Acid-Base Equilibria Notes

Stomach Acid & Heartburn

• Cells lining the stomach produce hydrochloric acid (HCl) for:

  • Killing unwanted bacteria.

  • Breaking down food.

  • Activating enzymes.

• Heartburn (acid reflux) occurs when stomach acid backs up into the esophagus, irritating tissues.

Curing Heartburn - изжога

• Mild cases can be alleviated by neutralizing acid in the esophagus.

  • Swallowing saliva (contains bicarbonate ions).

  • Taking antacids (contain hydroxide and/or carbonate ions).

GERD (Gastroesophageal Reflux Disease)

• Chronic heartburn is a persistent issue for some individuals.

• GERD involves chronic stomach acid leakage into the esophagus.

• In GERD, muscles separating the stomach from the esophagus don't close properly.

• Physicians diagnose GERD by measuring acidity levels in the esophagus over time using a pH sensor.

Properties of Acids

• Taste sour.

• React with "active" metals (e.g., Al, Zn, Fe) but not with less reactive metals (e.g., Cu, Ag, Au).

Example reaction: $2 Al + 6 HCl \rightarrow 2 AlCl3 + 3 H2$

• Corrosive.

• React with carbonates, producing carbon dioxide (CO2).

Example reaction with marble, baking soda, chalk, or limestone: $CaCO3 + 2 HCl \rightarrow CaCl2 + CO2 + H2O$

• Change the color of vegetable dyes (e.g., blue litmus turns red).

• React with bases to form ionic salts.

Common Acids

• Examples include hydrochloric acid (HCl), sulfuric acid (H2SO4), and nitric acid (HNO3).

Structures of Acids

• Binary acids have acid hydrogens attached to a nonmetal atom (e.g., HCl, HF).

• Oxyacids have acid hydrogens attached to an oxygen atom (e.g., H2SO4, HNO3).

• Carboxylic acids contain a COOH group (e.g., HC2H3O2, H3C6H5O7).

  • Only the first H in the formula is acidic (the H in the COOH group).

Properties of Bases

• Also known as alkalis.

• Taste bitter.

  • Alkaloids are plant products that are alkaline and often poisonous.

• Solutions feel slippery.

• Change the color of vegetable dyes (different color than acids; red litmus turns blue).

• React with acids to form ionic salts (neutralization).

• Example: Reaction of HCl(g) and NH3(g) resulting in a fog of NH4Cl(s).

Common Bases

• Examples include sodium hydroxide (NaOH), calcium hydroxide (Ca(OH)2), and ammonia (NH3).

Structure of Bases

• Most ionic bases contain hydroxide (OH−) ions (e.g., NaOH, Ca(OH)2).

• Some contain carbonate (CO3 2−) ions (e.g., CaCO3, NaHCO3).

• Molecular bases contain structures that react with H+ (mostly amine groups).

Indicators

• Chemicals that change color depending on the solution’s acidity or basicity.

• Many vegetable dyes are indicators (e.g., anthocyanins).

• Litmus:

  • Derived from Spanish moss.

  • Red in acid, blue in base.

• Phenolphthalein:

  • Found in laxatives.

  • Red in base, colorless in acid.

• Reference to a figure showing pH ranges for common acid-base indicators.

• Reference to a figure showing solutions containing three common acid-base indicators at various pH values.

Arrhenius Theory

• Bases dissociate in water to produce OH− ions and cations.

  • Ionic substances dissociate in water: $NaOH(aq) \rightarrow Na^+(aq) + OH^−(aq)$

• Acids ionize in water to produce H+ ions and anions.

• Molecular acids are not made of ions, so they cannot dissociate; they must be ionized by water.

• $HCl(aq) \rightarrow H^+(aq) + Cl^−(aq)$

Ionizable H is written in the front of the formula: $HC2H3O2(aq) \rightarrow H^+(aq) + C2H3O2^−(aq)$

• HCl ionizes in water, producing H+ and Cl– ions.

• NaOH dissociates in water, producing Na+ and OH– ions.

Hydronium Ion

• H+ ions produced by acids are so reactive they cannot exist in water; they are protons.

