Acid-base equilibria Chemistry 321, Summer 2014

Goals of this lecture Quantify acids and bases as analytes Measure [H + ] in solution  pH Control/stabilize [H + ] in solution  buffers Predict/control the form of acid or base in solution  ligands

Brønsted-Lowry definition of an acid A substance that ionizes in water to give hydrogen ions and a conjugate base acid (proton donor) hydronium ionconjugate base or, more simply: where K a is the acid dissociation constant

The acid dissociation constant [H + ] = [H 3 O + ]; we will use H + consistently instead of H 3 O + In the expression, the activities (e.g., a H + ) are replaced by concentrations (e.g., [H + ]) because the activity coefficients for all species are basically = 1 For a strong acid, K a is large (often  ∞ for acids that complete dissociate into ions) For a weak acid, K a is small and can be found in tables

Brønsted-Lowry definition of a base A substance that ionizes in water to give hydroxide ions and a conjugate acid For a strong base, K b is large For a weak base, K b is small

Measuring [H + ]: the pH scale Introduced in 1909 by Soren Sorensen at the Carlsberg Laboratory as a means of converting measured electrical potential of a solution into a hydrogen ion concentration (the technology of early pH meters), pH (potenz Hydrogen) is defined by the equation: pH = – log 10 [H + ]

The pH is based on the self-ionization of water The equilibrium expression K can be expressed in terms of species activities, or, assuming the species’ activities are close to one, in terms of concentrations:

Since the [H 2 O] is constant in most (dilute) solutions, that term can be multiplied by K and the resulting constant is the self-ionization constant of water. So, for pure water, [H + ] = [OH – ] = 1.0 × 10 –7 M, thus pH = 7.00 (2 significant figures). The pH of pure water In fact, is the pH of pure water at 25°C really 7.00?

Since the [H 2 O] is constant in most (dilute) solutions, that term can be multiplied by K and the resulting constant is the self-ionization constant of water. So, for pure water, [H + ] = [OH – ] = 1.0 × 10 –7 M, thus pH = 7.00 (2 significant figures). The pH of pure water In fact, is the pH of pure water at 25°C really 7.00? If the water has not been degassed, then CO 2 in the atmosphere will set up a bicarbonate equilibrium and decrease the pH (i.e., acidify the water).

Strong acids dissociate completely Therefore, [HA] (written on the outside of the bottle) = [H + ] i1 M0 M c–1 M+1 M e0 M1 M Thus, the pH of a 1 M HCl solution is – log (1) = 0.0

Strong bases dissociate completely Thus, a 1 M solution of NaOH will produce 1 M OH – ions. Using the water self-ionization equation, K w = [H + ] [OH – ], the [H + ] = 1 × 10 –14 M, which is a pH of Another way to get this is to calculate the pOH = - log 10 [OH – ], then use the expression pH + pOH = 14 to calculate the pH. In this case, pOH = 0.0, so pH = 14.0.

Weak acids and weak bases do not dissociate completely In this case, you cannot assume the concentration written on the bottle will be the [H + ]. Instead, [HA] + [A – ] = concentration on bottle.

A weak acid example Consider acetic acid (HAc), which dissociates into a hydrogen ion and an acetate ion (Ac – ). The K a for this dissociation is 1.8 × 10 –5 (at 25°C). A typical concern is the pH of a particular acetic acid solution: “What is the pH of a 0.20 M acetic acid solution?” Set up the equilibrium reaction equation and the ICE table: init0.20 M0 M change– x M+ x M equil0.20 – x Mx M

The calculation The approximation 0.20 – x ≈ 0.20 is useful because it allows the solving for x without using the quadratic formula. It is accurate because the acid will not dissociate to a large extent; in other words, [H + ] = [HCOO – ] << 0.20 M. Thus, x = M, which is indeed significantly less than 0.20 M. Since x = [H + ], then pH = In comparison, a strong acid of 0.20 M concentration has a pH of 0.70.

