Module 7.2: Acids and Bases

7.2.1 Definitions

I Arrhenius definition of acids and bases is the most specific.

  1. an Arrhenius acid must produce hydrogen ions (H+\textrm{H}^{+}) in water. Examples include HCl\textrm{HCl}, H2SO4\textrm{H}_{2} \textrm{SO}_{4}, HNO3\textrm{HNO}_{3}

  2. an Arrhenius base must produce hydroxide ions (OH\textrm{OH}^{-}) in water. Examples include NaOH\textrm{NaOH}, NH3\textrm{NH}_{3}

II Brønsted-Lowry definition of acids and bases is much broader.

  1. a Brønsted acid is a proton donor. Examples include Fe(H2O)63+\textrm{Fe}(\textrm{H}_{2}\textrm{O})_{6}^{3+}, water, etc.

  2. a Brønsted base is a proton acceptor. Examples include water, carbonate, etc.

  3. All Arrhenius acids and bases are Brønsted acids and bases.

  4. Bronsted acids and bases are not limited to aqueous reactions.

    1. Gas-phase reactions and reactions in nonaqueous solutions can also occur.

  5. Unless stated, assume all acid-base reactions are Brønsted acids and bases.

III Lewis definition of acids and bases is the broadest.

  1. a Lewis acid is an electron pair acceptor. Examples include BH3\textrm{BH}_{3}, metal ions such as in Ni(CO)4\textrm{Ni}(\textrm{CO})_{4}

  2. a Lewis base is an electron pair donor. Examples include F\textrm{F}^{-}, BF4\textrm{BF}_{4}^{-}, etc.

  3. All Brønsted acids and bases are Lewis acids and bases.

    1. A\textrm{A} in acid HA\textrm{HA} accepts the bonding electron pair of HA\textrm{H}-\textrm{A}; forming conjugate acid.

    2. H2O\textrm{H}_{2}\textrm{O} uses its lone pair to form a coordinate covalent bond with H+\textrm{H}^{+}.

7.2.2 Fundamentals

IV When an acid dissociates in water, a proton transfer reaction occurs;

HCl(aq)+H2O(l)H3O+(aq)+Cl(aq)\textrm{HCl} (\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \rightarrow \textrm{H}_{3}\textrm{O}^{+}(\textrm{aq}) + \textrm{Cl}^{-} (\textrm{aq})

  1. conjugate acid: an acid that results from protonating a base.

    1. In the reaction above, H3O+\textrm{H}_{3}\textrm{O}^{+} is the conjugate acid.

  2. conjugate base: a base that results from deprotonating an acid.

    1. In the reaction above, Cl\textrm{Cl}^{-} is the conjugate base.

  3. All acids in aqueous solution will produce hydronium ions; how much it produces (determined by the extent of the equilibrium) will determine its acid strength in aqueous media.

    1. If an acid does not fully deprotonate in water, or does not donate most of its protons to water molecules, it is considered a weak acid in water. In other words, weak acids are rather stable species in aqueous solution.

    2. If an acid fully dissociates in water, or donates most of its protons to water molecules, it is considered a strong acid in water. In other words, strong acids are unstable species in aqueous solution.

    3. Acid strength depends heavily on the solvent. This is because as the acid deprotonates, the conjugate acid of the solvent will determine the strength of the acid in solution

    4. All reactions like to go from unstable to stable; therefore, deprotonation of a stable, weak acid is unlikely to occur in comparison to an unstable, strong acid.

V Water is a special molecule; it can undergo autoprotolysis, donating a proton to another water molecule.

H2O(l)+H2O(l)H3O+(aq)+OH(aq)\textrm{H}_{2}\textrm{O} (\textrm{l}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \leftrightharpoons \textrm{H}_{3}\textrm{O}^{+} (\textrm{aq}) + \textrm{OH}^{-} (\textrm{aq})

  1. The equilibrium constant for this reaction, known as the autoprotolysis constant

    (Kw\textrm{K}_{w}) is Kw=[H3O+][OH]=1.0×1014@298K\textrm{K}_{w}=[\textrm{H}_{3}\textrm{O}^{+}][\textrm{OH}^{-}]=1.0 \times 10^{-14} \, @ 298 \textrm{K} (Eq. 112)

  2. Eq. 1 holds so much significance that it cannot be simply described with words

    1. The relation above illustrates that acidity and basicity are related

    2. If you can find the concentration of one species, you can find the other

    3. Note that temperature will affect the equilibrium constant

VI The logarithmic pX scale is frequently used everywhere:

  1. pK scale: defined as pK=logK\textrm{pK} = - \textrm{logK}

    1. Equilibrium constants (such as the acidity constant below) are often used in this scale to prevent using extremely large or small numbers.