• Instead, they react with water molecules to produce complex ions, mainly hydronium ion (H3O+).

$H^+ + H2O \rightarrow H3O^+$

Minor amounts of H+ exist with multiple water molecules, $H(H2O)n^+$

Arrhenius Acid–Base Reactions

• H+ from the acid combines with OH− from the base to make a molecule of H2O (H-OH).

• The cation from the base combines with the anion from the acid to make a salt.

  • acid + base → salt + water

  • $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

Problems with Arrhenius Theory

• Does not explain why molecular substances like NH3 dissolve in water to form basic solutions without containing OH– ions.

• Does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions without containing OH– ions.

• Does not explain why molecular substances like CO2 dissolve in water to form acidic solutions without containing H+ ions.

• Does not explain acid–base reactions that take place outside aqueous solutions.

Brønsted-Lowry Acid-Base Theory

• Redefines acids and bases based on what happens in a reaction (H+ transfer).

• Any reaction involving H+ transfer from one molecule to another is an acid–base reaction.

• Applies regardless of whether the reaction occurs in aqueous solution or if OH− is present.

• All reactions fitting the Arrhenius definition also fit the Brønsted-Lowry definition, but the latter encompasses more reactions.

Brønsted-Lowry Theory

• In a Brønsted-Lowry acid–base reaction, an H+ is transferred.

• The acid is an H+ donor.

• The base is an H+ acceptor.

  • The base structure must contain an atom with an unshared pair of electrons.

• In a Brønsted-Lowry acid-base reaction, the acid molecule gives an H+ to the base molecule.

  • $H–A + :B \rightleftharpoons :A– + H–B^+$

Brønsted-Lowry Acids

• Brønsted-Lowry acids are H+ donors.

  • Any material with H can potentially be a Brønsted-Lowry acid.

  • Molecular structure dictates which H is easier to transfer.

• When HCl dissolves in water, HCl is the acid because it transfers an H+ to H2O, forming H3O+ ions.

• Water acts as a base, accepting H+.

$HCl(aq) + H2O(l) \rightarrow Cl^–(aq) + H3O^+(aq)$

• acid base

Brønsted-Lowry Bases

• Brønsted-Lowry bases are H+ acceptors.

  • Any material with atoms containing lone pairs can potentially be a Brønsted-Lowry base.

  • Molecular structure dictates which atom is more willing to accept H+ transfer.

• When NH3 dissolves in water, NH3(aq) is the base because it accepts an H+ from H2O, forming OH–(aq).

• Water acts as an acid, donating H+.

$NH3(aq) + H2O(l) \rightleftharpoons NH_4^+(aq) + OH^–(aq)$

• base acid

Warning about Lone Pairs

• Chemists often omit drawing lone pair electrons on structures due to familiarity with common bonding patterns.

• It's crucial to recognize when an atom in a molecule has lone pair electrons.

• Practice drawing structures with lone pairs of electrons (examples: HClO, HCO3 −).

Amphoteric Substances

• Amphoteric substances can act as either an acid or a base.

  • They possess both a transferable H and an atom with lone pair electrons.

• Water acts as a base, accepting H+ from HCl.

$HCl(aq) + H2O(l) \rightarrow Cl^–(aq) + H3O^+(aq)$

• Water acts as an acid, donating H+ to NH3.

$NH3(aq) + H2O(l) \rightleftharpoons NH_4^+(aq) + OH^–(aq)$

Brønsted-Lowry Acid-Base Reactions (Reversible)

• One advantage of Brønsted-Lowry theory is its allowance for reversible reactions.

• $H–A + :B \rightleftharpoons :A– + H–B^+$

• The original base, after gaining an H+ (H-B+), can act as an acid in the reverse process.

• The original acid, after losing an H+ (:A-), can act as a base in the reverse process.

• $:A– + H–B^+ \rightleftharpoons H–A + :B$

Conjugate Pairs

• In a Brønsted-Lowry acid–base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process.

• Each reactant and the product it becomes is called a conjugate pair.

• The original base becomes its conjugate acid; the original acid becomes its conjugate base.

