16.5: Strong Acids and Bases (2023)

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    learning goals
    • Do you know the relationship between the strength of the acid or base and the magnitude of \(K_a\), \(K_b\), \(pK_a\) and \(pK_b\).
    • Understand the leveling effect.

    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, whereANDis the parent acid and A− is its conjugate base, it reads as follows:

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

    The equilibrium constant for this dissociation is the following:

    \[K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \label{16.5.2} \]

    As already mentioned, since water is the solvent, it has an activity of 1, so the term \([H_2O]\) in the equation \(\ref{16.5.2}\) actually \( \textit { a} _{H_2O}\), which is equal to 1.

    Here, too, \(H_3O^+\) can be written as \(H^+\) in equation \(\ref{16.5.3}\) for the sake of simplicity.

    \[HA_{(aq)} \rightleftharpoons H^+_{(aq)}+A^−_{(aq)} \label{16.5.3} \]

    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 hydronium ions, \(H_3O^ +\ ). The higher \(K_a\), the stronger the acid and the higher the concentration of \(H^+\) at equilibrium. Like all equilibrium constants, acid-base ionization constants are actually measured in terms of \(H^+\) or \(OH^−\) activities, so they have no units. The values ​​of \(K_a\) for various 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 \(AND\) \(K_a\) \(pK_a\) \(A^−\) \(K_b\) \(pK_b\)
    *The number in parentheses indicates the ionization step to which a polyprotic acid refers.
    iodic acid \(EY\) \(2 \ mal 10^{9}\) −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
    hydrogen sulfide \(HF\) \(6,3 \times 10^{−4}\) 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
    Piridina pura \(C_5H_5NH^+\) \(5,9 \times 10^{−6}\) 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
    Ammoniumion \(NH_4^+\) \(5,6 \times 10^{−10}\) 9.25 \(NH_3\) \(1,8 \times 10^{−5}\) 4,75
    agua \(H_2O\) \(1.0 \times 10^{−14}\) 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:

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

    Again, the activity of water is 1, so water does not appear in the expression for the equilibrium constant. The higher \(K_b\), the stronger the base and the higher the concentration of \(OH^−\) in equilibrium. The values ​​of \(K_b\) for different 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 the table \(\PageIndex{1}\).
    Hydroxid-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
    Dimethylamin \((CH_3)_2NH\) \(5,4 \times 10^{−4}\) 3.27 \((CH_3)_2NH_2^+\) \(1,9 \times 10^{−11}\) 10.73
    Methylamin \(CH_3NH_2\) \(4,6 \times 10^{−4}\) 3.34 \(CH_3NH_3^+\) \(2.2 \times 10^{−11}\) 10.66
    Trimethylamin \((CH_3)_3N\) \(6,3 \times 10^{−5}\) 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
    agua \(H_2O\) \(1.0 \times 10^{−14}\) 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 expression of the equilibrium constant for the ionization ofHCNIs the next:

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

    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:

    If you add the equations \(\ref{16.5.6}\) and \(\ref{16.5.7}\), you get
    Reaction equilibrium constants
    \(\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \) \(K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}\)
    \(\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}}\) \(K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}\)
    \(H_2O_{(l)} \rightleftHarpun H^+_{(aq)}+OH^−_{(aq)}\) \(K=K_a \times K_b=[H^+][OH^−]\)

    In this case, the sum of the reactions described by \(K_a\) and \(K_b\) is the autoionization equation of water, and the product of the two equilibrium constants is \(K_w\):

    (Video) 16.5 Strong Acids and Bases

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

    So 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 the 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 \(\ref{16.5.10}\), 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 it returns if

    \[pK_a + pK_b = 14,00 \label{16.5.16} \]

    The values ​​of \(pK_a\) and \(pK_b\) are given for various common acids and bases in the tables \(\PageIndex{1}\) and \(\PageIndex{2}\) and a in the tables E1 and E2 provide an additional extensive set of tables. Due to the use of negative logarithms, smaller values ​​of \(pK_a\) correspond to larger acid ionization constants and therefore stronger acids. For example, nitrous acid (\(HNO_2\)) with \(pK_a\) of 3.25 is an acid about a million times stronger than hydrocyanic acid (HCN) with \(pK_a\) of 9.21. On the other hand, smaller values ​​of \(pK_b\) correspond to ionization constants of larger bases and thus stronger bases.

