Chapter-7-Acid-Base-Equilbiria

Chapter Seven ACID–BASE EQUILIBRIA

Acid–base theories, p. 223
Acid–base equilibria in water, p. 225 (key equations: 7.11, 7.13, 7.19)
Weak acids and bases, p. 232
Salts of weak acids and bases, p. 234 (key equations: 7.27, 7.29, 7.32, 7.36, 7.39)
Buffers, p. 238 (key equations: 7.45, 7.58)
Polyprotic acids—α values, pp. 245, 248 (key equations: 7.74–7.77)
Using spreadsheets to prepare α vs. pH plots, p. 251
Salts of polyprotic acids, p. 255 (key equations: 7.96, 7.97, 7.99, 7.100)
Logarithmic concentration diagrams, p. 266
pH calculator programs, p. 269

7.1 The Early History of Acid–Base Concepts

The word acid derives from Latin acere, meaning sour.
Bases were referred to as alkali in early history and that word derives from Arabic al-qili, the ashes of the plant saltwort, rich in sodium carbonate.
In the mid-seventeenth century it was recognized that acids and bases (called alkali in early history) tend to neutralize each other (known as the Silvio-Tachenio theory) but the concepts were vague.
In 1664 Robert Boyle published in The Experimental History of Colours that extracts of certain plants such as red roses and Brazil wood changed color reversibly as the solution was made alternately acidic and basic.
In 1675, Boyle objected to the vagueness of the Silvio–Tachenio theory and largely because of his efforts; a set of definitions emerged about acids that sought to incorporate their known properties: acids taste sour, cause limestone to effervesce, turn blue plant dyes to red, and precipitate sulfur from alkaline solutions.
Alkalies are substances that are slippery to the touch and can reverse the effect of acids.
Antoine-Laurent Lavoisier formed his own opinion of how acids come to be.
Based primarily on his observations on combustion and respiration, in which carbon is converted to carbon dioxide (the acidic nature of carbon dioxide dissolved in water was already obvious), he named the gas recently (1774) discovered by Joseph Priestley, so essential for combustion or respiration, as oxygen (from Greek, meaning acid former), since he surmised it was what created the acidic product.
Alessandro Volta announced the electric pile—an early type of battery—in 1800.
Through electrolysis, Humphry Davy discovered several new elements.
In 1807 he electrolyzed fused potash and then soda—substances that many thought to be elements—and isolated potassium and sodium.
He also similarly isolated magnesium, calcium, strontium, and barium.
Davy recognized that these alkali and alkaline earth metals combine with oxygen and form already known oxides that are highly basic, which challenged Lavoisier’s theory that oxygen was the acidifying element.
He went on to establish that hydrochloric acid, not “oxymuriatic acid” as Lavoisier called it, was acidifying; by electrolysis he isolated hydrogen and one other element, chlorine (that he so named in 1810), which until then was believed to be a compound containing oxygen.
Rather than oxygen, Davy suggested in 1815 that hydrogen may be the acidifying element.
In 1838, Justus von Liebig identified an acid as a compound of hydrogen where the hydrogen can be replaced by a metal.

7.2 Acid–Base Theories—NotAllAre Created Equal

Several acid–base theories have been proposed to explain or classify acidic and basic properties of substances.
The Arrhenius theory is applicable only to water. Other theories are more general and are applicable to other solvents or even the gas phase.

Arrhenius Theory——H+ and OH−

Arrhenius introduced a dramatically new theory that an acid is any substance that ionizes (partially or completely) in water to give hydrogen ions (which associate with the solvent to give hydronium ions, $H3O^+$ .): $HA + H2O \rightleftharpoons H_3O^+ + A^-$
A base ionizes in water to give hydroxide ions.
Weak (partially ionized) bases generally ionize as follows: $B + H_2O \rightleftharpoons BH^+ + OH^-$
Strong bases such as metal hydroxides (e.g., NaOH) dissociate as $M(OH)_n \rightarrow M^{n+} + nOH^-$
This theory is obviously restricted to water as the solvent.