More on Brønsted-Lowry Acid–Base Reactions

• General reaction: $H–A + :B \rightleftharpoons :A– + H–B^+$

  • acid base conjugate conjugate

  • base acid

• Example 1: $HCHO2 + H2O \rightleftharpoons CHO2^– + H3O^+$

  • acid base conjugate conjugate

  • base acid

• Example 2: $H2O + NH3 \rightleftharpoons HO^– + NH_4^+$

  • acid base conjugate conjugate

  • base acid

Conjugate Pairs (Specific Examples)

• In the reaction $H2O + NH3 \rightleftharpoons HO^– + NH_4^+$

• H2O and HO– constitute an acid/conjugate base pair.

• NH3 and NH4 + constitute a base/conjugate acid pair.

Practice – Conjugate Acids

• Write the formula for the conjugate acid of the following:

  • H2O → H3O+

  • NH3 → NH4 +

  • CO3 2− → HCO3 −

  • H2PO4 1− → H3PO4

Practice – Conjugate Bases

• Write the formula for the conjugate base of the following:

  • H2O → HO−

  • NH3 → NH2 −

  • CO3 2− → cannot be an acid because it does not have an H

  • H2PO4 1− → HPO4 2−

Example: Identifying Brønsted-Lowry Acids and Bases

• Reaction: $H2SO4 + H2O \rightleftharpoons HSO4^– + H_3O^+$

  • acid base conjugate conjugate

  • base acid

• When H2SO4 becomes HSO4 , it loses an H+, so H2SO4 must be the acid, and HSO4  is its conjugate base.

• When H2O becomes H3O+, it accepts an H+, so H2O must be the base, and H3O+ is its conjugate acid.

Example: Identifying Brønsted-Lowry Acids and Bases

• Reaction: $HCO3^– + H2O \rightleftharpoons H2CO3 + HO^–$

  • base acid conjugate conjugate

  • acid base

• When HCO3  becomes H2CO3, it accepts an H+, so HCO3  must be the base, and H2CO3 is its conjugate acid.

• When H2O becomes OH , it donates an H+, so H2O must be the acid, and OH is its conjugate base.

Practice – Identifying Acids, Bases, and Conjugates

• Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the following reaction:

• $HSO4^− (aq) + HCO3^− (aq) \rightleftharpoons SO4^{2−} (aq) + H2CO_3(aq)$

• Base Conjugate Acid Acid Conjugate Base

Practice – Equations with Water as Monoprotic Acid

• Write equations for the following reacting with water and acting as a monoprotic acid; label the conjugate acid and base

• $HBr + H2O \rightleftharpoons Br^− + H3O^+$

  • Acid Base Conj. Conj.

  • base acid

• $HSO4^− + H2O \rightleftharpoons SO4^{2−} + H3O^+$

  • Acid Base Conj. Conj.

  • base acid

Practice – Equations with Water as Monoprotic Accepting Base

• Write equations for the following reacting with water and acting as a monoprotic-accepting base; label the conjugate acid and base

• $I^− + H_2O \rightleftharpoons HI + OH^−$

  • Base Acid Conj. Conj.

  • acid base

• $CO3^{2−} + H2O \rightleftharpoons HCO_3^− + OH^−$

  • Base Acid Conj. Conj.

  • acid base

Comparing Arrhenius and Brønsted-Lowry

• Brønsted–Lowry theory examples:

  • $HCl(aq) + H2O(l) \rightarrow Cl^−(aq) + H3O^+(aq)$

  • $HF(aq) + H2O(l) \rightleftharpoons F^−(aq) + H3O^+(aq)$

  • $NaOH(aq) \rightarrow Na^+(aq) + OH^−(aq)$

  • $NH3(aq) + H2O(l) \rightleftharpoons NH_4^+(aq) + OH^−(aq)$

• Arrhenius theory examples:

  • $HCl(aq) \rightarrow H^+(aq) + Cl^−(aq)$

  • $HF(aq) \rightleftharpoons H^+(aq) + F^−(aq)$

  • $NaOH(aq) \rightarrow Na^+(aq) + OH^−(aq)$

  • $NH4OH(aq) \rightleftharpoons NH4^+(aq) + OH^−(aq)$

Arrow Conventions

• Chemists use two kinds of arrows in reactions to indicate the degree of completion of the reactions

