http://www.cdli.ca/courses/ http://www.cbhs.k12.nf.ca/adrianyoung/ Unit 2 Acids and Bases http://www.cdli.ca/courses/ http://www.cbhs.k12.nf.ca/adrianyoung/

Topics Properties / Operational Definitions Acid-Base Theories pH & pOH calculations Equilibria (Kw, Ka, Kb) Indicators Titrations STSE: Acids Around Us

Operational Definitions An Operational Definition is a list of properties, or operations that can be performed, to identify a substance. See p. 550 for operational definitions of acids and bases

Operational Definitions (Properties – see p. 550) Acids pH < 7 taste sour react with active metals (Mg, Zn) to produce hydrogen gas Bases pH > 7 taste bitter no reaction with active metals feel slippery Sour – citric acid (lemons, limes, grapefruit), vinegar, sour candy Bitter – unflavoured milk of magnesia, coffee, unsweetened chocolate

Operational Definitions Acids blue litmus turns red react with carbonates to produce CO2 gas Bases red litmus turns blue no reaction with carbonates

Operational Definitions Acids conduct electric current neutralize bases to produce water and a “salt” Bases conduct electric current neutralize acids to produce water and a “salt” any ionic compound any ionic compound

Acid-Base Theories 1. Arrhenius Theory (p. 549 ) acid – any substance that dissociates or IONIZES in water to produce H+ ions ie. an acid must contain H+ ions

Arrhenius Theory eg. HCl(aq) → H+(aq) + Cl-(aq) H2SO4(aq) → H+(aq) + HSO4-(aq) H+(aq) + SO42-(aq)

Arrhenius Theory base – any substance that dissociates in water to produce OH- ions ie. a base must contain OH- ions

Arrhenius Theory eg. NaOH(aq) → Ca(OH)2(aq) → Na+(aq) + OH-(aq) Ca+(aq) + 2 OH-(aq)

Arrhenius Theory Which is an Arrhenius acid? a) KOH c) CH4 b) HCN d) CH3OH Which is a Arrhenius base?

Limitations of Arrhenius theory (p.551) H+ cannot exist as an ion in water. The positive H+ ions are attracted to the polar water molecules forming HYDRONIUM ions or H3O+(aq) H+(aq) + H2O(l) → H3O+(aq)

Limitations of Arrhenius theory CO2 dissolves in water to produce an acid. NH3 dissolves in water to produce a base. Neither of these observations can be explained by Arrhenius theory

Limitations of Arrhenius theory Some acid-base reactions can occur in solvents other than water. Arrhenius theory can explain only aqueous acids or bases.

Limitations of Arrhenius theory Arrhenius theory is not able to predict whether certain species are acids or bases. eg. NaHSO4 H2PO4- HCO3- Arrhenius theory needs some work

To be used when Arrhenius is inadequate Acid-Base Theories To be used when Arrhenius is inadequate 2. Modified Arrhenius Theory (p. 552) acid – any substance that reacts with water to produce H3O+ ions eg. HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

Modified Arrhenius Theory base – any substance that reacts with water to produce OH- ions eg. NH3(aq) + H2O(l) )→ NH4+(aq) + OH-(aq) pp. 558, 559 #’s 1, 3, 8, & 9

3. Brønsted-Lowry Theory (p. 553) Acid-Base Theories 3. Brønsted-Lowry Theory (p. 553) acid – any substance from which a proton (H+) may be removed ie. an acid is a substance that loses a proton (H+)

Brønsted-Lowry Theory base – any substance that can remove a proton (H+) from an acid. ie. a base is a substance that gains a proton (H+) In BLT , an acid-base reaction requires the transfer of a proton (H+) from an acid to a base.

Brønsted-Lowry Theory conjugate acid base eg. HCN(aq) + NH3(aq) → ← CN-(aq) + NH4+(aq) conjugate base acid

Brønsted-Lowry Theory What is a conjugate acid-base pair?? (p. 554) Two particles (molecules or ions) that differ by one proton are called a conjugate acid-base pair. The conjugate base forms when an acid loses a proton. The conjugate acid forms when a base gains a proton (H+).