Theory of Solvent Systems——Solvent Cations and Anions

In 1905, Franklin was working in liquid $NH_3$ as a solvent and noticed the similarity with acid-base behavior in water.
In 1925, Germann, working with liquid $COCl_2$ as solvent observed the similarities as well and formulated a general solvent system concept of acids and bases.
This theory recognizes the ionization of a solvent to give a cation and an anion; for example, $2H2O \rightleftharpoons H3O^+ + OH^-$ or $2NH3 \rightleftharpoons NH4 ^+ + NH_2 ^-$ .
An acid is defined as a solute that yields the characteristic cation of the solvent, while a base is a solute that yields the characteristic anion of the solvent.
Thus, $NH4Cl$ (which produces ammoniated $NH4 ^+$ , i.e., $[NH4(NH3)^+])$ , and $Cl^-$ ) is a strong acid in liquid ammonia (similar to $HCl$ in water:
$HCl + H2O \rightarrow H3O^+ + Cl^-$ )
$NaNH_2$ is a strong base in ammonia (similar to $NaOH$ in water); both of these compounds ionize to give the characteristic solvent cation and anion, respectively.
Ethanol ionizes as follows: $2C2H5OH \rightleftharpoons C2H5OH2 ^+ + C2H_5O^-$ .
Hence, sodium ethoxide, $NaOC2H5$ , is a strong base in this solvent.

Brønsted—Lowry Theory——Taking and Giving Protons

The theory of solvent systems is suitable for ionizable solvents, but it is not applicable to acid–base reactions in nonionizable solvents such as benzene or dioxane.
In 1923, Brønsted and Lowry independently described what is now known as the Brønsted–Lowry theory.
This theory states that an acid is any substance that can donate a proton, and a base is any substance that can accept a proton.
Thus, we can write a “half-reaction” $acid = H^+ + base$ $(7.1)$
The acid and base of a half-reaction are called conjugate pairs.
Free protons do not exist in solution, and there must be a proton acceptor (base) before a proton donor (acid) will release its proton.
That is, there must be a combination of two half-reactions.
Another way to look at it is that an acid is an acid because it can lose a proton. However, it cannot exhibit its acidic behavior unless there is a base present to accept the proton.
Some acid–base reactions in different solvents are illustrated in Table 7.1.
- In the first example, acetate ion is the conjugate base of acetic acid and ammonium ion is the conjugate acid of ammonia.
- The first four examples represent ionization of an acid or a base in a solvent, while the others represent a neutralization reaction between an acid and a base in the solvent.
It is apparent from the above definition that a substance cannot act as an acid unless a base is present to accept the protons.
Thus, acids will undergo complete or partial ionization in basic solvents such as water, liquid ammonia, or ethanol, depending on the basicity of the solvent and the strength of the acid.
But in neutral or “inert” solvents, ionization is insignificant.
However, ionization in the solvent is not a prerequisite for an acid–base reaction, as in the last example in the table, where picric acid reacts with aniline.

Lewis Theory——Taking and Giving Electrons

In 1923, G. N. Lewis introduced the electronic theory of acids and bases.
In the Lewis theory, an acid is a substance that can accept an electron pair and a base is a substance that can donate an electron pair.
The latter frequently contains an oxygen or a nitrogen as the electron donor.
Thus, nonhydrogen-containing substances are included as acids.
Examples of acid–base reactions in the Lewis theory are as follows:
$H^+ (solvated) + :NH3 \rightleftharpoons H:NH3$
$R2O: + AlCl3 \rightleftharpoons Cl3Al:OR2$

