore this term in both the numerator and the denominator of Eq. 19. The expression for fCa simplifies to the familiar one we use when the autoprotolysis of water has an insignificant effect on pH (Section 10.10) ... 

STRATEGY The solution contains N02 , a base, so we expect the pH to be i higher than that of nitrous acid alone. The K+ ion has no protons to donate j and cannot accept a proton, so it has no measurable effect on the pH of the ] solution. Identify the proton transfer equilibrium and use it to find the pH by means of an equilibrium table. Consider the initial molarity of HN02 (before reaction with water) to be 0.500 mol-L1. Because nitrite ions have also been added to the solution, set their initial molarity equal to the molar-5 ity of added salt (each KN02 formula unit supplies one N02 anion). Then proceed as described in Toolbox 10.1. Because the concentrations of the 1 added ions are much higher than 10-7 mol-L 1, we assume that we can S ignore the contribution to the pH from the autoprotolysis of water. 

The solution of this equation is x = 2.2 x 10-3, which is only 2.2% of 0.100 and only 0.44% of 0.500. The approximations are valid, and the. equilibrium molarity of H30+ ions is 2.2 X 10-3 mol-L 1. This concentra- tion is much larger than that contributed by the autoprotolysis of water so neglect of the autoprotolysis is also valid. It follows that the pH of the solution is... 

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