  2. pH scale: defined as pH=log[H3O+\textrm{pH} = - \textrm{log}[\textrm{H}_{3}\textrm{O}^{+}]

    1. At 298 K, when pH<7\textrm{pH} \lt 7, the solution is considered acidic

    2. At 298 K, when pH=7\textrm{pH} = 7, the solution is considered neutral

    3. At 298 K, when pH>7\textrm{pH} \gt 7, the solution is considered basic

  3. pOH scale: defined as pOH=log[OH\textrm{pOH} = - \textrm{log}[\textrm{OH}^{-}]

    1. Very similar to pH scale, except reverse. When pOH<7\textrm{pOH} \lt 7, the solution is basic.

    2. From Eq. 112, we can state that pH+pOH=14\textrm{pH} + \textrm{pOH} = 14 at 298 K.

    3. Autoprotolysis is endothermic; as temperature increases, Ka\textrm{K}_{a} will decrease. However, as temperature increases, water does not become acidic or basic!

VII Acidity constant (Ka\textrm{K}_{a}) determines the amount of deprotonation of an acid in solution

  1. for a general reaction HA(aq)+H2O(l)A(aq)+H3O+(aq)\textrm{HA}(\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \leftrightharpoons \textrm{A}^{-}(\textrm{aq}) + \textrm{H}_{3}\textrm{O}^{+}(\textrm{aq}) the acidity constant is equal to: Ka=[H3O+][A][HA]\textrm{K}_{a} = \frac{[\textrm{H}_{3}\textrm{O}^{+}][\textrm{A}^{-}]}{[\textrm{HA}]} (Eq. 113)

    1. A strong acid will have a very large acidity constant

    2. A weak acid will have a very small acidity constant

VIII Basicity constant (Kb) determines the amount of protonation of a base in solution

  1. for a general reaction B(aq)+H2O(l)HB+(aq)+OH(aq)\textrm{B}(\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \leftrightharpoons \textrm{HB}^{+}(\textrm{aq}) + \textrm{OH}^{-}(\textrm{aq}) the acidity constant is equal to: Ka=[H3O+][A]HA\textrm{K}_{a} = \frac{[\textrm{H}_{3}\textrm{O}^{+}][\textrm{A}^{-}]}{\textrm{HA}} (Eq. 114)

    1. A strong base will have a very large basicity constant

    2. A weak base will have a very small basicity constant

IX Multiplying the acidity constant and basicity constant results in the autoprotolysis constant

  1. This allows for the conversion between acidity constant and basicity constant

    1. recall that the protonation of a base results in a conjugate acid

    2. similarly, the deprotonation of an acid results in a conjugate base

    3. Use either Ka=KwKb\textrm{K}_{a} = \frac{\textrm{K}_{w}}{\textrm{K}_{b}} or Kb=KwKa\textrm{K}_{b} = \frac{\textrm{K}_{w}}{\textrm{K}_{a}}

  2. The pK scale is used for acidity and basicity constants as well (especially pKa\textrm{pK}_{a} for organic chemistry)

    1. pKa+pKb=14\textrm{pK}_{a} + \textrm{pK}_{b} = 14

X The four factors used to compare the strength of two binary acids:

  1. Atom. What element is the negative charge of the conjugate base on?

    1. if two elements are on the same period: electronegativity dominates. The more electronegative atom will be more comfortable (stable) bearing the negative charge.

    2. if two elements are on the same group: size dominates. The larger size of orbitals allows a wider spread of negative charge, and will be more comfortable (stable) with the negative charge.

  2. Resonance. If the two elements are the same, check if any has resonance.

    1. resonance will not only lower the energy of the overall molecule, but “spread” the negative charge over many atoms. This is known as delocalization.

    2. For oxoacids, the more oxygen atoms in a molecule leads to more delocalization of charge between those oxygen atoms.

    3. For oxoacids of chlorine in particular, pKa85n\textrm{pK}_{a} ≈ 8 - 5\textrm{n}, where n is the number of oxygen atoms attached to chlorine. This is known as Bell's rule. (Ref. Figure 77)

  3. Induction. If there are electronegative elements attached to the atom with the negative charge, it may withdraw electron density away from the negative charge of the conjugate base, stabilizing it.

