Net Ionic Equations & Solubility

Interpreting Net Ionic Equations and Spectator Ions

• A net ionic equation shows the chemical species that undergo reaction
• Spectator ions appear on both sides of the reaction and do not participate chemically
• To formulate the net ionic equation, eliminate spectator ions
• Given AgNO3 and NaCl are put in water: Ag+ + NO3– + Na+ + Cl– → NO3– + Na+ + AgCl (s)
• Na+ and NO3– are spectator ions
• The net ionic equation for the reaction is: Ag+ + Cl– → AgCl (s)
• Multiply charged species are generally insoluble, except for sulfates like CuSO4
• Nitrates, chlorates, and ammonium salts are always soluble
• Alkali metals and halogens are generally soluble, except for silver (Ag) salts like AgCl

Estimating Solubility from Ksp

• Salts dissociate to varying degrees in water which is measured by molar solubilitiy and Ksp
• Ksp represents the equilibrium between a solid salt and its dissolved ions and depends on the number of ions that dissociate
• Molar solubility is the amount of solute that exists in a saturated solution at equilibrium
• A small Ksp (large negative exponent) indicates low compound solubility
• The relationship between Ksp and molar solubility values depends on the number of ions that dissociate

Ksp and Molar Solubility Dissociation Table

• AB (2 ions): Ksp = [A][B], Ksp = x²
• AB2 or A2B (3 ions): Ksp = [A][B]2, Ksp = 4x³
• AB3 or A3B (4 ions): Ksp = [A][B]3, Ksp = 27x4
• A2B3 or A3B2 (5 ions): Ksp = [A]2[B]3, Ksp = 108x5
• Comparing Ksp values to rank solubility requires considering the number of ions in solution
• Estimate solubility by taking the root of the Ksp, where the root is the number of ions in the salt
• CaCO3: 2 ions form → Ksp = x² = 10-10, so x = 10-5
• Ag2CO3: 3 ions form → Ksp = x3 = 10-13, so x = 10-4.33
• Since 10-4.33 is larger than 10-5, Ag2CO3 exhibits greater molar solubility and is more soluble

Calculating Molar Solubility from Ksp

• Calculating molar solubility involves exact math rather than estimation
• Divide by the coefficient 4, 27, or 108 first

Molar Solubility Equations by Number of Ions

• 2 Ions: Ksp = x²
• 3 Ions: Ksp = 4x³
• 4 Ions: Ksp = 27x4
• 5 Ions: Ksp = 108x5
• Ag2CO3 has 3 ions: 2 Ag+ and 1 CO32-, so set Ksp = 4∙10-12 = 4x³ then solve for x: x=(3√4∙10−12)/4=3√10−12=𝟏𝟎−𝟒

Common Ion Calculation

• A soluble salt is added to an insoluble salt with a common ion
• By Le Chatelier's principle, the dissolution of the soluble salt shifts left reducing solubility

Steps to Solve Common Ion Problems

• Identify the common ion and the insoluble salt (IS)
• IS has a provided Ksp
• The common ion shows up in both compounds
• Write the Ksp expression for the insoluble salt (Ksp = [Mg2+][OH–]2 = 1.2 x 10-11)
• Plug in the concentration of the common ion directly, for example: Ksp = Mg2+2 = 1.2 x 10-11
• Solve for the concentration of the other ion for the new molar solubility
• In the example described, [Mg2+] = 1.2 x 10-9 is the new molar solubility

Selective Precipitation

• There are three ions in water and a mystery anion X¯ is added
• The ion precipitating first is the one with the lowest molar solubility
• The farther apart Ksp values are for the salts (same # ions), the easier it is to selectively precipitate
• If 0.1 M Au⁺, 0.1 M Cu⁺, and 0.1 M Ag+are in solution, the X concentration required for precipitation depends on the salt
• AuX with Ksp = 10-20 requires [X] = 10-19 M
• CuX: Ksp = 10-12 requires [X] = 10-11 M
• AgX: Ksp = 10-10 → [X¯] = 10-9 M
• AuX precipitates first as it has the smallest Ksp and requies the lowest X concentration

