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learning goals
- You know the relationship between the strength of the acid or base and the size of \(K_a\), \(K_b\), \(pK_a\) and \(pK_b\).
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the original acid and A− is its conjugate base, is:
\[HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}\]
The equilibrium constant for this dissociation is as follows:
\[K=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.2}\]
The equilibrium constant for this reaction is the acid ionization constant \(K_a\), also called the acid dissociation constant:
\[K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.3}\]
Thus, numerical values of K and \(K_a\) differ as a function of water concentration (55.3 M). Here, too, \(H_3O^+\) can be written as \(H^+\) in the equation \(\ref{16.5.3}\) for simplicity. Note, however, that free \(H^+\) does not exist in aqueous solutions and that in all acid ionization reactions, a proton is donated to \(H_2O\) to form \(H^3O^+ \) . The greater \(K_a\), the stronger the acid and the greater the concentration of \(H^+\) at equilibrium. Like all equilibrium constants, acid-base ionization constants are actually measured in terms of the \(H^+\) or \(OH^−\) activities that give rise to them.no unit. \(K_a\) values for several common acids are given in the table \(\PageIndex{1}\).
| acid | \(E\) | \(K_a\) | \(pK_a\) | \(A^−\) | \(K_b\) | \(pK_b\) |
|---|---|---|---|---|---|---|
| *The number in parentheses indicates the ionization step to which a polyprotic acid refers. | ||||||
| iodiric acid | \(HELLO\) | \(2\mal 10^{9}\) | −9,3 | \(eu^-\) | \(5.5 \times 10^{−24}\) | 23.26 |
| sulfuric acid (1)* | \(H_2SO_4\) | \(1 \times 10^{2}\) | −2,0 | \(HSO_4^−\) | \(1 \times 10^{−16}\) | 16,0 |
| nitric acid | \(HNO_3\) | \(2.3 \times 10^{1}\) | −1,37 | \(NO_3^−\) | \(4.3 \times 10^{−16}\) | 15.37 |
| hydronium ions | \(H_3O^+\) | \(1.0\) | 0,00 | \(H_2O\) | \(1.0 \times 10^{−14}\) | 14h00 |
| sulfuric acid (2)* | \(HSO_4^−\) | \(1.0 \times 10^{−2}\) | 1,99 | \(SO_4^{2−}\) | \(9.8 \times 10^{−13}\) | 12.01 |
| hydrosulfuric acid | \(HF\) | \(6.3 \times 10^{−4}\) | 3.20 | \(F^−\) | \(1.6 \times 10^{−11}\) | 10,80 |
| nitric acid | \(HNO_2\) | \(5.6 \times 10^{−4}\) | 3.25 | \(NO2^-\) | \(1.8 \times 10^{−11}\) | 10.75 |
| formic acid | \(HCO_2H\) | \(1.78 \times 10^{−4}\) | 3.750 | \(HCO_2−\) | \(5.6 \times 10^{−11}\) | 10.25 |
| benzoic acid | \(C_6H_5CO_2H\) | \(6.3 \times 10^{−5}\) | 4.20 | \(C_6H_5CO_2^−\) | \(1.6 \times 10^{−10}\) | 9,80 |
| acetic acid | \(CH_3CO_2H\) | \(1.7 \times 10^{−5}\) | 4,76 | \(CH_3CO_2^−\) | \(5.8 \times 10^{−10}\) | 9.24 |
| pyridinium | \(C_5H_5NH^+\) | \(5.9 \times 10^{−6}\) | 5.23 | \(C_5H_5N\) | \(1.7 \times 10^{−9}\) | 8.77 |
| Hypochlorous acid | \(HOCl\) | \(4.0 \times 10^{−8}\) | 7.40 | \(OCl^−\) | \(2.5 \times 10^{−7}\) | 6,60 |
| hydrogen cyanide | \(HCN\) | \(6.2 \times 10^{−10}\) | 9.21 | \(CN^−\) | \(1.6 \times 10^{−5}\) | 4,79 |
| ammonia | \(NH_4^+\) | \(5.6 \times 10^{−10}\) | 9.25 | \(NH_3\) | \(1.8 \times 10^{−5}\) | 4,75 |
| water | \(H_2O\) | \(1.0 \times 10^{−14}\) | 14h00 | \(OH^−\) | \(1,00\) | 0,00 |
| Acetylene | \(C_2H_2\) | \(1 \times 10^{−26}\) | 26,0 | \(HC_2^−\) | \(1 \times 10^{12}\) | −12,0 |
| Ammonia | \(NH_3\) | \(1 \times 10^{−35}\) | 35,0 | \(NH_2^−\) | \(1 \times 10^{21}\) | −21,0 |
Weak bases react with water to produce the hydroxide ion as shown in the general equation below, where B is the original base and BH+ is its conjugate acid:
\[B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}\]
The equilibrium constant for this reaction is the fundamental ionization constant (KB), also called the base dissociation constant:
\[K_b=\dfrac{[BH^+][OH^−]}{[B]} \label{16.5.5}\]
Again, the concentration does not appear in the equilibrium constant expression. The larger \(K_b\), the stronger the base and the higher the concentration of \(OH^−\) in equilibrium. Values of \(K_b\) for different common weak bases are given in table \(\PageIndex{2}\).