    1. CF3-\textrm{CF}_{3} groups are heavily influenced by induction and withdraw lots of electron density away from the atom with the negative charge.

  4. Orbitals. When considering the VB Theory, sp\textrm{sp} orbitals have their electrons considerably closer to the nucleus (at a lower energy) than sp2\textrm{sp}^{2} or sp3\textrm{sp}^{3} orbitals, stabilizing the negative charge.

    1. Coulomb's Law states that the closer the distance between the two charges, the stronger the attraction is.

    2. as p orbitals are higher in energy than s orbitals, the sp\textrm{sp} orbitals are much lower in energy in comparison to the sp3\textrm{sp}^{3} orbitals.

XI Dissolving some salts may affect the pH of the solution

  1. salts containing conjugate acids or bases from weak acids or bases will affect the pH.

    1. examples are F\textrm{F}^{-}, HCO3\textrm{HCO}_{3}^{-}, PO43\textrm{PO}_{4}^{3-}, NH4+\textrm{NH}_{4}^{+}, etc.

  2. salts containing highly charged metal cations may form acidic solutions.

    1. These cations act as a Lewis acid and water acts as a Lewis base.

    2. The highly charged metal cations withdraw electron density away from the hydrogen atoms, making the species more acidic. The more capable the Lewis acid is at stabilizing the negative charge, the stronger the acid will be.

    3. Some examples are Fe3+\textrm{Fe}^{3+}, Cr3+\textrm{Cr}^{3+}, Al3+\textrm{Al}^{3+}, Cu2+\textrm{Cu}^{2+}, etc.

    4. A general reaction of aquated cations acting as Bronsted acids is shown below:

      [M(OH2)6]n+(aq)+H2O(l)[M(OH2)5(OH)](n1)+(aq)+H3O+(aq)[\textrm{M}(\textrm{OH}_{2})_{6}]^{n+} (\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \leftrightharpoons [\textrm{M}(\textrm{OH}_{2})_{5}(\textrm{OH})]^{(n-1)+}(\textrm{aq}) + \textrm{H}_{3}\textrm{O}^{+}(\textrm{aq})

  3. Oxides can be acidic, basic, and amphoteric.

    1. an acidic oxide is often molecular and acts as Lewis acids; Examples include CO2\textrm{CO}_{2}, SO2\textrm{SO}_{2}.

    2. a basic oxide is an ionic compound that forms hydroxide ions by reacting with water

    3. an amphoteric oxide is a substance that reacts with both acids and bases; Examples include Al2O3\textrm{Al}_{2}\textrm{O}_{3}.

7.2.3 Calculations

XII Formal concentration (F): the concentration of an acid before any deprotonation occurs

  1. For a monoprotic weak acid, [HA]0[\textrm{HA}]_{0} is the same thing as formal concentration

  2. moles deprotonated + moles protonated must equal the formal concentration

XIII fraction of dissociation (α\alpha): the amount of acid that is deprotonated in solution:

α=[A][A]+[HA]=xF\alpha = \frac{[\textrm{A}^{-}]}{[\textrm{A}^{-}] + [\textrm{HA}]} = \frac{\textrm{x}}{\textrm{F}} (Eq. 115)

  1. If you multiply α\alpha by 100, you get percentage deprotonation.

  2. If α\alpha is small (less than 5%), it is considered as weak electrolytes (acids).

  3. Weak acids tend to dissociate further when their concentration is low.

  4. For strong acids, we often approximate that F[H+]\textrm{F} \approx [\textrm{H}^{+}], as there are so little protonated species.

