Acid-Base Equilibria

• Acid-base equilibria involve weak acids which only partially dissociate in aqueous solutions.
• An acid-dissociation constant is the equilibrium constant for the ionization of an acid.
• The H₂O is omitted from the expression.

Table 16.2 Weak Acids in Water at 25°C

• Chlorous acid (HClO₂) has a structural formula of H-O-Cl-O and a conjugate base of ClO₂⁻, with a Kₐ of 1.0 × 10⁻².
• Hydrofluoric acid (HF) has a structural formula of H-F and a conjugate base of F⁻, with a Kₐ of 6.8 × 10⁻⁴.
• Nitrous acid (HNO₂) has a structural formula of H-O-N=O and a conjugate base of NO₂⁻ , with a Kₐ of 4.5 × 10⁻⁴.
• Benzoic acid (C₆H₅COOH) has a structural formula of H-O-C with a benzene ring attached to the carbon and a conjugate base of C₆H₅COO⁻, with a Kₐ of 6.3 × 10⁻⁵.
• Acetic acid (CH₃COOH) has a structural formula of H-O-C-C-H with O and H attached to the second carbon and a conjugate base of CH₃COO⁻, with a Kₐ of 1.8 × 10⁻⁵.
• Hypochlorous acid (HOCl) has a structural formula of H-O-Cl and a conjugate base of OCl⁻ , with a Kₐ of 3.0 × 10⁻⁵.
• Hydrocyanic acid (HCN) has a structural formula of H-C≡N and a conjugate base of CN⁻, with a Kₐ of 4.9 × 10⁻¹⁰.
• Phenol (HOC₆H₅) has a structural formula of H-O with a benzene ring attached to the oxygen and a conjugate base of C₆H₅O⁻, with a Kₐ of 1.3 × 10⁻¹⁰.
• The magnitude of Kₐ indicates the tendency of the acid to ionize in water
• The larger the Kₐ value, the stronger the acid.

Calculating Kₐ from Measured pH (Formic Acid Example)

• A student prepared a 0.10 M solution of formic acid (HCOOH).

• The solution had a measured pH of 2.38 at 25 °C.

• Initial concentration of formic acid is 0.10 M.

• Prior to dissociation, there is no H⁺ or HCOO⁻ present.

• Use pH to find the quantity of H⁺ dissociated:

  10⁻².³⁸ = 0.00417 M

• Amount of HCOOH lost is equivalent to the amount of HCOO⁻ gained.

• The change in concentration is -4.17 × 10⁻³ for HCOOH and +4.17 × 10⁻³ for both H⁺ and HCOO⁻

• Equilibrium concentrations are determined: 0.09583 for HCOOH, and 4.17 × 10⁻³ for H⁺ and HCOO⁻.

• Substitute equilibrium concentrations into the Kₐ expression: Kₐ = (4.17×10⁻³)(4.17×10⁻³) / (0.10) = 1.74×10⁻⁴

• The amount of HCOOH that ionizes is small when compared with the initial concentration of the acid.

• The initial concentration of formic acid is used in the Kₐ expression.

Calculating Kₐ from Measured pH (Niacin Example)

• Niacin, one of the B vitamins, has a specific molecular structure.

• A 0.020 M solution of niacin has a pH of 3.26.

• Initial concentration of formic acid is 0.020 M.

• Prior to dissociation, there is no H⁺ or HCOO⁻ present.

• Use pH to find the quantity of H⁺ dissociated:

  10⁻³.²⁶ = 0.000550 M = 5.50 × 10⁻⁴ M = [H⁺].

• The amount of HCOOH lost and HCOO⁻ gained is also equal [H⁺].

• Equilibrium concentrations are determined

• Change in concentration (M) -5.50 × 10⁻⁴ for HA, +5.50 × 10⁻⁴ H+, +5.50 × 10⁻⁴

• Equilibrium: 0.01945 for HA, 5.50 × 10⁻ for H+, 5.50 × 10⁻⁴

• Substitute equilibrium concentrations in Kₐ expression: Kₐ = ((5.50×10⁻⁴)(5.50×10⁻⁴)/0.020) =1.51×10⁻⁵

Percent Ionization

• Percent ionization measures acid strength.