Acid/Base Theory

Arrhenius Theory

• Acids and bases are dissociation products for water: 2 H2O → H3O+ + OH–
• H3O+ (hydronium ion) is the acid
• OH– (hydroxide ion) is the base

Brønsted–Lowry Theory

• Built around the proton, H+
• A Brønsted acid is a proton (H+) donor such as HCOOH, HCl, or NH4+
• These acids becomes their conjugate bases ( HCOO–, Cl–, and NH3) when donating a proton
• A Brønsted base is a proton (H+) acceptor for example: HCOO–, Cl–, and NH3

Lewis Theory

• Build around the unbonded electron pair
• A Lewis base is an electron donor with unpaired electrons
• :NH3 is a famous example
• A Lewis acid is an electron acceptor with empty orbitals that binds to Lewis bases
• Borane (BH3) is a famous example and has an empty p orbital: BH3 + :NH3 → H3B:NH3

Autoprotolysis of Water

• Describes the properties of pure water (H2O) without added acids or bases

Definition of Neutral Water

• Given H2O (l) ⇌ H+ (aq) + OH– (aq), Kw = [H+][OH–] where [H+] = [OH–]
• At room temperature (25°C), Kw = 10-14, therefore [H+] = [OH–] = 10-7 M
• Dissociation of water is negligible where [H+] = [OH–] and pH = pOH

Temperature Dependence

• Water dissociation is endothermic due to the energy required to break a bond indicated by: heat + H2O (l) ⇌ H+ (aq) + OH– (aq)
• According to Le Chatelier’s Principle and the van’t Hoff equation: as T↑, Kw↑ and as T↓, Kw↓
• Water is still neutral because [H+] = [OH–]
• pH = 7 only at 25°C, but increases or decreases with temperature changes

Autoprotolysis of Water Values By Temp

• 0°C: Kw = 10-16 , [H+] (M) = 10-8, pH = pOH = 8
• 25°C: Kw = 10-14, [H+] (M) = 10-7 pH = pOH = 7
• 100°C: Kw = 10-12, [H+] (M) = 10-6 pH = pOH = 6

Navigating the Acid/Base Terrain

Conversions Between pH, pOH, [H+], and [OH–]

• pH = -log[H+]
• pOH = -log[OH–]
• Kw = [H+][OH–] = 10-14 at 25°C
• pKw = pH + pOH = 14 at 25°C

Conversions Between pKa, pKb, Ka, and Kb

• pKa = -log(Ka)
• pKb = -log(Kb)
• Kw = KaKb = 10-14 at 25°C
• pKw = pKa + pKb = 14 at 25°C

Ranking Acidity and Basicity based on K Values

• "The larger the Ka (= smaller pKa), the stronger the acid"
• "The larger the Kb (= smaller pKb), the stronger the base"
• Kw = KaKb, implies that "the stronger the acid, the weaker the conjugate base” (and vice versa)

Equilibrium Acid/Base Calculation: Strong Case

• Equilibrium calculations do not require a RICE table because there is no reaction and the system is already at equilibrium

Steps for Strong Acids and Bases Problems

• Memorize the 7 strong acids and 6 strong bases by formula and name
• Remove spectator ions leaving only H+ or OH-
• Convert to pH to answer the question

Strong Acids (Memorize)

• HCl (Hydrochloric acid)
• HBr (Hydrobromic acid)
• HI (Hydroiodic acid)
• HNO3 (Nitric acid)
• H2SO4 (Sulfuric acid)
• HClO3 (Chloric acid)
• HClO4 (Perchloric acid)

Strong Bases (Memorize)

• LiOH (Lithium hydroxide)
• NaOH (Sodium hydroxide)
• KOH (Potassium hydroxide)
• RbOH (Rubidium hydroxide)
• Ba(OH)2 (Barium hydroxide)
• Sr(OH)2 (Strontium hydroxide)

Equilibrium Acid/Base Calculation: Weak Case

• Weak acid/base calculation involves simply dumping a weak acid or base into water and finding the pH

Steps:

• Remove spectator ions
• Identify compound as strong or weak acid or base and its form: H+, OH-, HA, A-, BH+, or B
• There is no neutralization happening in the "simple" case, use these:
  • Weak Acid: [H+] = (KaCa)1/2
  • Weak Base: [OH-] = (KbCb)1/2
• Convert to pH, pOH, [H+], or [OH-] as requested to answer the question