XIV To calculate the pH of a weak acid:

HA(aq)+H2O(l)A(aq)+H3O+(aq)\textrm{HA}(\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{l}) \leftrightharpoons \textrm{A}^{-}(\textrm{aq}) + \textrm{H}_{3}\textrm{O}^{+}(\textrm{aq})

If we plug the equilibrium concentrations into our acidity constant equation,

Xa=x2Fx\textrm{X}_{a} = \frac{\textrm{x}^{2}}{\textrm{F} - \textrm{x}}

We can solve in terms of x, which is equal to our hydronium cation concentration, by approximating that F>>x\textrm{F} \gt \gt \textrm{x} (If α\alpha is fairly large, this approximation is invalid. 5% is a good cutline). Our general equation for the proton concentration for a monoprotic acid is then:

[H+]=F×Ka[\textrm{H}^{+}] = \sqrt{\textrm{F} \times \textrm{K}_{a}} (Eq. 116)

  1. A similar equation can be derived for weak bases. [OH]=F×Kb[\textrm{OH}^{-}] = \sqrt{\textrm{F} \times \textrm{K}_{b}} (Eq. 117)

  2. Note that both of these equations are invalid in the case of very dilute acids or very weak acids, as it does not take autoprotolysis into account.

XV A polyprotic acid is a species that can donate more than one proton.

  1. Conversely, a polyprotic base is a species that can accept more than one proton.

  2. With each successive deprotonation of polyprotic acids, the acidity constant decreases as it is much harder to remove a proton from a negatively charged species (as it would make it more negative)

    1. similar to how ionization energy increases with each step
  3. If Ka1\textrm{K}_{a1} and Ka2\textrm{K}_{a2} are more than three magnitudes apart, the deprotonation occurring from Ka2\textrm{K}_{a2} is negligible and can be calculated like a monoprotic acid.

  4. If Kb1\textrm{K}_{b1} and Kb2\textrm{K}_{b2} are more than three magnitudes apart, the protonation occurring from Kb2\textrm{K}_{b2} is negligible and can be calculated like a monoprotic base.

XVI amphiprotic: a species that can act as either a Bronsted acid or base. Examples include H2O\textrm{H}_{2}\textrm{O}, HCO3\textrm{HCO}_{3}^{-}, HS\textrm{HS}^{-}

  1. when amphiprotic species like HCO3\textrm{HCO}_{3}^{-} or HS\textrm{HS}^{-} are added into water, the pH of the solution can be approximated as:

    pH=12(pKa1+pKa2)\textrm{pH} = \frac{1}{2} (\textrm{pK}_{a1} + \textrm{pK}_{a2}) (Eq. 118)

    1. Condition: F>>KwKa2\textrm{F} \gt \gt \frac{\textrm{K}_{w}}{\textrm{K}_{a2}}, F>>Ka1\textrm{F} \gt \gt \textrm{K}_{a1} (negates deprotonation and protonation)

XVII When Ka1\textrm{K}_{a1} and Ka2\textrm{K}_{a2} are less than three magnitudes apart, a series of equilibrium constants must be solved.

  1. Sometimes, it is extremely convenient to calculate the composition of the polyprotic acid at a certain pH. The equations for a diprotic acid, such as carbonic acid, is shown below:

    f(H2A)=[H3O+]2H\textrm{f}(\textrm{H}_{2}\textrm{A}) = \frac{[\textrm{H}_{3}\textrm{O}^{+}]^{2}}{\textrm{H}}

    f(HA)=[H3O+]Ka1H\textrm{f}(\textrm{HA}^{-}) = \frac{[\textrm{H}_{3}\textrm{O}^{+}] \textrm{K}_{a1}}{\textrm{H}}

    f(A2)=Ka1Ka2H\textrm{f}(\textrm{A}^{2-}) = \frac{\textrm{K}_{a1} \textrm{K}_{a2}}{\textrm{H}}

    where

    H=[H3O+]2+[H3O+]Ka1+Ka1Ka2\textrm{H} = [\textrm{H}_{3}\textrm{O}^{+}]^{2} + [\textrm{H}_{3}\textrm{O}^{+}] \textrm{K}_{a1} + \textrm{K}_{a1} \textrm{K}_{a2}

  2. This relation is very useful for calculating the composition of a solution very quickly

  3. The fractions are graphed for carbonic acid below; we can see that maximum value of the intermediate species (bicarbonate) is when pH is between the two pKas\textrm{pK}_{a}\textrm{s}.

