Acids & Bases

Acids vs. Bases Acids Taste Sour Are Corrosive Bases Taste Bitter Feel Slippery

Litmus Def: An Acid-Base Indicator Acids Turn blue litmus PINK Bases Turn red litmus BLUE

Reactions Acids Become less acidic when combined with a base Bases Become less basic when combined with an acid NEUTRALIZATION!

Examples Acids HCl  H + + Cl - HCl Citric Acid Acetic Acid (Vinegar) H 2 SO 4 Bases NaOH  Na + + OH - Hydroxides: LiOH, NaOH, KOH, etc. NH 3 (ammonia)

1. Arrhenius Acids - Produce hydrogen (H+) ions. Bases - Produce hydroxide (OH-) ions. + - Svante Arrhenius

2. Bronsted-Lowry Acids - Proton (H+) donors. Bases - Proton (H+) acceptors. Johannes Bronsted Proton Thomas Lowry

3. Lewis Acids - Electron pair acceptors. Bases - Electron pair donors. GN Lewis Electron

Conductivity Demo Strong vs. Weak Acids

Dissociation STRONG Acids Dissociate completely Ex. HCl  H + + Cl - H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ Cl - ace - WEAK Acids Dissociate partially Ex. HC 2 H 3 O 2  H + + C 2 H 3 O 2 - ace -

Terms Amphoteric - can behave as an acid OR a base [ex: water] Conjugate Acid - the species formed when a proton is added to a base Conjugate Base - the remaining part of the acid Hydronium - H 3 O + (hydrated H + ) Polyprotic Acids - Contain more than 1 proton (H + ) [ex: H 2 SO 4, H 3 PO 4 ]

Problems 1.HBr + H 2 O  H3O+ H3O+ + Br – A B CACB 2. HF + H 2 O  H3O+ H3O+ + F–F– A B CACB 3. H 2 CO 3 + H 2 O  H3O+ H3O+ + HCO 3 – A B CA CB HCO 3 – + H 2 O  H3O+ H3O+ + CO 3 2– A BCACB

Problems 4.HNO 3 + H 2 O  H3O+ H3O+ + NO 3 – AB CA CB 5. H 3 PO 4 + H 2 O  H3O+ H3O+ + H 2 PO 4 – A BCACB H 2 PO 4 – + H 2 O  H3O+ H3O+ + HPO 4 2– A B CA CB HPO 4 2– + H 2 O  H3O+ H3O+ + PO 4 3– ABCACB

Acid and Base Strength pH scale - measures the strength of an acid or base – Strong Acid | Weak Acid | N | Weak Base | Strong Base Strong Acids & Bases are considered dangerous (burn skin, eat away at materials Weak Acids & Bases are generally safer (used in foods and baking products)

Examples of Acids & Bases Strong AcidsStrong Bases Weak AcidsWeak Bases HCl HNO 3 (nitric) H 2 SO 4 HF HNO 2 (nitrous) HC 2 H 3 O 2 (acetic) NaOH KOH LiOH NH 3 (ammonia) CH 3 NH 2

pH Calculations Kw Kw Constant 1.Since water is an amphoteric substance, it can behave as an acid OR a base 2.The constant for the dissociation of water is referred to as K w. At 25°C, this value is equal to 1.0 x 10 –14. Therefore, in any solution... [H + ][OH – ] = 1.0 x 10 – 14 3.Types of Solutions 1.[H + ]>[OH – ]: acidic 2.[H + ]=[OH – ]: neutral 3.[H + ]<[OH – ]: basic

pH Calculations K w Constant 1.Since water is an amphoteric substance, it can behave as an acid OR a base 2.The constant for the dissociation of water is referred to as K w. At 25°C, this value is equal to 1.0 x 10 –14. Therefore, in any solution... [H + ][OH – ] = 1.0 x 10 – 14 3.Types of Solutions 1.[H + ]>[OH – ]: acidic 2.[H + ]=[OH – ]: neutral 3.[H + ]<[OH – ]: basic

pH Problems 1)1.0 x 10 –6 M H+:H+: 2)3.2 x 10 –12 M H+:H+: 3)7.9 x 10 –5 M OH – :

pH Problems 4)1.0 x 10 –7 M OH – : 5)3.14 M H+:H+:

pH Calculations Equations [H + ][OH – ] = 1.0 x 10 –14 pH = –log [H + ] = 10 –pH pOH = –log [OH – ] = 10 –pOH pH + pOH = 14 Variables [H + ]: Hydronium Ion Molarity Concentration [OH – ]: Hydroxide Ion Molarity Concentration pH; pOH