| Base | \(B\) | \(K_b\) | \(pK_b\) | \(BH^+\) | \(K_a\) | \(pK_a\) |
|---|---|---|---|---|---|---|
| *As in table \(\PageIndex{1}\). | ||||||
| hydroxide ion | \(OH^−\) | \(1.0\) | 0,00* | \(H_2O\) | \(1.0 \times 10^{−14}\) | 14h00 |
| ion phosphate | \(PO_4^{3−}\) | \(2.1 \times 10^{−2}\) | 1,68 | \(HPO_4^{2−}\) | \(4.8 \times 10^{−13}\) | 12h32 |
| Dimethylamine | \((CH_3)_2NH\) | \(5.4 \times 10^{−4}\) | 3.27 | \((CH_3)_2NH_2^+\) | \(1.9 \times 10^{−11}\) | 10.73 |
| Methylamine | \(CH_3NH_2\) | \(4.6 \times 10^{−4}\) | 3.34 | \(CH_3NH_3^+\) | \(2.2 \times 10^{−11}\) | 10.66 |
| trimethylamine | \((CH_3)_3N\) | \(6.3 \times 10^{−5}\) | 4.20 | \((CH_3)_3NH^+\) | \(1.6 \times 10^{−10}\) | 9,80 |
| Ammonia | \(NH_3\) | \(1.8 \times 10^{−5}\) | 4,75 | \(NH_4^+\) | \(5.6 \times 10^{−10}\) | 9.25 |
| pyridine | \(C_5H_5N\) | \(1.7 \times 10^{−9}\) | 8.77 | \(C_5H_5NH^+\) | \(5.9 \times 10^{−6}\) | 5.23 |
| Aniline | \(C_6H_5NH_2\) | \(7.4 \times 10^{−10}\) | 9.13 | \(C_6H_5NH_3^+\) | \(1.3 \times 10^{−5}\) | 4,87 |
| water | \(H_2O\) | \(1.0 \times 10^{−14}\) | 14h00 | \(H_3O^+\) | \(1.0^*\) | 0,00 |
There is a simple relationship between the magnitude of \(K_a\) for an acid and \(K_b\) for its conjugate base. For example, consider the ionization of hydrocyanic acid (\(HCN\)) in water to produce an acidic solution and the reaction of \(CN^−\) with water to produce a basic solution:
\[HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}\]
\[CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}\]
The equilibrium constant expression for HCN ionization is as follows:
\[K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}\]
The equivalent expression for the reaction of cyanide with water is:
\[K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}\]
If we add the equations \(\ref{16.5.6}\) and \(\ref{16.5.7}\), we get the following (recall that the equilibrium constant for the sum of two reactions is the product of equilibrium constants for individual reactions):
\[\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}\]
\[\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}\]
\[H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+][OH^−]\]
In this case, the sum of the reactions described by \(K_a\) and \(K_b\) is the equation for the autoionization of water, and the product of the two equilibrium constants is \(K_w\):
\[K_aK_b = K_w \label{16.5.10}\]
Therefore, if we know \(K_a\) for an acid or \(K_b\) for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid-base pair.