    1. This justifies Eq. 118 as well.

7.2.4 Leveling Effect and Buffers

XVIII Leveling effect: the strength of strong acids is limited (leveled) by the basicity of the solvent

  1. in water, H3O+\textrm{H}_{3}\textrm{O}^{+} is the strongest acid. At dilute concentrations, different strong acids (H2SO4\textrm{H}_{2}\textrm{SO}_{4}, HCl\textrm{HCl}, HNO3\textrm{HNO}_{3}) will all be equally strong (all deprotonates completely)

  2. in water, OH\textrm{OH}^{-} is the strongest base. At dilute concentrations, different strong bases (NaOH\textrm{NaOH}, EtMgBr\textrm{EtMgBr}, NaNH2\textrm{NaNH}_{2}) will all be equally strong (all protonates completely)

  3. The leveling effect only occurs in amphoteric solvents with Bronsted-Lowry acid or base properties

  4. For solvents that are more basic than water (such as NH3\textrm{NH}_{3}):

    1. some weak aqueous acids will be strong acids in these basic solvents.

      NH3(l)+CH3COOH(am)NH4+(aq)+CH3COO(am)\textrm{NH}_{3} (\textrm{l}) + \textrm{CH}_{3}\textrm{COOH} (\textrm{am}) \rightarrow \textrm{NH}_{4}^{+} (\textrm{aq}) + \textrm{CH}_{3}\textrm{COO}^{-} (\textrm{am})

    2. some strong aqueous bases will be weak bases in these basic solvents.

      NH3(l)+NaOH(am)NH2(am)+Na+(aq)+H2O(am)\textrm{NH}_{3} (\textrm{l}) + \textrm{NaOH} (\textrm{am}) \leftrightharpoons \textrm{NH}_{2}^{-} (\textrm{am}) + \textrm{Na}^{+} (\textrm{aq}) + \textrm{H}_{2}\textrm{O} (\textrm{am})

    3. In general, if pKa(HA)>>pKa(solvent)\textrm{pK}_{a} (\textrm{HA}) \gt \gt \textrm{pK}_{a} (\textrm{solvent}), the leveling effect will occur.

      1. meaning, all acids with much greater pKa\textrm{pK}_{a} will appear equally as strong.

    4. In general, if pKa(BH)>>pKb(solvent)\textrm{pK}_{a} (\textrm{BH}) \gt \gt \textrm{pK}_{b} (\textrm{solvent}), the leveling effect will occur.

      1. meaning, all bases with much greater pKb will appear equally as strong.

XIX Buffer solutions resist changes in pH when a small amount of strong acid or base is added.

  1. A buffer solution consists of a final mixture of either a:

    1. weak acid and its conjugate base

    2. weak base and its conjugate acid

    3. addition of strong acid to an excess of weak base also creates a buffer.

    4. addition of strong base to an excess of weak acid also creates a buffer.

  2. Buffers provide a weaker acid than H3O+\textrm{H}_{3}\textrm{O}^{+} and a weaker base than OH\textrm{OH}^{-} that will be protonated/deprotonated instead of water.

    1. the strong acid will protonate a weak base (in the buffer) instead of water and form a weak acid, leading a smaller increase in H3O+\textrm{H}_{3}\textrm{O}^{+} to be present in solution.

    2. the strong base will deprotonate a weak acid (in the buffer) instead of water and form a weak base, leading a smaller increase in OH\textrm{OH}^{-} to be present in solution.

  3. Buffer solutions allows us to control the pH of a solution against “stress” (this could be added acid or base), making the system overall “stable”

    1. our bodies rely on complex buffer systems to ensure that the pH of our blood does not vary too heavily.

XX Buffers have a capacity (β\beta) in which they lose their ability to resist once a certain amount of strong acid or base is added

  1. when pH=pKa\textrm{pH} = \textrm{pKa}, or when [HA]=[A-][\textrm{HA}] = [\textrm{A-}], the buffer has maximum capacity.

  2. Buffers act most effectively when the pH\textrm{pH} is within a range of ±1\pm 1 unit of pKa\textrm{pK}_{a}.

XXI The Henderson-Hasselbalch equation is a specific rearrangement of the acidity constant

pH=pKa+log[A][HA]\textrm{pH} = \textrm{pK}_{a} + \textrm{log} \frac{[\textrm{A}^{-}]}{[\textrm{HA}]} (Eq. 119)

  1. the pH of a buffer can be calculated by making the approximation that

    [HA][HA]initial[\textrm{HA}] \approx [\textrm{HA}]_{\textrm{initial}} and [A][A]initial[\textrm{A}^{-}] \approx [\textrm{A}^{-}]_{\textrm{initial}}.

    1. this approximation becomes invalid in extremely dilute solutions or with very weak acids

  2. This equation can be also used to approximate the pH of a mixture solution