Is the criterion 6

In Example 6D.4, the pH of 0.15 m NH₄Cl(aq) is found to be 5.04. However, the contribution to the pH from the autoprotolysis of water was ignored. Repeat the calculation of the pH of this solution, taking into account the autoprotolysis of water.

Example 6D.4

You are working in the emergency room of a hospital where a patient suffering from influenza has developed metabolic alkalosis, a condition in which the pH of the blood is too high. You have available a stock solution of ammonium chloride, which is used to lower the pH of the blood of patients suffering from alkalosis, but you need to know its pH. Estimate the pH of 0.15 m NH₄Cl(aq) at 25°C.

ANTICIPATE Because NH₄⁺ is a weak acid and Cl^– is neutral, you should expect pH

PLAN Treat the solution as that of a weak acid, using an equilibrium table as in Toolbox 6D.1 to calculate the composition and hence the pH. First, write the chemical equation for proton transfer to water and the expression for K_a. Obtain the value of Ka from K_b for the conjugate base by using Eq. 4a in Topic 6C (Ka = K_w/K_b). The initial concentration of the acidic cation is equal to the concentration of the cation that the salt would produce if it retained all its acidic protons.

What should you assume? Assume that (1) the extent of deprotonation is so small that the change in concentration of NH₄⁺ is insignificant, and (2) the autoprotolysis of water does not affect the pH significantly. Check these assumptions at the end  of the calculation.

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Toolbox 6D.1 HOW TO CALCULATE THE PH OF A SOLUTION OF A WEAK ACID CONCEPTUAL BASIS Because the proton transfer equilibrium is established as soon as a weak acid is dissolved in water, the concentrations of acid, hydro- nium ion, and conjugate base of the acid must jointly correspond to the acidity constant of the acid. Any of these quantities can be cal- culated by setting up an equilibrium table like that in Toolbox 51.1. PROCEDURE Step 1 Write the chemical equation and the expression for K, for the proton transfer equilibrium. Set up a table with columns labeled by the acid (HA), H,O*, and the conjugate base of the acid (A). In the first row below the headings, show the initial concentration of each species. For this step, assume that no acid molecules have been deprotonated. In the second row, write the changes in the concentration needed for the reaction to reach equilibrium. Assume that the concentra- tion of the acid decreases by x mol-L¹ as a result of deprotonation. The reaction stoichiometry gives us the other changes in terms of x. In the third row, write the equilibrium concentration for each substance by adding the change in its concentration (line 2) to its initial concentration (line 1). initial concentration change in concentration equilibrium concentration Acid, HA [HA] initial -X [HA] initial- H₂O+ 0 +x X Conjugate base, A 0 +x Although a change in concentration may be positive (an increase) or negative (a decrease), the value of the concentration itself must always be positive. Step 2 Substitute the equilibrium concentrations into the expression for K₂. Step 3 Solve for the value of x, which (from line 3) gives [H₂O*]. The calculation of x can often be simplified, as shown in Toolbox 51.1, by ignoring changes of less than 5% of the initial concentra- tion of the acid. A simple way to anticipate that the approximation can be used is to compare the values of K₂ and the initial con- centration of weak acid. If the numerical value of [HA]Initial is at least two orders of magnitude greater than the value of K₂ (is more than 10² times as large), then the approximation is probably valid. However, at the end of the calculation, check that x is consistent with the approximation, by calculating the percentage of acid de- protonated. If this percentage is greater than about 5%, then the exact expression for K, must be solved for x. An exact calculation requires solving a quadratic equation. If the pH is greater than 6 (but less than 7), the acid is so dilute or so weak that the autopro- tolysis of water contributes significantly to the pH. In such cases, use the procedures described in Topic 6F, which take autoprotoly- sis into account. The contribution of the autoprotolysis of water in an acidic solution can be ignored only when the calculated H₂O* concentration is substantially (about 10 times) higher than 107 mol-L, corresponding to a pH of 6 or less. Step 4 Calculate the pH from pH = -log [H₂O*]. Although pH should be calculated to the number of significant figures appropriate to the data, the answers are often considerably less reliable than that. One reason for this poor reliability is that in- teractions between the ions in solution are not taken into account. This procedure is illustrated in Examples 6D.1 and 6D.2.