pH Calculations a)pH = –log [H + ] pH = –log (9.18 x 10 –11 M H+)H+) pH = 10.0; basic b)[OH – ] = 10 –pOH [OH – ] = 10 –7.8 [OH – ] = 1.6 x 10 –8 M OH – ; acidic c)pH + pOH = 14  pOH = 14 - pH pOH = pOH = 9.44; acidic

pH Calculations d)pOH = –log [OH – ] pH = –log (3.18 x 10 –5 M OH – ) pOH = 4.5 pH + pOH = 14  pH = 14 - pOH pH = pH = 9.5; basic e)pOH = –log [OH – ] pOH = –log (4.55 x 10 –10 M OH – ) pOH = 9.34; acidic (Note: pH = 4.66)

pH Calculations f)[H + ] = 10 –pH [H + ] = 10 –8.9 [H + ] = 1.3 x 10 –9 M H + ; basic g)pH + pOH = 14  pH = 14 - pOH pH = pH = 1.5; acidic

pH Calculations h)pH = –log [H + ] pH = –log (5.8 x 10 –8 M H+)H+) pH = 7.2; slightly basic pH + pOH = 14  pOH = 14 - pH pOH = pOH = 6.8

Calculating pH for SA K w Constant 1.Since strong acids completely dissociate into ions... [Conc. Of Acid] = [Conc. Of H + ] {For a Monoprotic SA} To calculate the pH, simply take the –log of the acid’s concentration (pH = –log[H + ] ). 2.Problems: Calculate the pH of the following SA solns. a.0.10 M HCl: pH = –log[0.10 M H + ] = 1.0 b M HNO 3 : pH = –log[ M H + ] = 3.0 c M HCl: pH = –log[ M H + ] = 1.6

Neutralization Reactions Lakeside Town: Acid from a factory leaks into the lake How can they fix this problem? A.What are Neutralization Reactions? 1.Definition: An acid and a base combine to form water and a salt. (HA + BOH  H 2 O + BA) 2.The salt is formed from the cation of the base and the anion of the acid. 3.The pH “neutralizes”.

Neutralization Reactions B.Writing Acid-Base Equations: Predict the products of the following neutralization reactions. Be sure to write the correct formulas for the salts. 1.HCl + KOH  H 2 O + KCl 2.HNO 3 + Ca(OH) 2  H 2 O + Ca(NO 3 ) 2 3.HF + LiOH  H 2 O + LiF 4.H 2 SO 4 + NaOH  H 2 O + Na 2 SO 4 5.HBr + Al(OH) 3  H 2 O + AlBr 3

Neutralization Reactions C.Titrations & Calculations: Predict the products of the following neutralization reactions. Be sure to write the correct formulas for the salts. 1.Definition: The slow addition of one substance to another in order to determine an unknown concentration. 2.Equipment: Buret (Titrant) & Flask (Analyte) 3.Common Titration Indicators a.Phenolphthalein: Clear (acid) / Dark Pink (base) b.Bromothymol Blue: Yellow (acid) / Blue (base)

Neutralization Reactions 4.Problems: Note: This is Stoichiometry. Not Dilution! Write a balanced equation. a.A 25.0 mL solution of 0.10 M HCl is titrated with 0.20 M NaOH until the indicator changes color. What volume of NaOH was used? HCl + NaOH  H 2 O + NaCl

Neutralization Reactions b.How many grams of Ca(OH) 2 are needed to neutralize 4.5 mL of 2.00 M HCl? Ca(OH) HCl  2 H 2 O + CaCl 2

Neutralization Reactions c.A 100. mL sample of an unknown solution of HNO 3 is placed in a flask and titrated with 0.25 M KOH. A color change occurs once 34.2 mL of KOH have been added. What must have been the molarity of the HNO 3 solution? HNO 3 + KOH  H 2 O + KNO 3

Neutralization Reactions d.What volume of 3.0 M H 2 SO 4 is needed to react completely with 1.5 L of 0.50 M NaOH? H 2 SO NaOH  2 H 2 O + Na 2 SO 4

Neutralization Reactions e.A sample of vinegar containing acetic acid is placed in a flask and titrated with M NaOH. A color change occurs after 13.6 mL of NaOH have been added. What was the mass of acetic acid in the vinegar? HC 2 H 3 O 2 + NaOH  H 2 O + NaC 2 H 3 O 2

Neutralization Reactions Book Problems: Page 507: , 73b, 74b Write balanced equations for each.