    16.5: Strong Acids and Bases (1)

    The relative strengths of some common acids and their conjugate bases are shown graphically in Figure \(\PageIndex{1}\). The conjugate acid-base pairs are listed in order (from top to bottom) of increasing acidity, corresponding to decreasing values ​​of \(pK_a\). This ordering corresponds to decreasing the strength of the conjugate base or increasing the values ​​of \(pK_b\). Below left in the figure \(\PageIndex{2}\) are the usual strong acids; In the top right corner 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 displayed:

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

    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. Hence the ionization equilibrium is almost entirely to the right, represented by a single arrow:

    \[HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17} \]

    In contrast, acetic acid is a weak acid and water is a weak base. Consequently, aqueous solutions of acetic acid mainly contain 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:

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

    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 scale is also on the left side:

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

    (Video) 16.5 pH of Strong Acids

    All acid-base balances favor the side with the weaker acid and weaker base. Therefore, the proton binds to the stronger base.

    Example \(\PageIndex{1}\): butyrate and dimethylammonium ions
    1. Calculate \(K_b\) and \(pK_b\) of the 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.
    2. Calculate \(K_a\) and \(pK_a\) of the dimethylammonium ion (\((CH_3)_2NH_2^+\)). The ionization constant of the base \(K_b\) 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 \(\ref{16.5.10}\). The \(pK_a\) and \(pK_b\) for an acid and its conjugate base are related, as in the equations \(\ref{16.5.15}\) and \(\ref{16.5.16}\) shown. Use the relations pK = −log K and K = 10−pK(Equations \(\ref{16.5.11}\) and \(\ref{16.5.13}\)) for conversion between \(K_a\) and \(pK_a\) or \(K_b\) and \(pK_b \).

    Solution:

    We get \(pK_a\) for butyric acid and are asked to calculate \(K_b\) and \(pK_b\) for its conjugate base, the butyric ion. Since the specified value of \(pK_a\) applies to a temperature of 25°C, we can use the equation \(\ref{16.5.16}\): \(pK_a\) + \(pK_b\) = pKC= 2:00 p.m. insertion of \(pK_a\) and solving for \(pK_b\),

    \[4.83+pK_b=14.00 \nonnumber\]

    \[pK_b=14.00−4.83=9.17 \no number\]

    Como \(pK_b = −\log K_b\), \(K_b\) es \(10^{−9.17} = 6,8 \times 10^{−10}\).

    In this case we get \(K_b\) for a base (dimethylamine) and are asked to calculate \(K_a\) and \(pK_a\) for its conjugate acid, the dimethylammonium ion. Since the given initial quantity is \(K_b\) instead of \(pK_b\), we can use equation \(\ref{16.5.10}\): \(K_aK_b = K_w\). replacing \(K_b\) and \(K_w\) at 25°C and solving for \(K_a\),

    \[K_a(5,4 \times 10^{−4})=1,01 \times 10^{−14} \no number\]

    \[K_a=1,9 \times 10^{−11} \no number\]

    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:

    \[pK_b=−\log(5.4 \times 10^{−4})=3.27 \no number\]

    \[pKa+pK_b=14,00 \nonúmero\]

    \[pK_a=10.73 \no number\]

    \[K_a=10^{−pK_a}=10^{−10,73}=1,9 \times 10^{−11} \no number\]

    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.

    (Video) 16.5 strong acids and bases

    responder
    • \(K_a = 1.4 \times 10^{−4}\) for lactic acid;
    • \(pK_b\) = 10,14 e
    • \(K_b = 7.2 \times 10^{−11}\) for the lactation

    A video on how to calculate pH in strongly acidic or strongly basic solutions:Calculation of pH in solutions of strong acids or strong bases[Youtube]

    Solutions of strong acids and bases: the leveling effect

    You will notice in table \(\PageIndex{1}\) that acids like \(H_2SO_4\) and \(HNO_3\) lie above the hydronium ion, which means they have \(pK_a\) values ​​less than zero and they are stronger acids than the ion \(H_3O^+\). Recall from Chapter 4 that in virtually all oxoacids the acidic proton is attached to one of the oxygen atoms in the oxoanion. Therefore, nitric acid must be correctly written as \(HONO_2\). Unfortunately, however, formulas for oxo acids are almost always written with hydrogen on the left and oxygen on the right, resulting in \(HNO_3\). In fact, the six common strong acids we first encountered in Chapter 4 have \(pK_a\) values ​​less than zero, meaning they have a greater tendency to lose a proton than \(H_3O^+ \ ). On the other hand, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton transfer equilibria for these strong acids are far to the right, and the addition of any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of \(H_3O ^+\ ) and the conjugate base of the acid .