7.3 Acid–Base Equilibria in Water

When an acid or base is dissolved in water, it will dissociate, or ionize, the amount of ionization being dependent on the strength of the acid or the base.
A “strong” electrolyte is completely dissociated, while a “weak” electrolyte is partially dissociated.
Table 7.2 lists some common electrolytes, some strong and some weak.
Hydrochloric acid is a strong acid, and in water, its ionization is complete: $HCl + H2O \rightarrow H3O^+ + Cl^-$ $(7.2)$
The proton $H^+$ exists in water as a hydrated ion, the hydronium ion, $H_3O^+$ .
Higher hydrates probably exist, particularly $H9O4 ^+$ .
The hydronium ion is written as $H_3O^+$ for convenience and to emphasize Brønsted behavior.
Acetic acid is a weak acid, which ionizes only partially in water (a few percent): $HOAc + H2O \rightleftharpoons H3O^+ + OAc^-$ $(7.3)$
We can write an equilibrium constant for this reaction: $Ka^\circ = \frac{a{H3O^+} \cdot a{OAc^-}}{a{HOAc} \cdot a{H_2O}}$
Salt cations or anions may also partially react with water after they are dissociated.
For example, acetate ion is formed from dissociated acetate salts, to give $HOAc$ .
The activity can be thought of as representing the effective concentration of an ion (described in Chapter 6).
The effects of protons in reactions are often governed by their activities, and it is the activity that is measured by the widely used pH meter (Chapter 13).
In dilute solutions, the activity of water remains essentially constant, and is taken as unity at standard state. Therefore, Equation 7.4 can be written as $Ka^\circ = \frac{a{H3O^+} \cdot a{OAc^-}}{a_{HOAc}}$
Pure water ionizes slightly, or undergoes autoprotolysis: $2H2O \rightleftharpoons H3O^+ + OH^-$ $(7.6)$
The equilibrium constant for this is $Kw^\circ = \frac{a{H3O^+} \cdot a{OH^-}}{a{H2O}^2}$
Again, the activity of water is constant in dilute solutions (its concentration is essentially constant at ∼55.5 M), so $Kw^\circ = a{H3O^+} \cdot a{OH^-}$
Calculations are simplified if we neglect activity coefficients. This simplification results in only slight errors for dilute solutions, and we shall use molar concentrations in all our calculations.
Simplified equations for the above reactions are:
- $HCl \rightarrow H^+ + Cl^-$ $(7.9)$
- $HOAc \rightleftharpoons H^+ + OAc^-$ $(7.10)$
- $Ka = \frac{[H^+][OAc^-]}{[HOAc]}$ $(7.11)$
- $H_2O \rightleftharpoons H^+ + OH^-$ $(7.12)$
- $Kw = [H^+][OH^-]$ $(7.13)$
$K_w$ is exactly $1.00 \times 10^{-14}$ at 24◦C and even at 25◦C, to a smaller number of significant figures, it is still accurately represented as $1.0 \times 10^{-14}$ .
The product of the hydrogen ion concentration and the hydroxide ion concentration in aqueous solution is always equal to $1.0 \times 10^{-14}$ at room temperature:
$[H^+][OH^-] = 1.0 \times 10^{-14}$ $(7.14)$
In pure water, then, the concentrations of these two species are equal since there are no other sources of $H^+$ or $OH^-$ except $H_2O$ dissociation: $[H^+] = [OH^-]$
Therefore, $[H^+][H^+] = 1.0 \times 10^{-14}$ $[H^+] = 1.0 \times 10^{-7} M \equiv [OH^-]$
If an acid is added to water, we can calculate the hydroxide ion concentration if we know the hydrogen ion concentration from the acid.
But when the hydrogen ion concentration from the acid is very small, $10^{-6}$ M or less, the contribution to $[H^+]$ from the ionization of water cannot be neglected

Example 7.1

A $1.0 \times 10^{-3}$ M solution of hydrochloric acid is prepared. What is the hydroxide ion concentration?

Solution

Since hydrochloric acid is a strong electrolyte and is completely ionized, the $H^+$ concentration is $1.0 \times 10^{-3}$ M. Thus, $(1.0 \times 10^{-3})[OH^-] = 1.0 \times 10^{-14}$ $[OH^-] = 1.0 \times 10^{-11} M$

7.4 ThepHScale

The concentration of $H^+$ or $OH^-$ in aqueous solution can vary over extremely wide ranges, from 1 M or greater to $10^{-14}$ M or less.
To construct a plot of $H^+$ concentration against some variable would be very difficult if the concentration changed from, say, $10^{-1}$ M to $10^{-13}$ M. This range is common in a titration.
It is more convenient to compress the acidity scale by placing it on a logarithm basis.
The pH of a solution was defined by Sørenson as $pH = -log[H^+]$ $(7.15)$
pH is really –log $a_{H^+}$ .
This is what a pH meter (glass electrode) measures—see Chapter 13.
The minus sign is used because most of the concentrations encountered are less than 1 M, and so this designation gives a positive number.
In general, $pAnything = - log Anything$ , and this method of notation will be used later for other numbers that can vary by large amounts, or are very large or small (e.g., equilibrium constants).

Example 7.2

Calculate the pH of a $2.0 \times 10^{-3}$ M solution of HCl.

Solution

HCl is completely ionized, so $[H^+] = 2.0 \times 10^{-3} M$
$pH = - log(2.0 \times 10^{-3}) = 3 - log 2.0 = 3 - 0.30 = 2.70$

A similar definition is made for the hydroxide ion concentration: $pOH = -log[OH^-]$ $(7.16)$
Equation 7.13 can be used to calculate the hydroxyl ion concentration if the hydrogen ion concentration is known, and vice versa.
The equation in logarithm form for a more direct calculation of pH or pOH is $- log K_w = -log[H^+][OH^-] = -log[H^+] - log [OH^-]$ $(7.17)$
$pK_w = pH + pOH$ $(7.18)$
At 25◦C, $14.00 = pH + pOH$ $(7.19)$

Example 7.3

Calculate the pOH and the pH of a $5.0 \times 10^{-2}$ M solution of NaOH at 25◦C.