Brønsted-Lowry Theory conjugate acid base conjugate base acid

Brønsted-Lowry Theory conjugate acid base eg. NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq) H2O(l) + H2O(l) → ← conjugate base acid ←

Brønsted-Lowry Theory - an amphoteric substance can be either an acid or a base these include WATER and negative ions that contain at least one hydrogen atom eg. H2O, HCO3-(aq), H2PO4-(aq)

Brønsted-Lowry Theory p.557 #’s 1 – 7 p. 558 #’s 8, 9 p. 559 #’s 2, 4-7, 10,11 Quiz - Tuesday Mar. 1

Strength of Acids and Bases Strong acids produce more H+ ions OR more H3O+ ions than weak acids with the same molar concentration A strong acid is an acid that ionizes or dissociates 100% in water eg. HCl(aq)→ Strong acids react 100% with water (BLT) eg. HCl(aq) + H2O(l) → H+(aq) + Cl-(aq) Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf H3O+(aq) + Cl-(aq)

Strength of Acids and Bases NOTE: The equilibrium symbol, º , is NOT used for strong acids because there is NO REVERSE REACTION. Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases A weak acid is an acid that ionizes or dissociates LESS THAN 100% eg. HF(aq) Weak acids react less than 100% with water eg. HF(aq) + H2O(l) Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases For weak acids, an equilibrium is established between the original acid molecule and the ions formed. DO NOT confuse the terms strong and weak with concentrated and dilute. Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases eg. Classify the following acids: 0.00100 mol/L HCl(aq) strong and dilute 12.4 mol/L HCl(aq) strong and concentrated 10.5 mol/L CH3COOH(aq) weak and concentrated Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases monoprotic – acids that contain or lose one proton diprotic – acids that contain or lose two protons polyprotic – acids that have more than one proton Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases A strong base is a base that dissociates 100% in water, or reacts 100% with water, to produce OH- ion. The only strong bases are hydroxide compounds of most Group 1 and Group 2 elements eg. NaOH(s) → Ca(OH)2(s) → Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Strength of Acids and Bases A weak base is a base that reacts less than 100% in water to produce OH- ion. eg. S2-(aq) + H2O(l) º HS-(aq) + OH-(aq) Football analogy: - SA like the QBack that delivers the ball; - WA like the QB that holds onto it - SB is like the wide receiver that always catches the ball:100% completion - WB is like the wide receiver that tends to drop the ball http://chemweb.stanford.edu/spring2004/chem33/Outreach/football.pdf

Writing Acid-Base Equations (BLT) Step 1: List all the molecules/ions present in the solution ionic compounds form cations and anions strong acids exist as hydronium ion and the anion (conjugate base) for weak acids use full formula of the compound (i.e. un-ionized molecule) always include water in the list.

Writing Acid-Base Equations (BLT) Step 2: Identify the strongEST acid and the strongEST base from Step 1. Step 3: Write the equation for the reaction by transferring a proton from the strongest acid to the strongest base.

Writing Acid-Base Equations (BLT) Step 4: Determine the type of reaction arrow to use in the equation. Stoichiometric (100%) reactions occur between: Hydronium (H3O+) and bases stronger than nitrite (NO2-) hydroxide (OH-)and acids stronger than hypochlorous acid (HOCl)

Writing Acid-Base Equations (BLT) Step 5: Determine the position of the equilibrium by comparing the strengths of both acids in the equation. The favoured side is the side with the weaker acid! 

Writing Acid-Base Equations (BLT) Sample problems: Write the net ionic equation for the acid-base reaction between: - aqueous sodium hydroxide (NaOH(aq)) and hydrochloric acid (HCl(aq)).

H3O+(aq) + OH-(aq) → 2 H2O(l) species present Na+(aq) OH-(aq) H3O+(aq) Cl-(aq) H2O(l) strongest acid strongest base H3O+(aq) + OH-(aq) H2O(l) + H2O(l) OR H3O+(aq) + OH-(aq) → 2 H2O(l)

Writing Acid-Base Equations (BLT) Sample problems: Write an equation for the acid-base reaction between nitrous acid (HNO2(aq)) and aqueous sodium sulfite (Na2SO3(aq)).

