Exam 3 Chem 2

Strong Acids

Hydrochloric Acid- HCl

Hydrobromic Acid- HBr

Hydroiodic Acid- HI

Nitric Acid- HNO3

Perchloric Acid- HClO4

Sulfuric Acid- H2SO4

Chloric Acid- HClO3

Buffer stops working when..

• If too much acid is added → all A⁻ is used up

• If too much base is added → all HA is used up

Strong Bases

Group 1:

lithium hydroxide- LiOH

sodium hydroxide- NaOH

potassium hydroxide- KOH

rubidium hydroxide- RbOH

cesium hydroxide- CsOH

Group 2:

calcium hydroxide- Ca(OH)2

strontium hydroxide- Sr(OH)2

barium hydroxide- Ba(OH)2

ammonium NH4 charge

⁺

acetate C₂H₃O₂

⁻

arsenate AsO₃

³⁻

carbonate CO₃

²⁻

chromate CrO₄

²⁻

chlorate ClO₃

⁻

nitrate NO₃

⁻

nitrite NO₂

⁻

perchlorate ClO₄

⁻

phosphate PO₄

³⁻

sulfate SO₄/sulfite SO₃

²⁻

Silver Ag and Zinc Zn

• + and 2+

If Q < Ksp= more soluble

Solution is unsaturated and will not form a precipitate

• Change in G is negative (spontaneous, reaction moves forward with high reactant concentration to reach equilibrium)

If Q > Ksp (lower Ksp)

Solution is saturated and will form a precipitate

Ksp= less soluble

• Positive change in G

Decreasing pH increases…

the solubility of salts with weak bases (react with H+ to make a weak acid)

pH formula

pH = −log[H⁺]

Bronsted-Lowry

• Acids will always donate a proton, while bases will always accept a proton

• This means that when you are looking at an acid-base reaction, the starting material WITH an H will be your acid, and it will not have an H in the products because it DONATES it

• The starting material WITHOUT an H will be your base, and it will have an H in the products because it ACCEPTS it.

• NH3+H2O→← NH4+ + OH-

Strong Acid + Strong Base Graph

Equivalence point at pH = 7 (neutral). The salt formed (e.g. NaCl) does not hydrolyze — neither ion reacts with water. The curve is symmetric and steep at the equivalence point.

Weak Acid + Strong Base Graph (look at equivalence point)

Equivalence point at pH > 7 because the conjugate base (e.g. CH₃COO⁻) is basic and hydrolyzes water. The ½ equivalence point is especially important: [acid] = [conjugate base], so the Henderson–Hasselbalch log term = 0, and pH = pKa.

Strong Acid + Weak Base graph

Equivalence point at pH < 7 because the conjugate acid (e.g. NH₄⁺) is acidic and donates a proton to water. At the ½ equivalence point, pH = pKa of the conjugate acid (= 14 − pKb of the base).

pH=pKa/pOH=pKb

Weak Acid-Strong Base titration: occurs at the half-equivalence point, the point in the buffer region where exactly half of the initial acid has been converted to its conjugate base, [A-]=[HA]

Strong Acid-Weak Base titration: occurs at the half-equivalence point, the point in the buffer region where exactly half of the weak base is neutralized, base [B] = [BH+] conj. acid

To find pH of weak base and strong acid use what formula?

Ka = Kw (1.0×10⁻¹⁴/)/Kb to find its pH

Lewis

• Acid: Electron Pair acceptor

• Base: Electron Pair donor

• BF3+ :NH3 (Base) → F3B-NH3

Arrhenius

• Acid: release H+

• Base: release OH-

• Acid Example: HCl (Hydrochloric acid)

  • When dissolved in water, it dissociates to release H+ ions.

• Base Example: NaOH (Sodium hydroxide)

  • When dissolved in water, it dissociates to release OH- ions.