• A single arrow indicates all the reactant molecules are converted to product molecules at the end

• A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products

Strong or Weak

• A strong acid is a strong electrolyte

  • Practically all the acid molecules ionize, →

• A strong base is a strong electrolyte

  • Practically all the base molecules form OH– ions, either through dissociation or reaction with water, →

• A weak acid is a weak electrolyte

  • Only a small percentage of the molecules ionize, ⇌

• A weak base is a weak electrolyte

  • Only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water, ⇌

Strong Acids

• $HCl \rightarrow H^+ + Cl^−$

• $HCl + H2O \rightarrow H3O^+ + Cl^−$

• 0.10 M HCl = 0.10 M H3O+

• The stronger the acid, the more willing it is to donate H

  • We use water as the standard base to donate H to.

• Strong acids donate practically all their H’s

  • 100% ionized in water

  • Strong electrolyte

• $[H_3O^+] = [strong acid]$

• [X] means the molarity of X

Weak Acids

• $HF \rightleftharpoons H^+ + F^−$

• $HF + H2O \rightleftharpoons H3O^+ + F^−$

• 0.10 M HF ≠ 0.10 M H3O+

• Weak acids donate a small fraction of their H’s

  • Most of the weak acid molecules do not donate H to water

  • Much less than 1% ionized in water

• [H_3O^+] << [weak acid]

Strengths of Acids & Bases

• Acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water

  • $HAcid + H2O \rightleftharpoons Acid^− + H3O^+$ $Base: + H_2O \rightleftharpoons HBase^+ + OH^−$

• The farther the equilibrium position lies toward the products, the stronger the acid or base

• The position of equilibrium depends on the strength of attraction between the base form and the H+

  • Stronger attraction means stronger base or weaker acid

General Trends in Acidity

• The stronger an acid is at donating H, the weaker the conjugate base is at accepting H

• Higher oxidation number = stronger oxyacid

  • H2SO4 > H2SO3; HNO3 > HNO2

• Cation stronger acid than neutral molecule; neutral stronger acid than anion

  • H3O^+ > H2O > OH^−; NH4^+ > NH3 > NH_2^−

  • Trend in base strength opposite

Acid Ionization Constant, Ka

• Acid strength measured by the size of the equilibrium constant when reacts with H2O

  • $HAcid + H2O \rightleftharpoons Acid^− + H3O^+$

• The equilibrium constant for this reaction is called the acid ionization constant, Ka

  • Larger Ka = stronger acid

Autoionization of Water

• Water is an extremely weak electrolyte, thus there must be a few ions present

• About 2 out of every 1 billion water molecules form ions through a process called autoionization

  • $H_2O \rightleftharpoons H^+ + OH^−$

  • $H2O + H2O \rightleftharpoons H_3O^+ + OH^−$

• All aqueous solutions contain both H3O+ and OH–

  • The concentration of H3O+ and OH– are equal in water

  • $[H_3O^+] = [OH^–] = 10^{−7}M @ 25 °C$

Ion Product of Water

• The product of the H3O+ and OH– concentrations is always the same number

• The number is called the Ion Product of Water and has the symbol Kw (= Dissociation Constant of Water)

  • $[H_3O^+] x [OH^–] = Kw = 1.00 x 10^{−14} @ 25 °C$

  • If you measure one of the concentrations, you can calculate the other

• As [H3O+] increases the [OH–] must decrease so the product stays constant

  • Inversely proportional

Acidic and Basic Solutions

• All aqueous solutions contain both H3O+ and OH– ions

• Neutral solutions have equal [H3O+] and [OH–]

  • $[H_3O^+] = [OH^–] = 1.00 x 10^{−7}$

• Acidic solutions have a larger [H3O+] than [OH–]

  • [H_3O^+] > 1.00 x 10^{−7}; [OH^–] < 1.00 x 10^{−7}

• Basic solutions have a larger [OH–] than [H3O+]

  • [H_3O^+] < 1.00 x 10^{−7}; [OH^–] > 1.00 x 10^{−7}

• Image included showing the relative concentrations of H+ and OH- in aqueous solutions at 25°C.

