Acids, Bases, and pH - Chapter 8

• Chapter 8 discusses acids, bases, and pH.

Explaining the Properties of Acids and Bases

• Acids and bases are defined by the ions they produce in water, according to the Arrhenius Theory.
• Acids dissociate in water to produce H+(aq) ions; HCl and H2SO4 are examples.
• Bases dissociate in water to produce OH-(aq) ions; NaOH and KOH are examples.
• Strong acids and bases fully dissociate in water.

Arrhenius Theory: Acid-Base Reactions

• Acid-Base reactions combine H+(aq) and OH-(aq) ions.
• Hydrochloric acid reacts with sodium hydroxide: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l), ΔH = -56 kJ.
• Total ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Cl-(aq) + Na+(aq) + H2O(l).
• Removing spectator ions, the net ionic equation is H+(aq) + OH-(aq) → H2O(l), ΔH = -56 kJ.
• All Arrhenius acids and bases produce the same net ionic equation and amount of energy per mole of water.

Hydronium Ions

• H+(aq) does not exist in water
• H+(aq) is attracted to water molecules and forms hydronium ions, H3O+(aq) via: H+(aq) + H2O(l) → H3O+(aq).
• Hydronium ions form hydrogen bonds with other water molecules, represented as [H(H2O)n]+, where n is usually 4 or 5.
• A single hydronium ion (H3O+) is used when writing equations.

Limitations to Arrhenius Theory

• Aqueous ammonia solutions are basic but don't contain hydroxide ions.
• Many aqueous salt solutions without hydroxide ions are also basic.
• Some reactions occur without any liquid solvent e.g., NH3(g) + HCl(g) → NH4Cl(s).

Brønsted-Lowry Theory: Acids and Bases as Proton Donors and Acceptors

• This theory overcomes the limitations of the Arrhenius theory.
• Acids are proton (H+) "donors" and bases are proton (H+) "acceptors".
• In aqueous solutions, H+ ions bond with water to form hydronium ions.
• The Brønsted-Lowry theory is not restricted to aqueous solutions.
• H+ ions can bond to the lone pair on an ammonia molecule, allowing liquid ammonia to act as a base.

Summary of Acid-Base Theories

• Arrhenius Theory defines acids as substances containing hydrogen that dissociate in water to form H+(aq).
• Arrhenius Theory defines bases as substances containing the hydroxide group that dissociate in water to form OH-(aq).
• Brønsted-Lowry Theory defines acids as substance from which a proton can be removed.
• Brønsted-Lowry Theory defines bases as substances that can accept a proton from an acid.

Conjugate Acid-Base Pairs

• Acetic acid is a weak acid that partially dissociates in water, favoring the reverse reaction at equilibrium.
• CH3COOH is the acid on the left of the equation, and CH3COO- (acetate ion) is the conjugate base.
• Water and hydronium form a conjugate acid-base pair.

Conjugate Acid-Base Pairs and Amphoteric Substances

• Ammonia is a weak base, forming few hydroxide ions when dissolved in water and the equilibrium lies to the left.
• Water acts as the acid in the presence of ammonia.
• Water's behavior varies, acting as an acid with stronger bases, and as a base with stronger acids i.e. amphoteric.
• Water can accept or donate H+.

Example: Identifying Conjugate Acid-Base Pairs

• Reaction: H3PO4(aq) + H2O(l) ⇌ H2PO4-(aq) + H3O+(aq).
• H3PO4(aq) acts as the ACID donating a proton.
• H2O(l) behaves as the BASE accepting the proton.
• H2PO4-(aq) is the CONJUGATE BASE.
• H3O+(aq) is the CONJUGATE ACID.

Molecular Structure and the Strength of Acids and Bases

• Most acids and bases are weak; strong acids and bases fully dissociate in water.
• Strong acids are categorized into binary acids (HX(aq), where X = Cl, Br, I, but not F) or hydrohalic acids.
• Strong acids are also categorized as oxoacids/oxyacids with oxygen atoms; where the acidic H is attached to an oxygen, like H2SO4 and HClO3 (chloric acid).
• The acid strength of hydrides increases across a period due to increased electronegativity (EN).
• Acid strength increases down a group (column) due to decreasing bond strength.