[OH-] → [H+] or [H+] → [OH-]

1 × 10^-14 / [OH-]= [H+] or 1 × 10^-14 / [H+]= [OH-]

[OH-] or [H+] → pH or pOH

-log[OH-]= pOH or -log[H+]= pH

pOH → pH or pH → pOH

14-pH= pOH or 14-pOH= pH

pOH or pH → [OH-] or [H+]

10^-pOH or 10^-pH

Autoionization of Water

H2​O(l)⇌H+(aq)+OH−(aq)

Kw​=[H+][OH−]=1.0×10−14 at 25°C

Neutral water

 [H+]=[OH−]=1.0×10−7

and your pH & pOH would be 7 

In a conjugate acid-base pair:

Ka​×Kb​=Kw​ (1.0×10−14)

pKa+pKb=pKw (14) 

Large Ka

Strong acid (nearly complete ionization)

Small Ka (< 1)

Weak acid (partial ionization)

Strong acids/bases

 Completely dissociate in water

Weak acids/bases

Partially dissociate; equilibrium exists between the undissociated species and its ions.

Water is not included in K because

•  pure liquid

Buffers resist pH change due to

• presence of both weak acid or base with its conjugate.

Adding acid/base shifts equilibrium but

 barely affects pH in buffer systems.

Salt Hydrolysis

• When ions in a salt react with water

  • Some salts do hydrolyze in water; others do not.

  • Only salts derived from weak acids or weak bases undergo hydrolysis because their conjugate ions can react with water

Salt hydrolysis: If a cation in reaction with water produces a weak base and H3O+…

pH= acidic, low

Salt hydrolysis: If a anion in reaction with water produces a weak acid and produces OH-…

pH= basic, high

Salt hydrolysis: If a cation or anion produces a strong acid or base…

no hydrolysis takes place

When a salt (metal and non-metal // ionic compound) dissociates it looks like

AB→A⁺+B⁻, and if A⁺ or B⁻ can react with water, hydrolysis occurs.

To see if it can occur, we combine A⁺ and B⁻ with H2O and see what is formed

Salt Hydrolysis examples

Weak Acid expression

Weak Base expression

Larger Ka and Kb

stronger acid or base

Ka/Kb are inversely proportional to…

•  pKa and pKb

• lower (smaller) pK_a indicates a stronger acid

• lower (smaller pK_b indicates a stronger base

Percent Ionization

• measures the ionization of an acid or base

• Stronger acid → larger Ka → higher % ionization

• More dilute → higher % ionization (Le Châtelier: equilibrium shifts right)

Percent Ionization of Acid

Percent Ionization of Base

Polyprotic Acids:

• Are capable of donating as many protons as they contain

• Typically, only the first donation of proton in used when solving for pH

• Each dissociation step has a DIFFERENT Ka value

• Ka1 > Ka2 > Ka3, understand that the first H will have the highest Ka value when donated

Amphoteric:

• Molecules that can act as both a base and as an acid

• Common Example: Water (H₂O)

• Most include polyprotic acids where there is a proton and a negative charge - HPO4 (2-) → acid= PO4 (3-) + base= H2PO4(-) and H2PO4(-) → acid= HPO4(2-) + base= H3PO4

Buffers

• Buffers help prevent changes in pH when adding something to the solution

• Consist of a conjugate weak acid and base pair

Buffer Capacity

• Buffer capacity refers to the amount of strong acid or base that can be added before a drastic change in pH occurs

• more moles of buffer= higher buffer capacity

Finding Buffer Capacity

1. When given choices, eliminate choices that are not buffers

   1. meaning, the pair must be a conjugate weak acid/base pair

2. Find the smallest number within each pair and cross out the larger one

3. Compare the smallest numbers. The largest number of this group has the largest buffer capacity.

4. Make sure add total concentration too (if smallest eliminate that choice)

5. Check ratio closest to 1:1 is ideal

6. Limiting amount is the smaller concentration

How to get pKa/PKb fro Ka and Kb

-log(Ka) or -log(Kb)

Titration curve

• ½ equivalence point→ [HA] = [A-]

• Equivalence point→ mol weak acid present=mol OH- added

• At the equivalence point, pH does not equal 7 due to the weak acid equilibrium in water ( A + H2O →← HA + OH- soln. becomes basic)

Henderson-Hasselbalch equation

• NEED TO HAVE A BUFFER TO USE THIS EQUATION (use pH= -log[H], etc.)

