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7.12: Relationship between Ka, Kb, pKa, and pKb

  • Page ID
    221518
  • Learning Objectives

    • To know the relationship between acid or base strength and the magnitude 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 parent acid and A− is its conjugate base, is as follows:

    \[HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}\]

    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 the numerical values of K and \(K_a\) differ by the concentration of water (55.3 M). Again, for simplicity, \(H_3O^+\) can be written as \(H^+\) in Equation \(\ref{16.5.3}\). Keep in mind, though, that free \(H^+\) does not exist in aqueous solutions and that a proton is transferred to \(H_2O\) in all acid ionization reactions to form \(H^3O^+\). The larger the \(K_a\), the stronger the acid and the higher the \(H^+\) concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of \(H^+\) or \(OH^−\), thus making them unitless. The values of \(K_a\) for a number of common acids are given in Table \(\PageIndex{1}\).

    Table \(\PageIndex{1}\): Values of \(K_a\), \(pK_a\), \(K_b\), and \(pK_b\) for Selected Acids (\(HA\) and Their Conjugate Bases (\(A^−\))
    Acid \(HA\) \(K_a\) \(pK_a\) \(A^−\) \(K_b\) \(pK_b\)
    *The number in parentheses indicates the ionization step referred to for a polyprotic acid.
    hydroiodic acid \(HI\) \(2 \times 10^{9}\) −9.3 \(I^−\) \(5.5 \times 10^{−24}\) 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 ion \(H_3O^+\) \(1.0\) 0.00 \(H_2O\) \(1.0 \times 10^{−14}\) 14.00
    sulfuric acid (2)* \(HSO_4^−\) \(1.0 \times 10^{−2}\) 1.99 \(SO_4^{2−}\) \(9.8 \times 10^{−13}\) 12.01
    hydrofluoric acid \(HF\) \(6.3 \times 10^{−4}\) 3.20 \(F^−\) \(1.6 \times 10^{−11}\) 10.80
    nitrous 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 ion \(C_5H_5NH^+\) \(5.9 \times 10^{−6}\) 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
    hydrocyanic acid \(HCN\) \(6.2 \times 10^{−10}\) 9.21 \(CN^−\) \(1.6 \times 10^{−5}\) 4.79
    ammonium ion \(NH_4^+\) \(5.6 \times 10^{−10}\) 9.25 \(NH_3\) \(1.8 \times 10^{−5}\) 4.75
    water \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 \(OH^−\) \(1.00\) 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 following general equation, where B is the parent 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 base ionization constant (Kb), also called the base dissociation constant:

    \[K_b=\dfrac{[BH^+][OH^−]}{[B]} \label{16.5.5}\]

    Once again, the concentration does not appear in the equilibrium constant expression.. The larger the \(K_b\), the stronger the base and the higher the \(OH^−\) concentration at equilibrium. The values of \(K_b\) for a number of common weak bases are given in Table \(\PageIndex{2}\).

    Table \(\PageIndex{2}\): Values of \(K_b\), \(pK_b\), \(K_a\), and \(pK_a\) for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
    Base \(B\) \(K_b\) \(pK_b\) \(BH^+\) \(K_a\) \(pK_a\)
    *As in Table \(\PageIndex{1}\).
    hydroxide ion \(OH^−\) \(1.0\) 0.00* \(H_2O\) \(1.0 \times 10^{−14}\) 14.00
    phosphate ion \(PO_4^{3−}\) \(2.1 \times 10^{−2}\) 1.68 \(HPO_4^{2−}\) \(4.8 \times 10^{−13}\) 12.32
    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 \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 \(H_3O^+\) \(1.0^*\) 0.00

    There is a simple relationship between the magnitude of \(K_a\) for an acid and \(K_b\) for its conjugate base. Consider, for example, 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 the ionization of HCN is as follows:

    \[K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}\]

    The corresponding expression for the reaction of cyanide with water is as follows:

    \[K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}\]

    If we add Equations \(\ref{16.5.6}\) and \(\ref{16.5.7}\), we obtain the following (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the 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 \(K_w\):

    \[K_aK_b = K_w \label{16.5.10}\]

    Thus if we know either \(K_a\) for an acid or \(K_b\) for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.