Solution

$[OH^-] = 5.0 \times 10^{-2} M$ $pOH = - log(5.0 \times 10^{-2}) = 2 - log 5.0 = 2 - 0.70 = 1.30$
$pH + 1.30 = 14.00$ $pH = 12.70$
or
$[H^+] = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-2}} = 2.0 \times 10^{-13} M$ $pH = - log(2.0 \times 10^{-13}) = 13 - log 2.0 = 13 - 0.30 = 12.70$

Example 7.4

Calculate the pH of a solution prepared by mixing 2.0 mL of a strong acid solution of pH 3.00 and 3.0 mL of a strong base of pH 10.00.

Solution

H $^+$ of acid solution $= 1.0 \times 10^{-3} M$
mmol H $^+$ $= 1.0 \times 10^{-3} M \times 2.0 mL = 2.0 \times 10^{-3} mmol$
pOH of base solution $= 14.00 - 10.00 = 4.00$ $[OH^-] = 1.0 \times 10^{-4} M$ $mmol OH^- = 1.0 \times 10^{-4} M \times 3.0 mL = 3.0 \times 10^{-4} mmol$
There is an excess of acid.
mmol H $^+$ $= 0.0020 - 0.0003 = 0.0017 mmol$
Total Volume $= (2.0 + 3.0) mL = 5.0 mL$
$[H^+] = 0.0017 mmol/5.0 mL = 3.4 \times 10^{-4} M$ pH $= - log 3.4 \times 10^{-4} = 4 - 0.53 = 3.47$

Example 7.5

The pH of a solution is 9.67. Calculate the hydrogen ion concentration in the solution.

Solution

$-log[H^+] = 9.67$ $[H^+] = 10^{-9.67} = 10^{-10} \times 10^{0.33}$ $[H^+] = 2.1 \times 10^{-10} M$

When $[H^+] = [OH^-]$ , then a solution is said to be neutral.
If [H^+] > [OH^-], then the solution is acidic.
If [H^+] < [OH^-], the solution is alkaline.
The hydrogen ion and hydroxide ion concentrations in pure water at 25◦C are each $10^{-7}$ M, and the pH of water is 7. A pH of 7 is therefore neutral.
Values of pH that are greater than this are alkaline, and pH values less than this are acidic.
The reverse is true of pOH values.
A pOH of 7 is also neutral.
Note that the product of $[H^+]$ and $[OH^-]$ is always $10^{-14}$ at 25◦C, and the sum of pH and pOH is always 14.
If the temperature is other than 25◦C, then $K_w$ is different from $1.0 \times 10^{-14}$ , and a neutral solution will have other than $10^{-7}$ M $H^+$ and $OH^-$ (see below).
A negative pH only means that the hydrogen ion concentration is greater than 1 M.
If the concentration of an acid or base is much less than $10^{-7}$ M, then its contribution to the acidity or basicity will be negligible compared with the contribution from water.
If the concentration of the acid or base is around $10^{-7}$ M, then its contribution is not negligible and neither is that from water; hence the sum of the two contributions must be taken.

Example 7.6

Calculate the pH and pOH of a $1.0 \times 10^{-7}$ M solution of HCl.

Solution

Equilibria:
$HCl \rightarrow H^+ + Cl^-$ $H2O \rightleftharpoons H^+ + OH^-$ $[H^+][OH^-] = 1.0 \times 10^{-14}$ $\ [H^+]{H2Odiss.} = [OH^-]{H2Odiss.} = x$ Since the hydrogen ions contributed from the ionization of water are not negligible compared to the HCl added, $[H^+] = C{HCl} + [H^+]{H2Odiss.}$
Then, $([H^+]_{HCl} + x)(x) = 1.0 \times 10^{-14}$ $(1.00 \times 10^{-7} + x)(x) = 1.0 \times 10^{-14}$ $x^2 + 1.00 \times 10^{-7}x - 1.0 \times 10^{-14} = 0$
Using the quadratic equation to solve [see Appendix B] or the use of Excel Goal Seek (Section 6.11),
$x = \frac{-1.00 \times 10^{-7} \pm \sqrt{1.0 \times 10^{-14} + 4(1.0 \times 10^{-14})}}{2} = 6.2 \times 10^{-8} M$
Therefore, the total $H^+$ concentration $= (1.00 \times 10^{-7} + 6.2 \times 10^{-8}) = 1.62 \times 10^{-7} M$ :
$pH = - log 1.62 \times 10^{-7} = 7 - 0.21 = 6.79$ $pOH = 14.00 - 6.79 = 7.21$
or, since $[OH^-] = x$ ,
$pOH = - log(6.2 \times 10^{-8}) = 8 - 0.79 = 7.21$
Note that, owing to the presence of the added $H^+$ , the ionization of water is suppressed by 38% by the common ion effect (Le Chatelier’s principle).