Acid Strength: HF Example

• HF is the weakest acid due to the strong H-F bond, indicating low bond strength.

Oxyacids: Impact of Oxygen Atoms on Acid Strength

• Acid strength increases with more oxygen atoms.
• Oxygen atoms are more electronegative than hydrogen, which increases the polarity of the bond between H and O.
• Increased polarity makes it easier for water to attract the acidic H+.

Monoprotic and Polyprotic Acids

• HCl, CH3COOH, and HF are monoprotic acids, having only one H to dissociate.
• H₂SO₄ is a polyprotic acid with more than one dissociable H.
• H₂SO₄ is a stronger acid than the hydrogen sulfate ion; more energy is required to remove a proton from a negatively charged ion.
• Dissociation of H₂SO₄: H₂SO₄(aq) + H₂O(l) ⇌ H₃O+(aq) + HSO₄-(aq) followed by HSO₄-(aq) + H₂O(l) ⇌ H₃O+(aq) + SO₄²⁻(aq).

Strong Bases

• Strong bases include oxides and hydroxides of alkali metals like KOH and NaOH.
• Also includes oxides and hydroxides of alkaline earth metals below beryllium, such as Ca(OH)2.
• BeO is a weak base with a strong bond that is not easily broken by water molecules.
• Strong basic oxides feature metal atoms with low electronegativity bonded ionically to oxygen.
• The bond between the metal and oxygen is easily broken by polar water molecules.
• Oxide ions react with water to form hydroxide ions: O²⁻(aq) + H₂O(l) → 2OH⁻(aq).

Example: Determining Acidity or Basicity of a Solution

• 25 ml of 1.40 mol/L nitric acid is mixed with 15.0 mL of 2.00 mol/L sodium hydroxide.
• HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l).
• Calculate moles of each reactant.
• nHNO3 = (1.40 mol/L) * (0.0250 L) = 0.0350 mol.
• nNaOH = (2.00 mol/L) * (0.0150 L) = 0.0300 mol.
• Limiting reagent impacts solution acidity.
• Excess HNO3 indicates acidic solution.
• Excess HNO3 = 0.0350 mol - 0.0300 mol = 0.0050 mol.
• Calculate the concentration of H3O+.
• [H3O+] = (0.0050 mol) / (0.0150 L + 0.0250 L) = 0.13 mol/L.
• Solution is acidic with [H3O+] = 0.13 mol/L.

The Ion Product Constant for Water

• Pure water contains a few ions due to the dissociation of water molecules.
• Dissociation reaction: 2H2O(l) ⇌ H3O+(aq) + OH−(aq).
• At 25°C, [H3O+] = [OH−] = 1.0 x 10-7 mol/L.
• Equilibrium constant expression: Kc=[H3O+][OH−]/[H2O]^2.
• Kw = Kc[H2O]2 = (1.0 × 10-7)2 = 1.0 × 10-14.

[H3O+] and [OH-] in Aqueous Solutions at 25°C

• [H3O+] equals the concentration of a strong acid, unless the solution is very dilute (1 x 10-7 mol/L).
• [OH-] equals the concentration of a strong base, unless the solution is very dilute (1 x 10-7 mol/L).
• Strong acids or bases suppress water dissociation.
• Kw can be used to find the concentration of one ion if the other is known.
• Acidic Solution: [H3O+] > 1 x 10-7 mol/L, [OH-] < 1 x 10-7 mol/L
• Neutral Solution: [H3O+] = [OH-] = 1 x 10-7 mol/L
• Basic Solution: [H3O+] < 1 x 10-7 mol/L, [OH-] > 1 x 10-7 mol/L

