Water, autoprotolysis

Water, autoprotolysis constant, AUTOPROTOLYSIS WATER, BOUND-... 

Autoprotolysis of water autoprotolysis, -> water and subentry -> heavy water... 

Many compounds that contain hydrogen can donate protons to a solvent sueh as water and so behave as aeids. Water itself undergoes ionie dissoeiation to a small extent by means of autoprotolysis the proeess is usually represented formally by the equilibrium... 

In addition, the liquid undergoes self-ionic dissociation to a greater extent than any other nominally covalent pure liquid (cf. BF3.2H2O, p. 198) initial autoprotolysis is followed by rapid loss of water which can then react with a further molecule of HNO3 ... 

NOj ions/ Addition of water to nitric acid at first diminishes its electrical conductivity by repressing the autoprotolysis reactions mentioned above. For example, at -10° the conductivity decrea.ses from 3.67 x 10 ohm cm to a minimum of 1.08 x 10" ohm" cm at 1.75 molal H2O (82.8% NjOs) before rising again due to the increasing formation of the hydroxonium ion according to the acid-base equilibrium... 

This value is compared with those for other acids and protonic liquids in Table 15.21 " the extent of autoprotolysis in H2SO4 is greater than that in water by a factor of more than lO " and is exceeded only by anhydrous H3PO4 and [HBp3(OH)] (p. 198). In addition to autoprotolysis, H2SO4 undergoes ionic selfdehydration ... 

The reaction is very fast in both directions, and so is always at equilibrium in water and in aqueous solutions. In every glass of water, protons from the hydrogen atoms are ceaselessly migrating between the molecules. This type of reaction, in which one molecule transfers a proton to another molecule of the same kind, is called autoprotolysis (Fig. 10.9). 

In dilute aqueous solutions (the only ones we consider in this chapter), the solvent, water, is very nearly pure, and so its activity may be taken to be 1. The resulting expression is called the autoprotolysis constant of water and is written Kw ... 

The concentrations of H30 + and OH are very low in pure water, which explains why pure water is such a poor conductor of electricity. To imagine the very tiny extent of autoprotolysis, think of each letter in this book as a water molecule. We would need to search through more than 50 books to find one ionized water molecule. The autoprotolysis reaction is endothermic (AH° = +56 kj-mol l), and so we can expect Kw to increase with temperature, and aqueous solutions to have higher concentrations of both hydronium and hydroxide ions at higher temperatures. Unless otherwise stated, all the calculations in this chapter will be for 25°C. 

FIGURE 10.9 As a result of autoprotolysis, pure water consists of hydronium ions and hydroxide ions as well as water molecules. The concentration of ions that results from autoprotolysis is only about 10 mol-L and so only about I molecule in ZOO million is ionized. The overlay shows only the ions. 

STRATEGY When Ba(OH)2 dissolves in water, it provides OH ions most hydroxides of Groups 1 and 2 can be treated as fully dissociated in solution. Decide from the chemical formula how many OH ions are provided by each formula unit and calculate the concentrations of these ions in the solution. To find the concentration of H,Oj ions, use the water autoprotolysis constant Kw = [H,0 1 [OH ]. 

The values of pH and pOH are related. To find that relation, we start with the expression for the autoprotolysis constant of water Kw = [H3Oh [Of I ]. Then we take logarithms of both sides ... 

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. 

We have to include the contribution of autoprotolysis to pH only when the concentration of strong acid or base is less than about 10 6 mol-L-1. To calculate the pH in such a case, we need to consider all species in solution. As an example, consider a solution of HCl, a strong acid. Other than water, the species present are H30+, OH, and Cl-. There are three unknown concentrations. To find them, we need three equations. 

Now consider a very dilute solution of a strong base, such as NaOH. Apart from water, the species present in solution are Na+, OH, and H30+. As we did for HCl, we can write down three equations relating the concentrations of these ions by using charge balance, material balance and the autoprotolysis constant. Because the cations present are hydronium ions and sodium ions, the charge-balance relation is... 

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. 

The calculation of pH for very dilute solutions of a weak acid HA is similar to that for strong acids in Section 10.18. It is based on the fact that, apart from water, there are four species in solution—namely, HA, A, H,0 +, and OH. Because there are four unknowns, we need four equations to find their concentrations. Two relations that we can use are the autoprotolysis constant of water and the acidity constant of the acid HA ... 

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. 

Many of the reactions that take place in water have analogous reactions in liquid ammonia, (a) Write the chemical equation for the autoprotolysis of NH,. (b) What are the formulas of the acid and base species that result from the autoprotolysis of... 

Heavy water is deuterium oxide, D20. The standard reaction Gibbs free energy for the autoprotolysis of pure deuterium oxide is +84.8 kj-mol 1 at 298 K. (a) If pD is defined analogously to pH, what is the pD of pure D20 at 298 K ... 

The pK for the autoprotolysis (more precisely, the autodeuterolysis, because a deuteron is being transferred) of heavy water (D20) is 15.136 at 20.°C and 13.8330 at 30.°C. Assuming AH° for this reaction to be independent of temperature, calculate A.Sr°for the autoprotolysis reaction. Suggest an interpretation of the sign. Suggest a reason why the autoprotolysis constant of heavy water differs from that of ordinary 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. 

Ammonia is a pungent, toxic gas that condenses to a colorless liquid at — 33°C. The liquid resembles water in its physical properties, including its ability to act as a solvent for a wide range of substances. Because the dipole moment of the NH3 molecule (1.47 D) is lower than that of the H20 molecule (1.85 D), salts with strong ionic character, such as KCI, cannot dissolve in ammonia. Salts with polarizable anions tend to be more soluble in ammonia than are salts with greater ionic character. For example, iodides are more soluble than chlorides in ammonia. Liquid ammonia undergoes much less autoprotolysis than water ... 

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