As with \(pH\), \(pOH\), and pKw, we can use negative logarithms to avoid exponential notation when writing ionization constants for acids and bases by defining \(pK_a\) as follows:
\[pKa = −\log_{10}K_a \label{16.5.11}\]
\[K_a=10^{−pK_a} \label{16.5.12}\]
and \(pK_b\) as
\[pK_b = −\log_{10}K_b \label{16.5.13}\]
\[K_b=10^{−pK_b} \label{16.5.14}\]
Likewise, Equation 16.5.10 that expresses the relationship between \(K_a\) and \(K_b\) can be written in logarithmic form as follows:
\[pK_a + pK_b = pK_w \label{16.5.15}\]
At 25°C this will
\[pK_a + pK_b = 14,00 \label{16.5.16}\]
\(pK_a\) and \(pK_b\) values are given for several common acids and bases in Table 16.5.1 and Table 16.5.2, respectively, and a larger data set is provided in Tables E1 and E2. Due to the use of negative logarithms, smaller values of \(pK_a\) correspond to higher acid ionization constants and therefore to stronger acids. For example, nitrous acid (\(HNO_2\)) with \(pK_a\) of 3.25 is about 1000 times more acidic than hydrocyanic acid (HCN) with \(pK_a\) of 9.21. On the other hand, smaller values of \(pK_b\) correspond to larger base ionization constants and therefore to stronger bases.
The relative strengths of some common acids and their conjugate bases are shown in Figure 16.5. The conjugate acid-base pairs are listed in order (from top to bottom) of increasing acidity, which corresponds to decreasing values of \(pK_a\). This order corresponds to decreasing the strength of the conjugate base or increasing the \(pK_b\) values. At the bottom left of Figure 16.5.2 are the usual strong acids; top right are the most common strong bases.Note the inverse relationship between the strength of the original acid and the strength of the conjugate base. Thus, the conjugate base of a strong acid is a very weak base and the conjugate base of a very weak acid is a strong base.
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid-base reaction by following a single rule: an acid-base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows shown:
\[\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \]
In an acid-base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that essentially completely ionizes in a dilute aqueous solution to produce \(H_3O^+\) and \(Cl^−\); only insignificant amounts of \(HCl\) molecules remain undissociated. Therefore, the ionization equilibrium is almost to the right, represented by a single arrow:
\[HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}\]
In contrast, acetic acid is a weak acid and water is a weak base. Consequently, aqueous acetic acid solutions contain mostly acetic acid molecules in equilibrium with a small concentration of \(H_3O^+\) and acetate ions, and the ionization equilibrium is far to the left, as shown by these arrows. :
\[ \ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- }\]
Likewise, in the reaction of ammonia with water, the hydroxide ion is a strong base and ammonia is a weak base, while the ammonium ion is a stronger acid than water. So this balance is also on the left side:
\[H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)}\]
All acid-base equilibria favor the side with the weaker acid and weaker base. Thus, the proton binds to the stronger base.
Example \(\PageIndex{1}\): butyrate and dimethylammonium ions
- Calculate \(K_b\) and \(pK_b\) from butyration (\(CH_3CH_2CH_2CO_2^−\)). The \(pK_a\) of butyric acid at 25 °C is 4.83. Butyric acid is responsible for the bad smell of rancid butter.
- Calculate \(K_a\) and \(pK_a\) of the dimethylammonium ion (\((CH_3)_2NH_2^+\)). The basic ionization constant \(K_b\) of dimethylamine (\((CH_3)_2NH\)) is \(5.4 \times 10^{−4}\) at 25 °C.
given: \(pK_a\) e \(K_b\)
asked: corresponding \(K_b\) and \(pK_b\), \(K_a\) and \(pK_a\)
Strategy:
The constants \(K_a\) and \(K_b\) are related as shown in equation 16.5.10. The \(pK_a\) and \(pK_b\) for an acid and its conjugate base are related as shown in Equation 16.5.15 and Equation 16.5.16. Use the relations pK = −log K and K = 10−pK (Equation 16.5.11 and Equation 16.5.13) to compare \(K_a\) and \(pK_a\) or \(K_b\) and \(pK_b \ to to convert.).