Equilibrium Acid/Base Calculation: Buffer

• A buffer solution contains a weak acid/base and its conjugate base/acid in water to resist pH changes

Simple Buffer Calculation Steps

• Step 1: Remove spectator ions
• Step 2: Identify compounds as strong or weak acid or base and its form: H+, OH-, HA, A-, BH+, or B and recognize they are conjugates
  • HA and A- or BH+ and B for example: HF and F- or NH4+ and NH3
• Step 3: There is no neutralization happening in the "simple" case, so solve the appropriate calculation for the [H+] and [OH-] concentrations.
  • [H+] = Ka(Ca/Cb) or [OH-] = Kb(Cb/Ca)
• Step 4: Convert to pH, pOH, [H+], or [OH-] as requested

Neutralization: Strong Acid/Strong Base Case

Neturalization Calculation Steps

• Step 1: Remove Spectator Ions
• Step 2: Identify if the compounds are strong or weak acid or base
• Step 3: Determine if they neturalize, if there is both an acid and a base, and at least one is strong
• Step 4: Set-up a RICE table and convert to moles, determine the limiting reagent

Neutralization Strong Acids and Bases

• Strong Acids with Strong Bases use neturalization for the folllwing reactions
  • H+ + OH- -> H2O -HA (aq) + OH- (aq -> A-(aq) + H2O (I)
  • Example: What is the pH when 100 mL of 1 M HCIO4 is titrated with 900 mL of 0.1 M NaOH
  • If you can do quick recognition, there will be more strong acid remaining, and the pH is low(0-2pH)
  • If not, the limiting reagent in the problem should be solved, and the mole difference/ratio to the 1L should be used determine if the H+ or OH- remains

Neutralization to Buffer Region

Neutralization Calculation steps

• Step 1: Remove spectator ions
• Step 2: Identify compounds that are strong or weak acids and bases; for buffers recongize the the compounds are conjugates
• Step 3: Confirm there is neturalization, if an acid + base remains with at least one being strong
• Step 4: Creating a RICE table with converted moles, and solve the limiting reagent problem
• Step 5: Indentify what type of remains: for the remaining moles for the approrpriate calculation
  • H+ + BH- <-> B + H2O (Weak Acid form-> H+ and A(weak Base form
• Step 6: Convert to pH, pOH, (H+), or (OH<->

Identifying features of a Titration Curve

Key items and features

• Analyte: the known item for concentration and/or equilibrium constant
• Titrant: is the known Solution you are adding to
  • Remove any spectator ions
  • for a reaction to be neturalized
• HA (aq) + OH (aq)->A-)aq) +H2p(l) A- (aq) + H+ (aq) -> HA

Weak Neutralization is only possible when the A- compared to HA can only be slightly change(and pH

• There is H+ added to be
• Set-up and RICE chart, confirm the 2 species left in solution are conjugates
• H=Ka* Ca/Cb -> plug and solve
• OR use calculator to determine the ka, pH

Polyprotic Acids, Equillibrium Expressions

• Is for the D and Triprotic acids, Know to find the equilibrium and K for with the followings

diprotic acids (H2A)

• H2A <-> HA- + H+
• HA --<-> A2- +H+ (pka1,pka2) (Ka3 is the ion association to always know hich Pka we can use

triprotic acids

• H2A <-> H2A -+ H+
• H2A <-> HA2 - H+
• HA-> A3 -+H +

Polyprotic Acids, Calculation

• If Polyprotic acid or its salts are thrown into solution you must find the pH

The following examples, follow these step to solve

• Remove Spectators and then ID the compound with is NaH2po4/becomes h2ap4 , which is and- because is a charge, is species of

• H2A - ->HA

• A2 and since since the HA-. the pear by pks there are 4 and 7, we can do pH is 0.5

Polyprotic Acid: Netrualization

• to determine the netrauzliaiton: find p if polyactic acid
• 2moles of Nh A and 1 more of n

Polyprotic Acid Alpha Diagrams

• Fractional Diagram- diagram of the alpha will need interpreation:

1. to create fraction vs pH
2. Identify Pck; Value