• The formula is:

  Percent ionization = (concentration ionized / original concentration) × 100%.

• The stronger the acid, the greater the percent ionization.

• The concentration of acid that ionizes equals the concentration of H⁺ that forms.

Calculating Percent Ionization (Formic Acid Example)

• A 0.10 M solution of formic acid (HCOOH) contains 4.2 × 10⁻³ M H⁺ (aq).

• The percentage of the acid that is ionized is calculated as:

  Percent ionization = (4.2×10⁻³) / 0.10 × 100% = 4.2%

Calculating Percent Ionization (Niacin Example)

• A 0.020 M solution of niacin has a pH of 3.26.

• To calculate the percent ionization of niacin, use pH = 3.26 to find [H⁺]

  10⁻³.²⁶ = 5.50 × 10⁻⁴ M = [H⁺].

• Calculate the percent ionization:

  (5.50×10⁻⁴)/(0.020) ×100% =2.75%

• Percent ionization decreases as concentration increases for weak acids.

• For any acid, the concentration of acid that ionizes is equal to concentration of H+ that forms.

• In an acetic acid solution, CH₃COOH molecules are less prone to dissociation as concentration increases.

Calculating pH

• If Kₐ and the initial concentration of weak acid are known, the concentration of H⁺ can be determined.

Using Kₐ to Calculate pH (HCN Example)

• Calculate the pH of a 0.20 M solution of HCN.
• Write the equilibrium equation: HCN ⇌ H⁺ + CN⁻
• Write the equilibrium-constant expression using the known Kₐ: Kₐ = [H⁺][CN⁻] / [HCN]
• Express equilibrium concentrations (ICE Table): represent H⁺ and CN⁻ released using x.
  • Initial concentration of HCN is 0.20 M.
  • Prior to dissociation, there is no H+ or CN⁻ present.
  • The Change of concentration is -x for HCN, +x for H+ and CN⁻,
  • Equilibrium concentrations are determined using rows 1 and 2: 0.20 - x for [HCN], x for [H+], x for [CN-]
• Substitute equilibrium concentrations into the equilibrium expression: Kₐ =(x²)/(0.20-x)=4.9×10⁻¹⁰
• Solve for x using the quadratic formula: Simplify by assuming x is small compared to the initial concentration of acid.
• In general, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula.

Using Kₐ to Calculate pH (Niacin Example)

• Given the Kₐ for niacin is 1.5 × 10⁻⁵, the pH of a 0.010 M solution of niacin may be calculated.
• Write the equilibrium equation: HA ⇌ H⁺ + A⁻
• Write the equilibrium-constant expression using the known Kₐ.
• Express equilibrium concentrations via an ICE Table, represent H⁺ and A⁻ released using x.
  • Initial concentration of niacin is 0.010 M.
  • Prior to dissociation no H⁺ or A⁻ is present
• The Change of concentration is -x for HA, +x for H+ and A⁻,
  • Equilibrium concentrations are determined to be: 0.010 - x, x, x
  • Substitute equilibrium concentrations in Kₐ expression ((x²) /0.010) =1.5 ×10⁻⁵. Simplify by assuming that “x” is much smaller than 0.010. If the value for “x” is still more than about 5% of the initial concentration, the quadratic formula should be used.

Calculating Percent Ionization:

• Need to determine H⁺ values to use in % ionization determination.
• Percentage of HF molecules ionized calculated

Using Kₐ to Calculate Percent Ionization (HF Example - part a)

• Given of 0.10 M HF solution
• Set up Kₐ expression: HF ⇌ H⁺+ F⁻
• Represent H⁺ and F⁻ released using x.
  • Initial concentration of HF is 0.10 M.
  • Initially, no H⁺ and F⁻
• Change of concentration: -x, +x and +x Equilibrium 0.10-x, x and x are the equilibrium concentrations.
• Substitute equilibrium concentrations to the equilibrium expression: (x²)/(0.10) = 6.8×10⁻⁴, if x<< 5% of [acid]
  • Because x is > than 5% of 0.10, the simplifying approximation is not appropriate
• Use equilibrium concentrations from ICE table.
• Quadratic equation used and the following formula: ax² + bx + c = 0