• Captures the buffer region

When to add and subtract from [A] and [HA]

• If you ADD a strong base to a buffer solution, then you add the mole amount to the base and subtract the mole amount from the acid

• If you ADD a strong acid to a buffer solution, then you add the mole amount to the acid and subtract the mole amount from the base 

• MAKE SURE TO USE STOICHIOMETRY 

  • If you add .2 mole of Ba(OH)2, you would add .4 moles of OH

When adding strong acid to a buffer:

• the acid will react with the base to form more of the conjugate acid

• When solving:

  • acid moles subtracts from the conjugate base

  • acid adds moles to the acid

When adding strong base to a buffer:

• Base will react with the acid to form more of the conjugate base

• When solving:

  • Base moles subtract from acid moles

  • Base moles add to conjugate base moles

Titrations

Determine the concentration of an unknown acid or base by reacting it with a known solution

Titrant

• solution of known concentration

Analyte

solution of unknown concentration

Equivalence Point

 moles of acid = moles of base

Phase 1 (Before titrant is added- before any of your second material added)

•  at this point all you have in your solution is water and the original acid or base. 

  • If it’s a strong acid or strong base:

    • Fully dissociated, just use concentration to find pH/pOH.

    • We can use the [H]->pH and [OH]->[pOH]->[pH]

If it’s a weak acid/base:

• Separate the weak acid as reactants into ions as products

• Set up an ICE table with Ka​ or Kb

• Your ICE table formula will be the dissociation of the weak acid or base

• Your initial amount of acid or base will be the given molarity, and your products will be 0

• From there, you can get your H concentration, then go to [H]->pH

4 Phases of a titration: Phase 2 (Buffer region)

• BUFFER REGION ONLY HAPPENS WHEN YOU ARE DEALING WITH A STARTING WEAK ACID or WEAK BASE

  • At this point, your weak acid or base will react with an ion from your strong acid / strong base

    • If it is a weak base and your titrant is a STRONG acid, then HA + OH- → A- + H2O

    • If it is a weak acid and your titrant is a STRONG base, then B + H+ → BH+ + H2O

  • Do a BCA table 

    • We use a BCA table here because the reaction goes to completion, unlike an ice table, where the reactant and products will be at equilibrium

    • get MOLE amounts, not molarity -> need mole amts for HH equation

    • Using a BCA table means that whichever reactant has LESS moles, you can just use that as your (x) value and subtract that from the other reactants and add it to the product’s side

    • You will see that when you do this step, your reactant of H or OH will be 0 and you will have left over HA and A- or B and BH

    • You can plug these values in for HH (25.00mL of 0.100M CH3COOH titrated with 0.100 M NaOH, Ka= 1.8×10^-5 (pKa=4.74))

Phase 3 (equivalence point)

• At this point in the reaction / table, all of your original acids and bases have been turned into their conjugates 

• If strong acid and strong base titration:

  • pH will be 7 

• If weak acid and strong base titration: A- + H2O →←HA + OH-

  • All original HA has been converted to A⁻

  • Now, all the A⁻ undergoes hydrolysis (reacts with water):

  • You would use an ICE table to find the amount of OH -> pOH -> pH

• If weak base and strong acid titration: BH+ + H2O →← B + H3O+

  • All original B has been converted to BH+

  • Now, all the BH undergoes hydrolysis (reacts with water):

  • You would use an ICE table to find the amount of OH -> pOH -> pH

• USE Kw (1.0×10^-14)= Ka x Kb

• 25 + 25 because 0.002500/0.1000= 0.02500 L= 25mL

Phase 4 (excess strong base or acid)

• pH is determined by the excess of strong acid or base in solution.

  • Calculate moles of titrant added – moles reacted = excess moles.

  • Divide by total volume → find [H⁺] or [OH⁻].

  • Convert to pH or pOH.

1. Estimate the pH at 10.00 mL NaOH added

0.01000L (10mL)×0.1000M=0.001000mol

Reaction:

• HA remaining: 0.002500mol−0.001000mol=0.001500

• A⁻ formed: 0.001000

Use Henderson–Hasselbalch:

pH=4.74+log⁡(0.00100/0.001500)= 4.56

After 30.00 mL NaOH added

This is past equivalence.

Excess OH⁻:

0.03000−0.02500=0.00500 L

0.00500L x 0.1000M=0.00500mol OH-

Total Volume:

25.00+30.00=55.00 mL=0.05500 L

[OH-]= 0.000500mol/0.05500L= 9.09×10^-3

(-log) → pOH= 2.04 → pH= 11.96

Molar Solubility (x)

Moles of a solid that dissolve in 1 L of water at saturation.