Soluble salts of weak acid conjugate base ions Consider the soluble salt sodium acetate (CH 3 COONa). It dissolves completely (up to its solubility, which is quite high) into sodium ions and acetate ions. The acetate ion is the conjugate base of acetic acid (a weak acid). Sodium acetate dissolved in water results in an alkaline (pH>7) solution. Consider the soluble salt sodium chloride (NaCl). It dissolves completely into sodium ions and chloride ions. The chloride ion is the conjugate base of hydrochloric acid (a strong acid). Sodium chloride dissolved in water does not change the pH of the water. How are the two scenarios different? NaCl(s)  Na + (aq) + Cl – (aq) CH 3 COONa (s)  Na + (aq) + CH 3 COO – (aq)

The pH of a salt solution In the case of sodium chloride, the generation of chloride ion will not induce any undissociated hydrochloric acid to form. Thus, there is no effect on the hydrogen ion concentration so the pH remains the same. The acetate ion generated in the dissolution of sodium acetate will then participate in the equilibrium above because there is always a hydrogen ion present in a water molecule. Thus, the acetate ion will react with the hydrogen ion to form undissociated acetic acid, effectively removing H + ions from solution, leaving hydroxide ions and making the solution alkaline.

The pH of a salt solution What is the pH of a 0.10 M sodium acetate solution? First, since sodium acetate is soluble, [CH 3 COO – ] = 0.10 M. Then, write the acetate equilibrium reaction: CH 3 COO – (aq) + H 2 O (l)CH 3 COOH (aq) + OH – (aq) Note that this will have a slightly different equilibrium constant: you generated a hydroxide ion, so the equilibrium expression will generate the base equilibrium constant, K b.

Next, set up the ICE table to show how the acetate equilibrium affects that concentration of acetate ion. – x+ x 0.10 – xxx Make the approximation that x << 0.10 so x = [OH – ] = 7.5 × 10 –6 M and pOH = 5.13, so pH = 8.87.

Buffers allow solutions to maintain a constant pH upon addition of acid or base Buffers are solutions that contain a weak acid and the salt of its conjugate base; for instance, an acetate buffer contains acetic acid and (usually) sodium acetate, which is highly soluble and thus is an excellent source of acetate ions. The buffer maintains its pH because of LeChatelier’s principle: when an acid is added, the equilibrium above shifts to the left (generating undissociated acid). When a base is added, it will react with the H +, and the equilibrium will shift to the right to restore H + concentration.

To make a buffer, use the Henderson-Hasselbalch equation Two physicians, Lawrence Henderson (1908) and Karl Hasselbalch (1916), studying metabolic acidosis, developed the equation that allowed the prediction of a buffer pH relying only on species concentrations and the pK a of the weak acid involved. where HA, the weak acid, dissociates into A –. [HA] 0 is the initial concentration of HA and [A – ] 0 is the initial concentration of A –.

A buffer example To make an acetate buffer of pH 5.00, use the equation to determine the initial concentrations of acetic acid and acetate ion. so so if the acetic acid is 0.10 M, then the concentration of acetate ion must be 0.18 M.

Choosing the right weak acid for making a buffer Could other weak acids be used to make a pH 5.00 buffer? Yes, as long as they have a pK a near Why? Calculate [A – ] 0 /[HA] 0. acidKapKa[A-] 0 /[HA] 0 acetic1.8 × salicylic1.0 × Note that salicylic acid would be a bad choice to use for the buffer since there’s hardly any undissociated salicylic acid at pH 5.00 – this means that this solution will not buffer against the addition of base very well.

Graphic behavior of weak acids Note a couple of things: The equivalence point pH is not 7 (it is higher than 7) Near the half- equivalence point, the pH is not affected by the addition of significant amounts of titrant (this is the buffering behavior of a weak acid solution)

Challenge problem Formic acid (HCOOH) is a weak acid with a K a = 1.8 × 10 –4, and dissociates into H + and the formate ion, HCOO –. The formate ion can also be introduced to a solution by dissolving a soluble salt like sodium formate, NaHCOO. What is the pH of a solution that is labeled 0.20 M NaHCOO? Note that you must take into account the formic acid equilibrium in your calculation.