Practice – [H+] vs. [OH−]

• Complete the table for [H+] vs. [OH−] as follows:

  • If:

    • [H+] = 10^0

    • [H+] = 10^-1

    • [H+] = 10^-3

    • [H+] = 10^-5

    • [H+] = 10^-7

    • [H+] = 10^-9

    • [H+] = 10^-11

    • [H+] = 10^-13

    • [H+] = 10^-14

  • Then:

    • [OH-] = 10^-14

    • [OH-] = 10^-13

    • [OH-] = 10^-11

    • [OH-] = 10^-9

    • [OH-] = 10^-7

    • [OH-] = 10^-5

    • [OH-] = 10^-3

    • [OH-] = 10^-1

    • [OH-] = 10^0

• Notes:

  • Acid: H+ > OH-

  • Base: OH- > H+

Example: Calculating [OH−] from [H3O+]

• Calculate the [OH] at 25 °C when the [H3O+] = 1.5 x 10−9 M, and determine if the solution is acidic, basic, or neutral

• Conceptual Plan: [H3O+] [OH]

• Relationships: $[H_3O^+][OH^-]=1.00x10^{-14}$

• Given: $[H_3O^+] = 1.5 x 10^{-9} M$

• Find: $[OH^−]$

• Solution:

  • $[OH^-]=\frac{1.00x10^{-14}}{[H_3O^+]}=\frac{1.00x10^{-14}}{1.5x10^{-9}}=6.7x10^{-6}M$

• Check:

  • The units are correct; the fact that the [H_3O^+] < [OH^−] means the solution is basic

Practice: Determining [H3O+] from [OH−]

• Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M

• Conceptual Plan: [OH] [H3O+]

• Relationships:$[H_3O^+][OH^-]=1.00x10^{-14}$

• Given: $[OH^−] = 2.5 x 10^{−9} M$

• Find: $[H_3O^+]$

• Solution: $[H_3O^+] = (1.00 x 10^{-14}) / (2.5 x 10^{-9}) = 4.0 x 10^{-6} M$

• Check: the units are correct; the fact that the [H3O+] > [OH] means the solution is acidic

Measuring Acidity: pH

• The acidity or basicity of a solution is often expressed as pH

• $pH = −log[H_3O^+]$

  • Exponent on 10 with a positive sign

  • $pH_{water} = −log[10^{−7}] = 7$

  • Need to know the [H3O+] concentration to find pH

• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral

• $[H_3O^+] = 10^{−pH}$

• Image included showing a digital pH meter.

Sig Figs & Logs

• When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number:

  • $log(2.0 x 10^6) = log(10^6) + log(2.0) = 6 + 0.30303… = 6.30303…$

• Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log

  • $log(2.0 x 10^6) = 6.30$

What Does the pH Number Imply?

• The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution

  • 1 pH unit corresponds to a factor of 10 difference in acidity

• Normal range of pH is 0 to 14

  • pH 0 is [H3O+] = 1 M, pH 14 is [OH–] = 1 M

  • pH can be negative (very acidic) or larger than 14 (very alkaline)

Example: Calculating pH from [OH-]

• Calculate the pH at 25 °C when the [OH] = 1.3 x 10−2 M, and determine if the solution is acidic, basic, or neutral

• Conceptual Plan: [OH ] [H3O+] pH

• Relationships:

  • $[H_3O^+][OH^-]=1.00x10^{-14}$

  • $pH = -log[H_3O^+]$

• Given: $[OH^−] = 1.3 x 10^{−2} M$

• Find: pH

• Solution:

  • $[H_3O^+]=\frac{1.00x10^{-14}}{[OH^-]}=\frac{1.00x10^{-14}}{1.3x10^{-2}}=7.69x10^{-13}M$

  • $pH = -log[H_3O^+]=-log(7.69x10^{-13})=12.11$

• Check:

  • pH is unitless; the fact that the pH > 7 means the solution is basic

Practice – Determine pH from [OH-]

