Weak acids simplifying assumptions

We now have four equations (6.35, 6.36, 6.37, and 6.38) and four unknowns ([HE], [E-], [H3O4], and [OH-]) and are ready to solve the problem. Before doing so, however, we will simplify the algebra by making two reasonable assumptions. Eirst, since HE is a weak acid, we expect the solution to be acidic thus it is reasonable to assume that... 

Strategy Acetic acid is a weak acid consequently, we expect the molarity of H30+ ions to be less than 0.10 moI-L-1 and, therefore, its pH to be greater than 1.0. To find the actual value, we set up an equilibrium table S with the initial molarity of acid equal to 0.10 mol-L 1 and allow the molarity of acid to decrease by x mol-L1 to reach equilibrium. Assume that the presence of acid dominates the pH and therefore that the autoprotolysis of 5 water need not be considered. We assume x is less than about 5% of the ini-j rial molarity of acid and simplify the expression for the equilibrium constant f by ignoring x relative to the initial molarity of the acid. This assumption i must be verified at the end of the calculation. 

In the following discussions of expressions relating hydrogen ion concentration to various points in the titration of a weak base with an acid, several simplifying assumptions are made ... 

From our calculations we can draw some conclusions. In a solution containing only a weak monoprotic acid, the concentration of H3O+ is equal to the concentration of the anion of the acid. Unless the solution is very dilute, such as less than 0.050 M, the concentration of nonionized acid is approximately equal to the molarity of the solution. When the value of for the weak acid is greater than 10 , then the extent of ionization will be large enough to make a significant difference between the concentration of nonionized acid and the molarity of the solution. In such cases we cannot make the simplifying assumption. 

We will hot constract a diprotic titration curve here, but if you want a good mental exercise, try it You just can t make the simplifying assumptions that we can usually use with monoprotic acids that are sufficiently weak or not too dilute. See your CD, Chapter 8, for auxiliary data for the spreadsheet calculation of the titration curve for 50.00 mL 0.1000 M H2C1O4 versus 0.1000 M NaOH. You can download that and enter the Kai and Kai values for other diprotic acids and see what their titration curves look like. Try, for example, maleic acid. For the calculations, we used the more exact equations mentioned above for the initial pH, the first buffet zone, and the first equivalence point. We did not use the quadratic equation for the second equivalence point since Cr04 is a quite weak base (Kbi = 3.12 X 10 ). See Ref. 8 for other examples of calculated titration curves. 

The first method for finding approximate roots to an equation is to modify the equation by making simplifying assumptions. As an example, let us consider an equation for the hydrogen-ion concentration in a solution of a weak acid in which the hydrogen ions from the ionization of water cannot be ignored. We must solve simultaneous equations for the ionization of the weak acid and ionization of water. Later in this chapter, we will derive the equation... 

Weak acids and weak bases do not ionize (or protolyze) completely in aqueous solution. The approach used to solve for the concentrations of solution components for weak acid or base solutions is similar to that used for strong acids and strong bases, but we are not able to make the simplifying assumption in theKa orKt, equilibrium equations that complete dissociation takes place. Typical calculations for weak acid and weak base systems are illustrated in the following example. 

We should expect the extent of dissociation to be small for a weak acid. So we can simplify the calculation if we assume that x is small enough that (0.10 - x) 0.10. With that assumption, we can rewrite the equation above so that we will not need to use the quadratic formula ... 

You follow the three steps for solving equilibrium problems that were introduced in Example 15.7. In the last step, you solve the equilibrium-constant equation for the equilibrium concentrations. The resulting equation is quadratic, but because the equilibrium concentration of a weak acid is usually negligibly different from its starting value, the equation simplifies so that it involves only the square of the unknown, which is easily solved by taking the square root. (You will need to check that this assumption is valid.)... 

Examples 16-8 and 16-9 present a common problem involving weak acids and weak bases calculating the pH of a solution of known molarity. The calculation invariably involves a quadratic equation, but very often we can make a simplifying assumption that leads to a shortcut that saves both time and effort. 

In this example we will apply the same techniques as we did in Example 16-8. We will work the problem twice to see that the simplifying assumption breaks down for weak acids and weak bases at very low concentrations. 

The usual simplifying assumption is that of treating a weak acid or weak base as though it remains essentially nonionized (so that c — x c). In general, this assumption will work if the concentration (molarity) of the weak acid, c, or that of the weak base, cg, exceeds the value of or by a factor of at least 100. That is,... 

We see that in this example, the simplifying assumption works. We also note that the solution is fairly basic for a relatively dilute solution of a salt of a weak acid and a strong base. 

An aqueous solution of two weak acids has a stoichiometric concentration, c, in each add. If one acid has a value twice as large as the other, show that the pH of the solution is given by the equation pH = —5 log (3c K ), where is the ionization constant of the weaker acid. Assume that the criteria for the simplifying assumption on page 757 are met. 

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