7.5 pHat Elevated Temperatures:BloodpH

At 100◦C, for example, $Kw = 5.5 \times 10^{-13}$ , and a neutral solution has $[H^+] = [OH^-] = \sqrt{5.5 \times 10^{-13}} = 7.4 \times 10^{-7} M$ $pH = pOH = 6.13$ $pKw = 12.26 = pH + pOH$
At 37◦C, $Kw = 2.5 \times 10^{-14}$ and $pKw = 13.60$ .
- The pH (and pOH) of a neutral solution is 13.60/2 = 6.80.
- The hydrogen ion (and hydroxide ion) concentration is $\sqrt{2.5 \times 10^{-14}} = 1.6 \times 10^{-7}$ M.
Since a neutral solution at 37◦C would have pH 6.8, a blood pH of 7.4 is more alkaline at 37◦C by 0.2 pH units than it would be at 25◦C.
This is important when one considers that a change of 0.3 pH units in the body is extreme.
The hydrochloric acid concentration in the stomach is about 0.1 to 0.02 M. Since $pH = - log[H^+]$ , the pH at 0.02 M would be 1.7.
It will be the same regardless of the temperature since the hydrogen ion concentration is the same (neglecting solvent volume changes), and the same pH would be measured at either temperature.
But, while the pOH would be $14.0 - 1.7 = 12.3$ at 25◦C, it is $13.6 - 1.7 = 11.9$ at 37◦C.
Not only does the temperature affect the ionization of water in the body and therefore change the pH of neutrality, it also affects the ionization constants of the acids and bases from which the buffer systems in the body are derived.
As we shall see later in the chapter, this influences the pH of the buffers, and so a blood pH of 7.4 measured at 37◦C will not be the same when measured at room temperature, in contrast to the stomach pH, whose value was determined by the concentration of a strong acid.

7.6 WeakAcidsandBases—What Is thepH?

We have limited our calculations so far to strong acids and bases in which ionization is assumed to be complete.
Weak acids (or bases) are only partially ionized.
While mineral (inorganic) acids and bases such as $HCl$ , $HClO4$ , $HNO3$ , and $NaOH$ are strong electrolytes that are totally ionized in water; most organic acids and bases, as found in clinical applications, are weak.
The ionization constant can be used to calculate the amount ionized and, from this, the pH.
The acidity constant for acetic acid at 25◦C is $1.75 \times 10^{-5}$ :
$\frac{[H^+][OAc^-]}{[HOAc]} = 1.75 \times 10^{-5}$ $(7.20)$
When acetic acid ionizes, it dissociates to equal portions of $H^+$ and $OAc^-$ by such an amount that the computation on the left side of Equation 7.20 will always be equal to $1.75 \times 10^{-5}$ :
$HOAc \rightleftharpoons H^+ + OAc^-$ $(7.21)$
If the original concentration of acetic acid is C and the concentration of ionized acetic acid species ( $H^+$ and $OAc^-$ ) is x, then the final concentration for each species at equilibrium is given by
$HOAc \rightleftharpoons H^+ + OAc^-$ $(C - x) \quad x \quad x$ $(7.22)$

Example 7.7

Calculate the pH and pOH of a $1.00 \times 10^{-3}$ M solution of acetic acid.