º HNO2(aq) Na+(aq) SO32-(aq) H2O(l) HNO2(aq) + SO32-(aq) species present HNO2(aq) Na+(aq) SO32-(aq) H2O(l) SB SA º HNO2(aq) + SO32-(aq) NO2-(aq) + HSO3 - (aq) Weaker Acid Stronger Acid Products favored

Write the Net Ionic Equation for each aqueous reaction below: Na2CO3(aq) and CH3COOH(aq) NH3(aq) and HNO2(aq) HNO3(aq) and RbOH H2SO4(aq) and K3PO4(aq) HF(aq) and NH4CH3COO(aq) CaCl2(aq) and PbSO4(aq) p. 564 #’s 10 &11 Acids & Bases #3

º H2O(l) Na+(aq) CH3COOH(aq) CO32-(aq) species present SB SA CH3COO-(aq) + HCO3-(aq) Stronger Acid Weaker Acid Products Favoured

º NH3(aq) HNO2(aq) H2O(l) HNO2(aq) + NH3(aq) NO2-(aq) + NH4 + (aq) species present NH3(aq) HNO2(aq) H2O(l) SA SB º HNO2(aq) + NH3(aq) NO2-(aq) + NH4 + (aq) Weaker Acid Products favored Stronger Acid

H3O+(aq) + OH-(aq) → 2 H2O(l) species present NO3-(aq) Rb+(aq) H3O+(aq) OH-(aq) H2O(l) strongest acid strongest base H3O+(aq) + OH-(aq) H2O(l) + H2O(l) OR H3O+(aq) + OH-(aq) → 2 H2O(l)

species present HSO4-(aq) K+(aq) PO43-(aq) H3O+(aq) H2O(l) SA SB H3O+(aq) + PO43-(aq) H2O(l) + HPO42-(aq)

º HF(aq) H2O(l) species present NH4+(aq) CH3COO-(aq) SA SB Weaker Acid HF(aq) + CH3COO-(aq) F-(aq) + CH3COOH(aq) Weaker Acid Products favored Stronger Acid

º Ca2+(aq) Cl-(aq) Pb2+(aq) SO42-(aq) H2O(l) strongest acid species present Ca2+(aq) Cl-(aq) Pb2+(aq) SO42-(aq) H2O(l) strongest base º H2O(l) + SO42-(aq) HSO4-(aq) + OH-(l) Weaker Acid Reactants favored Stronger Acid

NO!! Products are NOT always favoured Try these: CH3COOH(aq) + NH4F(aq) HCN(aq) + NaHS(aq)

Acid-Base Calculations Kw Ka Kb [H3O+] [OH-] pH pOH

Kw (Ionization Constant for water) With very sensitive conductivity testers, pure water shows slight electrical conductivity. PURE WATER MUST HAVE A SMALL CONCENTRATION OF DISSOLVED IONS

Kw Kw = K = Auto-Ionization of water H2O(l) + H2O(l) º H3O+(aq) + OH-(aq) Kw = K = [H3O+] [OH-] [H2O] [H2O]

Kw In pure water at 25 °C; [H3O+] = 1.00 x 10-7 mol/L [OH-] = 1.00 x 10-7 mol/L Calculate Kw at 25 °C.

GET REAL!! H2O(l) + H2O(l) º H3O+(aq) + OH- LCP: What happens if we add OH- ions (NaOH(aq)) to water? shift to the left [H3O+] ? [OH-] ? Does Kw change? GET REAL!!

Kw = [H3O+] [OH-] [H3O+] [OH-] 1.00 x 10-14 = [H3O+] [OH-] 0.00357 M 4.89 x 10-3 mol/L 12.5 M 1.50 mol/L 2.80 x 10-12 2.04 x 10-12 8.00 x 10-16 6.67 x 10-15

Calculations with Kw (p. 564 – 566) For strong acids and strong bases, the [H3O+] and [OH-] may be calculated using the solute concentration. eg. What is the [H3O+] in a 2.00 mol/L solution of HNO3(aq)? Ans: 2.00 mol/L [OH-] = ???