Solution:
We obtain \(pK_a\) for butyric acid and are asked to calculate \(K_b\) and \(pK_b\) for their conjugate base, the butyric ion. Since the given value of \(pK_a\) applies at a temperature of 25°C, we can use equation 16.5.16: \(pK_a\) + \(pK_b\) = pKw = 14.00. Insert \(pK_a\) and solve for \(pK_b\),
\[4,83+pK_b=14,00\]
\[pK_b=14,00−4,83=9,17\]
Como \(pK_b = −\log K_b\), \(K_b\) es \(10^{−9.17} = 6,8 \times 10^{−10}\).
In this case, we obtain \(K_b\) for a base (dimethylamine) and are asked to calculate \(K_a\) and \(pK_a\) for its conjugate acid, the dimethylammonium ion. Since the specified initial size is \(K_b\) instead of \(pK_b\), we can use Equation 16.5.10: \(K_aK_b = K_w\). Substitute the values of \(K_b\) and \(K_w\) at 25°C and solve for \(K_a\),
\[K_a(5.4 \ bad 10^{−4})=1.01 \ bad 10^{−14}\]
\[K_a=1,9 \times 10^{−11}\]
Because \(pK_a\) = −log \(K_a\), \(pK_a = −\log(1.9 \times 10^{−11}) = 10.72\). We could also have converted \(K_b\) to \(pK_b\) to get the same answer:
\[pK_b=−\log(5,4 \times 10^{−4})=3,27\]
\[pKa+pK_b=14,00\]
\[pK_a=10,73\]
\[K_a=10^{−pK_a}=10^{−10.73}=1.9 \leads 10^{−11}\]
Given one of these four quantities for an acid or a base (\(K_a\), \(pK_a\), \(K_b\) or \(pK_b\)), we can calculate the other three.
Exercise \(\PageIndex{1}\): lactic acid
Lactic acid (\(CH_3CH(OH)CO_2H\)) is responsible for the pungent taste and smell of sour milk; It is also believed to cause pain in tired muscles. Its \(pK_a\) is 3.86 at 25°C. Calculate \(K_a\) for lactic acid and \(pK_b\) and \(K_b\) for lactation.
- Responder
-
\(K_a = 1.4 \times 10^{−4}\) for lactic acid;
\(pK_b\) = 10.14 and \(K_b = 7.2 \times 10^{−11}\) for lactation
Summary
Two species that differ by only one proton form a conjugate acid-base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Likewise, the equilibrium constant for the reaction of a weak base with water is the basic ionization constant (Kb). For each conjugate acid-base pair, \(K_aK_b = K_w\). Lower \(pK_a\) values correspond to higher acid ionization constants and therefore stronger acids. On the other hand, smaller values of \(pK_b\) correspond to larger base ionization constants and therefore to stronger bases. At 25°C, \(pK_a + pK_b = 14.00\). Acid-base reactions always proceed in the direction in which the weakest acid-base pair is formed.
the central theses
- The values of Ka and Kb for a conjugate acid-base pair are related by KcBravura:\[K_aK_b = K_w \]
- The conjugate base of a strong acid is a very weak base and the conjugate base of a very weak acid is a strong base.
key equations
- Acid ionization constant: \[K_a=\dfrac{[H_3O^+][A^−]}{[HA]}\]
- Basic ionization constant: \[K_b=\dfrac{[BH^+][OH^−]}{[B]} \]
- Relationship between \(K_a\) and \(K_b\) of a conjugate acid-base pair: \[K_aK_b = K_w \]
- Definition of \(pK_a\): \[pKa = −\log_{10}K_a \nonumber\] \[K_a=10^{−pK_a}\]
- Definition of \(pK_b\): \[pK_b = −\log_{10}K_b \nonumber\] \[K_b=10^{−pK_b} \]
- Relationship between \(pK_a\) and \(pK_b\) of a conjugate acid-base pair:
\[pK_a + pK_b = pK_w \]
\[pK_a + pK_b = 14,00 \; \text{bei 25°C} \]
Credits and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser u.)Chem1 Virtual Textbook
(Video) Exercícios Ka e Kb