Example: Finding [H3O+] and [OH-] for Solutions

• For 2.5 mol/L nitric acid (HNO3), a strong acid:
• HNO3 completely dissociates in water, producing 1 mole of [H3O+] per mole of HNO3.
• [HNO3] = 2.5 mol/L = [H3O+].
• To find [OH-], use Kw = 1.0 × 10-14 = [H3O+][OH-].
• [OH-] = (1.0 × 10-14) / (2.5) = 4.0 × 10-15 mol/L.
• For 0.16 mol/L barium hydroxide (Ba(OH)2), a strong base:
  • Each mole of Ba(OH)2 produces 2 moles of OH-.
  • [OH-] = (2) * (0.16 mol/L) = 0.32 mol/L.
  • To find [H3O+], use Kw = 1.0 × 10-14 = [H3O+][OH-].
  • [H3O+] = (1.0 × 10-14) / (0.32) = 3.1 × 10-14 mol/L.

pH and pOH Scales

• pH scale is logarithmic.
  • pH = -log[H3O+]
  • pOH = -log[OH-]
• pH + pOH = 14
• They have no units.

Shampoo Acidity Example

• A shampoo's hydroxide ion concentration is 6.8 x 10-5 mol/L at 25°C.
  • This would make the shampoo basic
• [H3O+] calculation: Kw = 1.0 x 10-14 = [H3O+][OH-].
• 1.0 * 10^-14 = [H3O+] * 6.8e-5
• H = [H3O+] = (1.0 × 10-14)/ (6.8 ×10-5) = 1.5 × 10-10 mol/L.
• Determine pH: pH = -log[H3O+] = -log(1.5 × 10-10) = 9.8.
• Determine pOH: pOH = -log[OH-] = -log(6.8 × 10-5) = 4.2.
• pOH + pH = 14 validates calculations.

Finding [OH-] and [H3O+] from pH or pOH

• log is actually log₁₀ (log base 10).
• If log₁₀ a = b, then 10ᵇ = a. -For pH, if pH = 3, then [H3O+] = 10-3

Kw of Weak Acids

• Table 8.2 presents Some Acid Dissociation Constants for Weak Acids at 25°C.
• Acetic acid (CH3COOH) is 1.8 x 10-5
• Chlorous acid (HClO2) is 1.1 x 10-2
• Formic acid (HCOOH) is 1.8 x 10-4
• Hydrocyanic acid (HCN) is 6.2 x 10-10
• Hydrofluoric Acid (HF) is 6.6 x 10-4
• Hydrogen oxide (H2O) is 1.0 x 10-14
• Lactic acid (CH3CHOHCOOH) is 1.4 x 10-4
• Nitrous acid (HNO2) is 7.2 x 10-4
• Phenol (C6H5OH) is 1.3 x 10-10

Acid Dissociation Constant Ka

• Weak acids do not completely dissociate in water.
• The monoprotic acid will look like this: HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq).
• Equilibrium expression for the reaction Kc = [H3O+][A-] /[[HA][H2O].
• Dilute solutions [H₂O] is constant with: Kc[H2O]=Ka
• Ka = Kc[H2O] = [H3O+][A-] / [HA]
• Ka is known as the acid dissociation constant or the acid ionization constant.

Solving Equilibrium Problems Involving Acids and Bases

• Write the chemical equation and use an ICE table.
• Concentrations of [OH-] and [H3O+] in pure water are negligible in weak acid or base solutions.
• Let x represent the change in concentration for the smallest coefficient substance.
• Compare initial concentration ([HA]) with Ka (acid ionization constant).
• When [HA]/Ka > 500, x is negligible and can be ignored.
• When [HA]/Ka < 500, x may not be negligible.

Percentage Dissociation Example: Propanoic Acid

• Propanoic acid (CH3CH2COOH) is used to control of mold. It's a weak monoprotic acid.
• A 0.10 mol/L solution with a pH of 2.96.
• First writing what all molecules present are with an ice table: CH3CH2COOH(aq) + H2O(l) ⇌ CH3CH2COO⁻(aq) + H3O⁺(aq). Given [CH3CH2COOH] = 0.10 mol/L and pH = 2.96.