    Although \(K_a\) to \(HI\) is approximately 108 times greater than \(K_a\) to \(HNO_3\), the reaction of \(HI\) or \(HNO_3\) with water essentially yields a result stoichiometric solution of \(H_3O^+\) and I− or \(NO_3^−\). In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M \(H_3O^+\), regardless of the identity of the strong acid. This phenomenon is called the leveling effect: any species that is an acid stronger than the conjugate acid of water (\(H_3O^+\)) ​​levels off in aqueous solution to the strength of \(H_3O^+\ ) a because \(H_3O ^+\) is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the forces of acids such as e.gEYand HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acidity.

    One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, less likely to accept a proton than \(H_2O\). Conductivity measurements of 0.1 M solutions of HI and \(HNO_3\) in acetic acid show that HI is fully dissociated, but \(HNO_3\) is only partially dissociated and behaves like a weak acid in this solvent. This result tells us unequivocally that HI is a stronger acid than \(HNO_3\). The relative order of the strengths of the acids and the approximate values ​​of \(K_a\) and \(pK_a\) for the strong acids at the beginning of the \(\PageIndex{1}\) table were determined by measurements such as this and various non-aqueous solvents .

    emaqueous solutions, \(H_3O^+\) is the strongest acid and \(OH^−\) is the strongest base that can exist in equilibrium with \(H_2O\).

    The leveling effect also applies to solutions of strong bases: in aqueous solution, any base stronger than OH- will be balanced by the strength of OH-, since OH- is the strongest base that can exist in equilibrium with water. Salts such as \(K_2O\), \(NaOCH_3\) (sodium methoxide), and \(NaNH_2\) (sodium amide or sodium amide), whose anions are the conjugate bases of species listed in Table \( \PageIndex { 2}\), are all strong bases that essentially completely (and often violently) react with water by accepting a proton to give a solution of \(OH^−\) and the corresponding cation:

    \[K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \label{16.5.18} \]

    \[NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \label{16.5.19 } \]

    \[NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \label{1} \]

    Other examples you might come across are potassium hydride (\(KH\)) and organometallic compounds such as methyllithium (\(CH_3Li\)).

    Polyprotic acids and bases

    As you learned, polyprotic acids like \(H_2SO_4\), \(H_3PO_4\), and \(H_2CO_3\) contain more than one ionizable proton, and the protons are gradually lost. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Therefore, the strength of the acid decreases with the subsequent loss of protons and consequently \(pK_a\) increases. For example, consider \(H_2SO_4\):

    \[HSO^−_{4(aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \unnumbered\]

    The equilibrium in the first reaction is far to the right, consistent with \(H_2SO_4\) being a strong acid. In contrast, in the second reaction, significant amounts of \(HSO_4^−\) and \(SO_4^{2−}\) are present in equilibrium.

    For a polyprotic acid, the strength of the acid decreases and \(pK_a\) increases with the sequential loss of each proton.

    The hydrogen sulfate ion (\(HSO_4^−\)) is the conjugate base of \(H_2SO_4\) and the conjugate acid of \(SO_4^{2−}\). Like water, HSO4- can act as either an acid or a base depending on whether the other reactant is a stronger acid or a stronger base. On the other hand, the sulfate ion (\(SO_4^{2−}\)) is a polyprotic base capable of gradually accepting two protons:

    \[SO^{2−}_{4(aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^ - \Anzahl von nenhum\]

    \[HSO^{−}_{4(aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \label{16.6} \]

    As with any other conjugate acid-base pair, the strengths of conjugate acids and bases are related as \(pK_a\) + \(pK_b\) = pKw. For example, consider the conjugate acid-base pair \(HSO_4^−/ SO_4^{2−}\). From the table \(\PageIndex{1}\) we see that \(pK_a\) of \(HSO_4^−\) is 1.99. So \(pK_b\) of \(SO_4^{2−}\) is 14.00 − 1.99 = 12.01. Hence, sulfate is a fairly weak base while \(OH^−\) is a strong base, so the equilibrium shown in equation \(\ref{16.6}\) is on the left-hand side. The ion \(HSO_4^−\) is also a very weak base (\(pK_a\) of \(H_2SO_4\) = 2.0, \(pK_b\) of \(HSO_4^− = 14 − (−2, 0) = 16 \)), which agrees with what we expect for the conjugate base of a strong acid.

    Example \(\PageIndex{2}\)

    Predict whether the equilibrium is left or right for each reaction as written.

    • \(NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}\ )
    • \(CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)}\rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}\)

    given: balanced chemical equation

    Requested:equilibrium position

    Strategy:

    Identify the conjugate acid-base pairs in each reaction. Then use the tables \(\PageIndex{1}\) and \(\PageIndex{2}\) and the figure \(\PageIndex{2}\) to determine which is the strongest acid and base. balanceAlwaysfavors the formation of the weaker acid-base pair.