    Just as with \(pH\), \(pOH\), and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, 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

    \[pK_b = −\log_{10}K_b \label{16.5.13}\]

    \[K_b=10^{−pK_b} \label{16.5.14}\]

    Similarly, Equation 16.5.10, which 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 becomes

    \[pK_a + pK_b = 14.00 \label{16.5.16}\]

    The values of \(pK_a\) and \(pK_b\) are given for several common acids and bases in Table 16.5.1 and Table 16.5.2, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (\(HNO_2\)), with a \(pK_a\) of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a \(pK_a\) of 9.21. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases.

    The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.5. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of \(pK_a\). This order corresponds to decreasing strength of the conjugate base or increasing values of \(pK_b\). At the bottom left of Figure 16.5.2 are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent 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:

    \[\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \]

    In an acid–base reaction, the proton always reacts with the stronger base.

    For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce \(H_3O^+\) and \(Cl^−\); only negligible amounts of \(HCl\) molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as 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 solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of \(H_3O^+\) and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:

    \[ \ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- }\]

    Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:

    \[H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)}\]

    All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.

    Example \(\PageIndex{1}\): Butyrate and Dimethylammonium Ions

    1. Calculate \(K_b\) and \(pK_b\) of the butyrate ion (\(CH_3CH_2CH_2CO_2^−\)). The \(pK_a\) of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
    2. Calculate \(K_a\) and \(pK_a\) of the dimethylammonium ion (\((CH_3)_2NH_2^+\)). The base ionization constant \(K_b\) of dimethylamine (\((CH_3)_2NH\)) is \(5.4 \times 10^{−4}\) at 25°C.

    Given: \(pK_a\) and \(K_b\)

    Asked for: 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 relationships pK = −log K and K = 10−pK (Equation 16.5.11 and Equation 16.5.13) to convert between \(K_a\) and \(pK_a\) or \(K_b\) and \(pK_b\).

    Solution:

    We are given the \(pK_a\) for butyric acid and asked to calculate the \(K_b\) and the \(pK_b\) for its conjugate base, the butyrate ion. Because the \(pK_a\) value cited is for a temperature of 25°C, we can use Equation 16.5.16: \(pK_a\) + \(pK_b\) = pKw = 14.00. Substituting the \(pK_a\) and solving for the \(pK_b\),

    \[4.83+pK_b=14.00\]

    \[pK_b=14.00−4.83=9.17\]

    Because \(pK_b = −\log K_b\), \(K_b\) is \(10^{−9.17} = 6.8 \times 10^{−10}\).

    In this case, we are given \(K_b\) for a base (dimethylamine) and asked to calculate \(K_a\) and \(pK_a\) for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is \(K_b\) rather than \(pK_b\), we can use Equation 16.5.10: \(K_aK_b = K_w\). Substituting the values of \(K_b\) and \(K_w\) at 25°C and solving for \(K_a\),

    \[K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14}\]

    \[K_a=1.9 \times 10^{−11}\]

    Because \(pK_a\) = −log \(K_a\), we have \(pK_a = −\log(1.9 \times 10^{−11}) = 10.72\). We could also have converted \(K_b\) to \(pK_b\) to obtain the same answer:

    \[pK_b=−\log(5.4 \times 10^{−4})=3.27\]

    \[pKa+pK_b=14.00\]

    \[pK_a=10.73\]

    \[K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11}\]

    If we are given any 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 thought to produce soreness in fatigued muscles. Its \(pK_a\) is 3.86 at 25°C. Calculate \(K_a\) for lactic acid and \(pK_b\) and \(K_b\) for the lactate ion.

    Answer

    \(K_a = 1.4 \times 10^{−4}\) for lactic acid;

    \(pK_b\) = 10.14 and \(K_b = 7.2 \times 10^{−11}\) for the lactate ion

     

     

    Summary

    Two species that differ by only a proton constitute 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). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, \(K_aK_b = K_w\). Smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases. At 25°C, \(pK_a + pK_b = 14.00\). Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. 

    Key Takeaways

    • The Ka and Kb values for a conjugated acid–base pairs are related through the Kw value: \[K_aK_b = K_w \]
    • 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: \[K_a=\dfrac{[H_3O^+][A^−]}{[HA]}\]
    • Base 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:

    \[pK_a + pK_b = pK_w \]

    \[pK_a + pK_b = 14.00 \; \text{at 25°C} \]

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