Example continued...

• [CH3CH2COO-][H3O+] [CH3CH2COOH]
• Ka = (x)(x) / (0.10 - x)
• [H3O^+] = [CH3CH2COO^-] calculation: H = 10^-2.96 - 1.1*10-3 mol/L
• Ka = (1.1 × 10-3)2 / (0.10 – 1.1 × 10-3) = 1.2 × 10-5
• This Ka value is reasonable
• Use previous value to calculate percentage as well: 1.1 × 10-3 mol/L / 0.10 mol/L × 100 = 1.1 %

Calculating pH Example: Formic Acid

• Formic acid (HCOOH) is in ant sting.
• Find the pH of a 0.025 mol/L, given [HCOOH] = 0.025 mol/L
• Formic Acid: HCOOH(aq) + H2O(l) ⇌ HCOO⁻(aq) + H3O⁺(aq).
• Table set up and then acid dissociation constant: Kₐ = 1.8 x 10⁻⁴
• [HCOOH]/Kₐ = .025/ 1.8 x 10⁻⁴ = 139
• As previous ratio is less than 500 there is notable change with approximation and change is accounted for. - Solve for x

Example continued...

• Acid dissociation constant Kₐ = 1.8 x 10⁻⁴ where by using the previous ICE table we solve for : [H3O+][HCOO-]/ [HCOOH] with (x)² = .025-x = 1.8 x 10⁻⁴
• From that equation the final calculation is as follows: x² = (3.3 × 10-6)(1.7 × 10-3)
• x = 2.0 × 10-3
• Since x is the change in H3O we can plug it in and solve pH
• By solving for x we now get as follows: pH = -(2.0x10^-3) = 2.70

Polyprotic Acids

• All polyprotic acids except H2SO4 are weak.
• Only the first dissociation step is factored in: When calculating [H₂O+] with pH of a polyprotic acid we only use the first step.
• Equal ratio: The concentration of anions formed in the second level of dissociation of a polyprotic acid and it is equal to Ka2.
• For H2SO4 the second dissociation is weak.
• For less than the less than 1.0 mol/L the second dissociation count towards [H₂O+].
• H2SO4(aq) + H20(l) --> HSO4-(aq) + H3O+(aq) First level is 100%
• HSO4(aq) + H20(l) --> SO4(aq) + H3O+(aq) This is 1/Ka level where it is weak

Example continued...

• First Level Dissociation: By solving it with previous ICE table format we find there is low changes in the concentration so the overall is likely negligible. This is where H3O+ (aq) + H2PO4(aq) has now happened: H3PO4(aq) + H2O(l) --> H3O+ (aq) + H2PO4(aq)
• Second Level dissociation: Solving for Ka: (x²)/3.5 = 7.0 x 10 to solve gives x= 0.16 mol/L

Calculating pH Example

• (x)(x+0.16)) /(3.5) and now this allows to solve this equation: (x)(x + 0.16)/ 0.16 We now achieve values x = 6.3 x 10 to the -7. Once we solve that dissociation is very low given Ka so the pH is .80

Base Dissociation Constant K

• A weak dissociation water as follows: B(aq) + H₂O(l) --> HB+(aq) - OH-(aq): The constant can be written as follows: Kc=[HB+][OH-]/ [[B][H₂O] and K₁ = Kc[H₂O] = [HB+][OH-]/[B].

Example - Determining Kb

• Calculate the constant used to manufacture many medications: pyridine, C₅H₅N pH = 9.10 where C₅H₅N + H20 —> C₅H₅NH + OH

Continued

• Solve concentrations where it follows as follows: Kb = 10^-14 - (10^(4-9))*10^(-5)^2)/ * (4 - 9)/ 0.125 *10^-9

Acids and their Conjugate Bases

The first dissociation of many acids are as follows: CH₃COOH + H20 = CH3 + H30 The expression here is Ka = [CH3COO®][H3O+]/ [CH3COOH]

Example Continued

Now using the 2 level equation we use that it follows as the following (x)(x))/(0.152).