• Determine the pH @ 25 ºC of a solution that has [OH−] = 2.5 x 10−9 M

• Conceptual Plan: [OH ] [H3O+] pH

• Relationships:

  • $[H_3O^+][OH^-]=1.00x10^{-14}$

  • $pH = -log[H_3O^+]$

• Given: [OH] = 2.5 x 10−9 M

• Find: pH

• Solution:

  • $[H_3O^+]=\frac{1.00x10^{-14}}{[OH^-]}=\frac{1.00x10^{-14}}{2.5x10^{-9}}=4.0x10^{-6}M$

  • $pH = -log[H_3O^+]=-log(4.0x10^{-6})=5.40$

• Check:

  • pH is unitless; the fact that the pH < 7 means the solution is acidic

Practice – Determine [OH−] from pH

• Determine the [OH−] of a solution with a pH of 5.40

• Conceptual Plan: pH [H3O+] [OH]

• Relationships:

  • $[H_3O^+]=10^{-pH}$

  • $[H_3O^+][OH^-]=1.00x10^{-14}$

• Given: pH = 5.40

• Find: [OH−], M

• Solution:

  • $[H_3O^+]=10^{-5.40}=3.98x10^{-6}M$

  • $[OH^-]=\frac{1.00x10^{-14}}{3.98x10^{-6}}=2.51x10^{-9}M$

• Check:

  • because the pH < 7, [OH−] should be less than 1 x 10−7; and it is

pOH

• Another way of expressing the acidity/basicity of a solution is pOH

• $pOH = −log[OH^−]$, $[OH^−] = 10^{−pOH}$

  • $pOH_{water} = −log[10^{−7}] = 7$

  • Need to know the [OH] concentration to find pOH

• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral

• $pH + pOH = 14.0$

pH and pOH

• Complete the table for pH and pOH as follows:

  • If:

    • pH = 0

    • pH = 1

    • pH = 3

    • pH = 5

    • pH = 7

    • pH = 9

    • pH = 11

    • pH = 13

    • pH = 14

  • Then

    • pOH = 14

    • pOH = 13

    • pOH = 11

    • pOH = 9

    • pOH = 7

    • pOH = 5

    • pOH = 3

    • pOH = 1

    • pOH = 0

• Notes:

  • Acid: H+ > OH-

  • Base: OH- > H+

Relationship between pH and pOH

• $pH + pOH = 14.00$

  • at 25 °C

  • you can use pOH to find pH of a solution

Example: Calculating pH from [OH-] using pOH

• Calculate the pH at 25 °C when the [OH] = 1.3 x 10−2 M, and determine if the solution is acidic, basic, or neutral

• Conceptual Plan: [OH] pOH pH

• Relationships:

  • $pOH = -log[OH^-]$

  • $pH+pOH=14.00$

• Given: [OH] = 1.3 x 10−2 M

• Find: pH

• Solution:

  • $pOH = -log[OH^-]=-log(1.3x10^{-2})=1.89$

  • $pH = 14-pOH=14-1.89=12.11$

• Check: pH is unitless; the fact that the pH > 7 means the solution is basic

Practice – Determine pOH from [H3O+]

• Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2.5 x 10−9 M

• Conceptual Plan: [H3O+] pH pOH

• Relationships:

  • $pH = -log[H_3O^+]$

  • $pH+pOH=14.00$

• Given: $[H_3O^+] = 2.5 x 10^{−9} M$

• Find: pOH

• Solution:

  • $pH = -log[H_3O^+]=-log(2.5x10^{-9})=8.6$

  • $pOH = 14-pH=14-8.6=5.4$

• Check: pH is unitless; the fact that the pH < 7 means the solution is acidic

pK

• A way of expressing the strength of an acid or base is

A larger Ka value means that at equilibrium, the concentrations of the products (H₃O⁺ and A⁻) are higher relative to the concentration of the undissociated acid (HA). A higher concentration of H₃O⁺ ions indicates a greater extent of ionization, which is the defining characteristic of a stronger acid. It readily donates protons (H⁺) to water, forming more hydronium ions.