Using Kₐ to Calculate Percent Ionization (HF Example - part b)

• Calculate the percentage of HF molecules ionized in a) 0.10 M HF solution, b) a 0.010 M HF solution. For the 0.010 M HF solution: HF(aq) ⇌ H⁺(aq) + F⁻(aq).
• Express set up Kₐ: HF(aq) ⇌ H⁺(aq) + F⁻(aq) via the ICE table, and substitute equilibrium concentrations into the equilibrium expression,
• Find the [H] = [F] = 2.3 × 10⁻³ M. %H ionized = 23%

Practice Exercise

• In Practice Exercise 16.11, found that percent ionization of niacin Kₐ = 1.5 × 10⁻⁵ in a 0.020 M solution = 2.7%. Calculated percentage of niacin molecules ionized solution: a) 0.010 M, and b) 1.0 × 10⁻³ M
• Set up Kₐ a) HA(aq) ⇌ +H(aq) A⁻(aq) Representation =x
• Calculated and Substitued*
• Find concentration of H+ to be: 3.8x10⁻⁴M (3.9%)

Polyprotic Acids

• Acids that have more than one ionizable H atom and successive ionizations.
• It is always easier to remove the first proton from a polyprotic acid than to remove the second.
• Kₐ values become successively smaller as successive protons are removed.
• The determination of polyprotic acids is possible by considering Kₐ

Table 16.3. Acid-Dissociation Constants of Some Common Polyprotic Acids

• Ascorbic acid (H₂C₆H₆O₆) has Kₐ₁ of 8.0 × 10⁻⁵ and Kₐ₂ of 1.6 × 10⁻¹².
• Carbonic acid (H₂CO₃) has Kₐ₁ of 4.3 × 10⁻⁷ and Kₐ₂ of 5.6 × 10⁻¹¹.
• Citric acid (H₃C₆H₅O₇) has Kₐ₁ of 7.4 × 10⁻⁴, Kₐ₂ of 1.7 × 10⁻⁵, and Kₐ₃ of 4.0 × 10⁻⁷.
• Oxalic acid (H₂C₂O₄) has Kₐ₁ of 5.9 × 10⁻² and Kₐ₂ of 6.4 × 10⁻⁵.
• Phosphoric acid (H₃PO₄) has Kₐ₁ of 7.5 × 10⁻³, Kₐ₂ of 6.2 × 10⁻⁸, and Kₐ₃ of 4.2 × 10⁻¹³.
• Sulfurous acid (H₂SO₃) has Kₐ₁ of 1.7 × 10⁻² and Kₐ₂ of 6.4 × 10⁻⁸.
• Sulfuric acid (H₂SO₄) has a large Kₐ₁ and Kₐ₂ of 1.2 × 10⁻².
• Tartaric acid (H₂C₄H₄O₆) has Kₐ₁ of 1.0 × 10⁻³ and Kₐ₂ of 4.6 × 10⁻⁵. H₃PO₄ ⇌ H⁺ + H₂PO₄⁻, 7.5 × 10⁻³ H₂PO₄⁻⇌ H⁺+ HPO₄⁻², 6.2 × 10⁻⁸ HPO₄⁻²⇌ H⁺+ PO₄⁻³, 4.2 × 10⁻¹³
• Determination of the pH of polyprotic acids is possible by considering only Kₐ₁.

Calculating the pH of a Polyprotic Acid Solution:

• Given the solubility of CO₂ in water is 0.0037 M Set up K expression and subtstitute equal concentrations into the equeation.
• Calculate a -log(4.0 × 10⁻⁵) = 4.40 = pH.

Calculating the pH of a 0.020 M solution of oxalic acid (H₂C₂O₄)

• H₂C₂O₄(aq) ⇌ H⁺(aq) + C₂O₄²⁻(aq), Calculate using K for different acids and find H+ for the second dissociation.