The Henderson-Hasselbalch Equation

• Allows pH of a buffer solution to be found.
• pH = pKa + log([Base]/[Acid])
• The equation is commonly misused.
• Accurate when concentrations of acid and conjugate base are high, conditions which occur with buffers.

Kw of Weak Bases

• Table 8.3 presents Some Base Dissociation Constants at 25°C.
• Ethylenediamine (NH₂CH₂CH₂NH₂) is 5.2 x 10⁻⁴
• Dimethylamine ((CH₃)₂NH) is 5.1 x 10⁻⁴
• Methylamine (CH₃NH₂) is 4.4 x 10⁻⁴
• Trimethylamine ((CH₃)₃N) is 6.5 x 10⁻⁵
• Ammonia (NH3) is 1.8 x 10⁻⁵
• Hydrazine (N₂H₄) is 1.7 x 10⁻⁶
• Pyridine (C₅H₅N) is 1.4 x 10⁻⁹
• Aniline (C₆H₅NH₂) is 4.2 x 10⁻¹⁰
• Urea (NH₂CONH₂)₂ is 1.5 x 10⁻¹⁴

Buffer Solutions

• Buffered solution, or buffer, resists pH changes when small acids or bases are added.
  • Made from a weak acid & its salt (e.g., acetic acid & sodium acetate).
  • Made from a weak base & its salt (e.g., ammonia & ammonium chloride).

How do Buffer Solutions Work?

• HF reacts with OH⁻ to produce water and F⁻.
• F⁻ reacts with H⁺ to form HF and water.

Practical Example of a buffer Solution: Why Buffers are Important to the Body

• Weak acids like CH3COOH's molecules are undissociated with a high [CH3COOH].
• Salt of conjugate creates high (H₃COO] -H₂O+ / OH- are taken to keep the system level. H₃O+ / OH- have components that stop shifts from their direction like shifting the equil curve.

Buffers: An Everyday Example

• Weak acids form an equilibrium to counter and shift it (Chatelier's Equilibrium)
• Hydroxide has these compounds to shift or the direction of it.

Buffer Capacity

• Buffer capacity depends on the concentration of the weak acid/base & their conjugate. This means that the more concentration the more it resists the pH.
• The ratio hits 1 by volume and each compound hits maximum
• Blood contains pH range where cells are functional and are vital for life.

Buffers in the Blood

• The pH of blood needs to stay within a narrow pH of 7.4 ideally, which can have many changes.
• The pH is dependent between carbon and carbonates, that shift
• Carbon reacts by having carbonic acid for carbonate. -CO2 + 2H20 & HCO(3-) + H3O +
• For carbonate and for hydrogen shifting from the equation
• HCO(3-) + H20 & C03 + H3o + If it shifts it is for HCO3 + 2H2O + H(10
• If we apply ioniation it creates it for many equations H30 to create + 2h30 C02 shift the equilibrium.

Acid-Base Titration

• Performed to figure out the concentration . - Use a solution with a known solution known a titrant, to figure out the unknow.

Acids and Bases- An Important Tool to Titrate

• Allows the pH to be read from the solution when it reached a certain level from it
• At those ratios you get the equation from many solutions.

Titration: Strong Acid and Strong Base

• Start to see titration near the pH and have this slope that goes slowly up Just before the previous we see that goes fast from the previous point
• At the equivalence point it means that it matches moles in PH After much more base that it happens the level gets thrown to the next

Strong Acid - Strong Base

• They mix very well -You can measure during volume This is where there equation is: (H+(aq)) + (OH negative (aq)) = HO(l). We measure from level and equivalence.

Titration Curve- Weak Bases

• Titration happens from there being color change with acids from there being PH.
• PH - Unlike bases the bases can have some form it as a weak -Good base like 8.2 in pH is good to test here