Solution

$HOAc \rightleftharpoons H^+ + OAc^-$
The concentrations of the various species in the form of an ICE table are as follows:
$[HOAc] \quad [H^+] \quad [OAc^-]$
Initial $1.00 \times 10^{-3} \quad 0 \quad 0$
Change (x = mmol/mL HOAc ionized) $\ -x \quad +x \quad +x$
Equilibrium $1.00 \times 10^{-3} - x \quad x \quad x$
From Equation 7.20
$\frac{(x)(x)}{1.00 \times 10^{-3} - x} = 1.75 \times 10^{-5}$

If C{HA} > 100Ka, x can be neglected compared to $C_{HA}$ . The expression may be simplified by neglecting x compared with C ( $10^{-3}$ M in this case).
$\frac{x^2}{1.00 \times 10^{-3}} = 1.75 \times 10^{-5}$
$x = 1.32 \times 10^{-4} M \equiv [H^+]$
Therefore,
$pH = - log(1.32 \times 10^{-4}) = 4 - log 1.32 = 4 - 0.12 = 3.88$
$pOH = 14.00 - 3.88 = 10.12$

Example 7.8

The basicity constant $K_b$ for ammonia is $1.75 \times 10^{-5}$ at 25◦C. Calculate the pH and pOH for a $1.00 \times 10^{-3}$ M solution of ammonia.

Solution

$NH3 + H2O \rightleftharpoons NH4 ^+ + OH^-$ $(1.00 \times 10^{-3} - x) \quad x \quad x$ $\frac{[NH4 ^+][OH^-]}{[NH_3]} = 1.75 \times 10^{-5}$
The same rule applies for the approximation applied for a weak acid. Thus, $\frac{(x)(x)}{1.00 \times 10^{-3}} = 1.75 \times 10^{-5}$
$x = 1.32 \times 10^{-4} M = [OH^-]$
$pOH = - log 1.32 \times 10^{-4} = 3.88$ $pH = 14.00 - 3.88 = 10.12$

7.7 Saltsof WeakAcidsandBases—TheyAren’tNeutral

The salt of a weak acid, for example, $NaOAc$ , is a strong electrolyte and completely ionizes.
In addition, the anion of the salt of a weak acid is a Brønsted base, which will accept protons. It partially hydrolyzes in water (a Brønsted acid) to form hydroxide ion and the corresponding undissociated acid.
For example, $OAc^- + H_2O \rightleftharpoons HOAc + OH^-$ $(7.23)$
This ionization is also known as hydrolysis of the salt ion.
Because it hydrolyzes, sodium acetate is a weak base (the conjugate base of acetic acid).
We can write an equilibrium constant: $KH = Kb = \frac{[HOAc][OH^-]}{[OAc^-]}$ $(7.24)$
$K_H$ is called the hydrolysis constant of the salt and is the same as the basicity constant.
We will use $K_b$ to emphasize that these salts are treated the same as for any other weak base.
The value of $Kb$ can be calculated from $Ka$ of acetic acid and $Kw$ if we multiply both the numerator and denominator by $[H^+]$ : $Kb = \frac{[HOAc][OH^-]}{[OAc^-]} = \frac{[HOAc][OH^-][H^+]}{[OAc^-][H^+]}$ $(7.25)$
$Kb = \frac{Kw}{K_a} = \frac{1.0 \times 10^{-14}}{1.75 \times 10^{-5}} = 5.7 \times 10^{-10}$ $(7.26)$
The product of $Ka$ of any weak acid and $Kb$ of its conjugate base is always equal to $Kw$ :
$KaKb = K_w$ $(7.27)$
For any salt of a weak acid HA that hydrolyzes in water,
$A^- + H2O \rightleftharpoons HA + OH^-$ $(7.28)$ $\frac{[HA][OH^-]}{[A^-]} = \frac{Kw}{Ka} = Kb$ $(7.29)$
The pH of such a salt (a Brønsted base) is calculated in the same manner as for any other weak base.
When the salt hydrolyzes, it forms an equal amount of HA and $OH^-$ .
If the original concentration of A $^-$ is CA $^-$ , then
$A^- + H2O \rightleftharpoons HA + OH^-$ $(C{A^-} - x) \quad x \quad x$ $(7.30)$
The quantity x can be neglected compared to CA $^-$ if C{A^-}> 100Kb, which will generally be the case for such weakly ionized bases.
We can solve for the $OH^-$ concentration using Equation 7.30:
$\frac{[OH^-][OH^-]}{ C{A^-}} = \frac{Kw}{Ka} = Kb$ $(7.31)$
$[OH^-] = \sqrt{\frac{Kw}{Ka} \cdot C{A^-}} = \sqrt{ Kb \cdot C_{A^-}}$

Example 7.9

Calculate the pH of a 0.10 M solution of sodium acetate.

Solution

Write the equilibria
$NaOAc \rightarrow Na^+ + OAc^-(ionization)$
$OAc^- + H2O \rightleftharpoons HOAc + OH^-(hydrolysis)$ Write the equilibrium constant $$\frac{[HOAc][OH^-]}{[OAc^-]} = Kb = \frac{Kw}{Ka}