    Solution:

    The conjugate acid-base pairs are \(NH_4^+/NH_3\) and \(HPO_4^{2−}/PO_4^{3−}\). According to the tables \(\PageIndex{1}\) and \(\PageIndex{2}\), \(NH_4^+\) is a stronger acid (\(pK_a = 9.25\)) than \(HPO_4^{ 2) −}\) (pKa = 12.32), and \(PO_4^{3−}\) is a stronger base (\(pK_b = 1.68\)) than \(NH_3\ ) (\(pK_b = 4.75\ ) ). Hence the equilibrium is to the right, favoring the formation of the weaker acid-base pair:

    \[ \underset{\text{strong acid}}{NH^+_{4(aq)}} + \underset{\text{strong base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weak base}}{NH_{3(aq)}} +\underset{\text{weak acid}} {HPO^{2-}_{4 (aq )}} \no number \]

    The conjugate acid-base pairs are \(CH_3CH_2CO_2H/CH_3CH_2CO_2^−\) and \(HCN/CN^−\). According to the table \(\PageIndex{1}\), HCN is a weak acid (pKa = 9.21) and \(CN^−\) a moderately weak base (pKb = 4.79). However, propionic acid (\(CH_3CH_2CO_2H\)) is not listed in table \(\PageIndex{1}\). In such a situation, it is best to look for a similar compound whose acid-base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid (\(\ce{−CH_2CH_3}\) vs. \(\ce{−CH_3}\)), na and we might expect that the both compounds should have similar acid-base properties. In particular, we would expect the \(pK_a\) of propionic acid to be similar in size to the \(pK_a\) of acetic acid. (In fact, the \(pK_a\) of propionic acid is 4.87 compared to 4.76 for acetic acid, making propionic acid a slightly weaker acid than acetic acid.) The propionic acid must be a significantly stronger acid than the \(HCN\ ) . Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. Hence the equilibrium is to the right, favoring the formation of the weaker acid-base pair:

    \[ \underset{\text{strong acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{strong base}}{CN^-_{(aq)}}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \nonumber \]

    (Video) 16.5 Strong Acids and Bases
    Exercise \(\PageIndex{1}\)

    Predict whether the equilibrium is left or right for each reaction as written.

    1. \(H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}\)
    2. \(HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}\)
    Responder a

    links

    answer b

    links

    A video about polyprotic acids:polybasic acids[Youtube]

    Summary

    Acid-base reactions always involve two conjugate acid-base pairs. Every acid and every base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ only by a 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 (\(K_a\)). Likewise, the equilibrium constant for the reaction of a weak base with water is the ionization constant of the base (\(K_b\)). For each conjugate acid-base pair, \(K_aK_b = K_w\). Lower values ​​of \(pK_a\) correspond to higher acid ionization constants and therefore stronger acids. On the other hand, smaller values ​​of \(pK_b\) correspond to ionization constants of larger bases and thus stronger bases. At 25°C, \(pK_a + pK_b = 14.00\). Acid-base reactions always proceed in the direction in which the weaker acid-base pair is formed. No acid stronger than \(H_3O^+\) and no base stronger than \(OH^−\) can exist in aqueous solution, leading to the phenomenon known as the capping effect . Polyprotic acids (and bases) gradually lose (and gain) protons, with fully protonated species being the strongest acid and fully deprotonated species being the strongest base.

    key equations

    • Acid ionization constant: \[K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber \]
    • Basic ionization constant: \[K_b= \dfrac{[BH^+][OH^−]}{[B]} \nonumber \]
    • Relation between \(K_a\) and \(K_b\) of a conjugate acid-base pair: \[K_aK_b = K_w \nonumber \]
    • Definition von \(pK_a\): \[pKa = −\log_{10}K_a \nonumber\] \[K_a=10^{−pK_a}\nonumber\]
    • Definition von \(pK_b\): \[pK_b = −\log_{10}K_b \nonumber \] \[K_b=10^{−pK_b} \nonumber \]
    • Ratio between \(pK_a\) and \(pK_b\) of a conjugate acid-base pair: \[pK_a + pK_b = pK_w \nonumber \] \[pK_a + pK_b = 14.00 \; \text{at 25°C} \nonumber \]

    employees and tasks

    Videos

    1. 16.5 Practice Problems: Strong Acids
    (DonoChem)
    2. 16.5 Lecture Video Strong Acids and Bases
    (Webster Science)
    3. 16.5 Strong Acids and Bases
    (Johnathan Chase)
    4. 16.5 Strong Acids
    (Roxi Hulet)
    5. 16.4 pH Calculations for Strong Acids and Bases | General Chemistry
    (Chad's Prep)
    6. 16.5 Strong Bases, Weak Acids, and Weak Bases
    (Roxi Hulet)

    References

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