Calculations with Kw Ans: 4.00 mol/L eg. What is the [OH-] in a 2.00 mol/L solution of NaOH(aq)? Ans: 2.00 mol/L eg. What is the [OH-] in a 2.00 mol/L solution of Ca(OH)2(aq)? Ans: 4.00 mol/L [H3O+] = ???

Calculations with Kw Ans: 0.150 mol/L eg. What molar concentration of Al(OH)3(aq) is needed to obtain a [OH-] = 0.450 mol/L? Ans: 0.150 mol/L

What is the [H3O+] and [OH-] in: [H3O+] mol/L [OH-] mol/L 1.0 x 10-8 1.0 x 10-6 5.00 x 10-14 0.200 1.50 6.67 x 10-15 1.0 x 10-2 1.0 x 10-12

p. 566 #’s 12 - 15 solute [H3O+] [OH-] 0.680 mol/L HCl(aq) 1.50 M NaOH 0.0500 M Ca(OH)2(aq) _____ mol/L HClO4(aq) 0.450 M ____ mol/L Mg(OH)2(aq) 0.500 mol/L 0.680 1.47 x 10-14 6.67 x 10-15 1.50 1.00 x 10-13 0.100 0.450 2.22 x 10-14 0.250 2.00 x 10-14 p. 566 #’s 12 - 15

By what factor does the [H3O+] change when the pH value changes by 1? pH and pOH (See p. 568)

The [H3O+] changes by a factor of 10 (10X) for each pH changes of 1. pH and pOH (See p. 568)

pH and pOH FORMULAS pH = -log [H3O+] pOH = -log [OH-] [H3O+] = 10-pH [OH-] = 10-pOH

pH and pOH eg. What is the pH of a 0.0250 mol/L solution of HCl(aq)? What is the pOH of a 0.00087 mol/L solution of NaOH(aq)? What is the pH of a 1.25 mol/L solution of KOH(aq)? [H3O+] = 0.0250 mol/L pH = 1.602 [OH-] = 0.00087 mol/L pOH = 3.06 [OH-] = 1.25 mol/L [H3O+] = 8.00 x 10-15 mol/L pH = 14.097

Significant digits in pH values? The number of significant digits in a concentration should be the same as the number of digits to the right of the decimal point in the pH value. eg. In a sample of OJ the [H3O+] = 2.5 × 10−4 mol/L pH = 3.60 (See p. 568)

[H3O+] [OH-] pH pOH 0.0035 1.2 x 10-5 4.68 9.15 8.33 x 10-15 -1.10

[H3O+] [OH-] pH pOH 0.0035 1.2 x 10-5 4.68 9.15 8.33 x 10-15 -1.10 2.9 x 10-12 2.46 11.54 8.3 x 10-10 9.08 4.92 2.1 x 10-5 4.8 x 10-10 9.32 1.4 x 10-5 7.1 x 10-10 4.85 14.079 1.20 -0.079 7.9 x 10-16 13 15.10

pH, pOH and Kw p. 569 #’s 16 – 19 p. 572 #’s 20 – 25 Examine #23. Where is the energy term in this equation? H2O(l) + H2O(l) º H3O+(aq) + OH-(aq)

Dilutions When a solution is diluted the number of moles does not change. OR ninitial = nfinal CiVi = CfVf

eg. 400. 0 mL of water was added to 25 eg. 400.0 mL of water was added to 25.0 mL of HCl(aq) that had a pH of 3.563. Calculate the pH of the resulting solution. calculate [H3O+] dilution formula calculate pH

Before dilution: [H3O+] = 10-3.563 = 2.753 x 10-4 After dilution: (2.753 x 10-4) (25.0 mL) = (Cf)(425.0 mL) [H3O+] = 1.609 x 10-5 pH = -log (1.609 x 10-5) = 4.793 p.574 #’s 26 - 29