1. The solubility product constant (Ksp) of calcium fluoride (CaF₂) at 25 °C is 3.9 x 10^-11. Find the molar solubility (x) of CaF₂ in pure water.

Common Ion Effect

Adding a common ion shifts equilibrium (changes initial ICE values).
Applies to weak acids/bases and solubility.
Usually decreases solubility (shifts left).

Effect of pH on Solubility

If the anion is a weak acid’s conjugate base, it reacts with H₃O⁺ → removes ion → shifts right → increases solubility.

All Brønsted–Lowry acids are

• Lewis acids (not all Lewis acids are Brønsted–Lowry)

Complex Ion Equilibria

• Metal + Lewis base → Complex ion

• Some complexes react with water → acidic

• Kf = how strongly the complex forms

• Overall solubility: Koverall = Ksp × Kf → forming a complex increases solubility

K rules

1. When reactions are combined, their K values multiply:
   K1 × K2 × K3 = Kt (overall equilibrium constant)

2. If you reverse a reaction → invert K: (1/K) 

3. If you multiply or divide coefficients by a factor → raise K to that power: Knew =K^n

• Solubility (g/L): = grams of solute that dissolve per liter.

1. Relates to molar solubility through the compound’s molar mass, pH and Solubility

2. Insoluble bases dissolve better in acidic solutions (H₃O⁺ reacts with OH⁻)

3. Insoluble acids dissolve in basic solutions (OH⁻ neutralizes H₃O⁺ or acidic H⁺ ions).

4. Insoluble salts tend to dissolve better in acidic pH environments →  salts containing anions derived from weak acids are generally more soluble in acidic solutions.
   

Insoluble base in acid:

Mg(OH)2 (s)⇌ Mg2+ (aq) + 2OH−

Adding acid (↑ H₃O⁺) → H₃O⁺ + OH⁻ → 2H₂O

Removes OH⁻ → shifts equilibrium right → more dissolves.

Insoluble acid in base:

Al(OH)3 (s) + OH−(aq) ⇌ [Al(OH)4]−(aq)

Adding base (↑ OH⁻) forms complex ion → increases solubility

ΔH (enthalpy), ΔS (entropy), Temperature (T)

1.  energy change→ Negative = favorable

2. disorder→ Positive = favorable

3. temperature → Controls importance of entropy

Equation of free energy change + spontaneity

ΔG=ΔH−TΔS 

• ΔG < 0 → spontaneous

• ΔG > 0 → non-spontaneous

• ΔG = 0 → equilibrium

Enthalpy + Disorder Results at different temps.

More moles of gas

•  ↑ entropy

• Gas > Liquid > Solid

S_mixture​>S_{pure substance​}

·       S_{high temperature}>S_{low temperature}

·       S_{increasing mass}>S_{decreasing mass}

·       ·  More complex structures → greater entropy

·       ·  For allotropes: more ordered forms → lower entropy

Temperature Dependence of ΔG

• Determines spontaneity at different temperatures

Threshold temperature (when ΔG = 0):

T=ΔH/ΔS

ΔS universe​

ΔS system​+ΔS surroundings​

Phase Change Formula

• ΔS=ΔH/T

• T must be in Kelvin

• Units: J/mol·K

ΔS^∘

∑S^∘products​−∑S^∘​reactants

• First Law of Thermodynamics

• Second Law of Thermodynamics

• Third Law of Thermodynamics

1. energy of the universe is constant, but the various forms of energy can be interchanged in physical and chemical processes.

2. in any spontaneous process, there is always an increase in entropy.

3. a perfect crystal at absolute zero (0 Kelvin) has an entropy value of zero.

Gibbs Free Energy

• measures the maximum usable energy in a system to perform work at constant temperature and pressure, and it predicts whether a reaction is spontaneous.

ΔG rxn∘

∑nΔGf∘(products)−∑mΔGf∘(reactants)

Standard conditions: 298 K, 1 atm, 1 M

• ΔGf∘ of pure elements = 0

ΔG (not at equilibrium)

• ΔG = ΔG∘ +RTln(Q)

• Q=[products]​/[reactants]

Same form as equilibrium constant K

Large K

more negative ΔG° (product-favored)