Autoprotolysis contribution

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 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-]. 

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. 

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 rest of this chapter is a variation on a theme the use of equilibrium constants to calculate the equilibrium composition of solutions of acids and bases. We begin with solutions of acids, bases, and salts, explore the contribution of the autoprotolysis of the solvent to the pH, which is significant in very dilute solutions, and see how to handle the complications of acids that can donate more than one proton. Although the applications are varied, the techniques are all very similar and are based on the material in Chapter 9. 

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... 

Step 5 Use an equilibrium table to find the H30+ ion concentration and convert to pH. Assume that the contribution of the autoprotolysis of water to the pH is insignificant. Alternatively, if the concentrations are large compared with the concentration of hydronium ions, the Henderson- Hasselbalch equation may be used to determine the pH. 

The low autoprotolysis constant of liquid ammonia (10 at — 50°C) suggests that the strongly basic properties of ammonia in comparison with water are more than counterbalanced by its feeble acidic properties. Once more, a relatively low dielectric constant of 22 contributes to association to a minor extent. 

Since the autoprotolysis constant of acetonitrile is small, the SH2 contribution from dissociation of the solvent may be neglected. 

Note This value of [H30 " ] is not much different from the value for pure water, 1.0x10 mol L therefore, it is at the lower limit of safely ignoring the contribution to [H3O " ] from the autoprotolysis of water. The exercise should be solved by simultaneously considering both equilibria. 

We neglect as it is often the case the contribution to the concentration of water and assume that all HCl molecules are dissociated which is why from autoprotolysis of... 

It may be seen from the calculations that the contribution to the concentration from the autoprotolysis of water is negligible compared to the calculated x. Thus, the contribution from the autoprotolysis of water may in practise be omitted when x is in the range of 10 mol/L or larger. 

Again we have for the completeness added the contribution to the HsO concentration for the autoprotolysis of water, even though this is negligible compared to the calculated x-value. In this case the contribution from the autoprotolysis may be neglected. 

Again the contribution from autoprotolysis of water is added although in may in practice be neglected. 

In this fifth chapter we have looked at central part of the aqueous chemistry being acid/base chemistry. We initially saw how a H ion from the acid to the base is transferred and how acid strength is defined analogously to the principles of equilibrium we say earlier in chapter 4. further the pH scale was defined and we saw how water molecules may be able to react with itself in the process of autoprotolysis. The autoprotolysis of water contributes to the HsO and OH" concentration but may in most cases be neglected unless the calculated values of HsO and OH" concentrations are in the order of 10 M or less. 

From Fig. 48, one can deduce that the pH must be left of the pH-coordinate of Pi, as on the right side the sum cqh- + hb- + 2cg2- is always larger than the equilibrium concentration of HsO" " (in other words, the HsO" " results from the protolysis of H2B and HB, and also from the autoprotolysis of water). The pH is mainly given by the reaction equation (37), and the other two reactions have only a smaller contribution to the overall HsO concentration. To the left of the pH-coordinate of Pi, the equilibrium concentration of OH ions (cf. P2) is so small that it can be neglected in the charge balance, and one can write ... 

The desired quantity is = [H+].[A ]/[AH] at equilibrium, where ([AH] [A ]) = Cah°- The strictly correct treatment takes into account the amount of [H+] contributed by the autoprotolysis of water, but here we assume that [A ] = [H+] << Cah° (in fact this introduces negligible systematic error in this case), so that [H+]VCah°. 

The most reUable way to estimate the pH of a solution of a weak acid is to consider the contributions from deprotonation of the acid and autoprotolysis of water to the total concentration of hydronium ion in solution (see Further information 4.1). Autoprotolysis may be ignored if the weak acid is the main contributor of hydronium ions, a condition that is satisfied if the acid is not very weak and is present at not too low a concentration. Then we can estimate the pH of a solution of a weak acid and calculate either of these fractions by using the following strategy ... 

The initial molar concentrations of the species, ignoring any contributions to the concentration of H3O or OH from the autoprotolysis of water. 

A much smaller contribution to the ionic composition of the sprayed solution than any of the above effects is from the autoprotolysis of the solvent. In practical analytical solutions, this effect is rarely greater than it is in water where the maximum concentration of ions formed is about 1 x 10 M and that only in a neutral solution. 

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