Acids & Bases Chemistry. Ms. Siddall..

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Presentation transcript:

Acids & Bases Chemistry. Ms. Siddall.

Standard 5a: Properties Acid Base Taste Sour (lemon) Bitter (soap) Touch Like water Like soap Reaction with metal Vigorous No reaction Conductivity Conducts electricity Litmus Paper Blue  red Red  blue Neutralization Acid + base  salt + water e.x. HCl + NaOH  NaCl + H2O e.x. HCl + Mg  H2(g) + MgCl2

Summary 1 Write the balanced equation for the reaction of aluminum with HCl to form hydrogen and aluminum chloride. A substance turns red litmus blue and does not react with metal. What is it?

Standard 5b: Brønsted-Lowry Definition Acids donate hydrogen ions (Hydrogen ion = H+ = proton) e.x. HBr + H2O  H3O+ + Br- Bases accept hydrogen ions e.x. NH3 + H2O  NH4+ + OH-

Summary 2 Write the balanced equation for the reaction of HF (hydrofluoric acid) with water.

Strong acid ionization weak acid ionization

Standard 5c: Dissociation Acids: Strong acid = strong electrolyte Strong electrolyte = 100% dissociation e.x. HNO3 + H2O  H3O+ + NO3- in solution (H3O+ = hydronium ion) You MUST know these strong acids: HCl, HBr, HI, HNO3, H2SO4 0% 100%

Summary 3 Which of the following are strong acids? H2SO4, HSO3, HClO2, HBr, HNO3, HNO2

Standard 5c: continued Weak acids = weak electrolytes Weak electrolyte = Partial dissociation (= equilibrium) e.x. HCN + H2O  H3O+ + CN- In solution All other acids are weak acids Examples: acetic acid HC2H3O2, carbonic acid H2CO3, HCN, H3PO4 ~97% ~3%

Summary 4 At equilibrium does a weak acid solution contain mostly products or mostly reactants? At equilibrium does a strong acid solution contain mostly products or mostly reactants?

Conjugate acids & bases: Standard 5c: continued Conjugate acids & bases: e.x. HF + H2O  H3O+ + F- acid Donates hydrogen ion Conjugate base Different from acid by a proton Base Accepts hydrogen ion Conjugate acid Different from base by a proton CB A B CA

Summary 5 HNO3 + H2O  H3O+ + NO3- Label the acid (A), base (B), conjugate acid (ca), and conjugate base (cb) in the following reaction: HNO3 + H2O  H3O+ + NO3-

Conjugate acids & base pairs: e.x. HF + H2O  H3O+ + F- Acid and Conjugate base pair = HF & F- A strong acid has a weak conjugate base A weak acid has a strong conjugate base Base and conjugate acid pair = H2O & H3O+ A strong base has a weak conjugate acid A weak base has a strong conjugate acid

Summary 6 HF + H2O  H3O+ + F- Identify the acid and conjugate base. Label each as ‘weak’ or ‘strong’ Identify the base and conjugate acid. Label each as ‘weak’ or ‘strong’

pH measures the concentration (amount) of H3O+ ions in solution Standard 5d: pH scale Definition: pH = -log[H3O+] pH measures the concentration (amount) of H3O+ ions in solution

Summary 7 Write the equation for HCl reacting with water to make a hydronium ion and a chloride ion Is the concentration of hydronium ions in this solution high or low?

acidic neutral basic pH0 pH1 pH2 pH3 pH4 pH5 pH6 pH7 pH8 pH9 pH10 pH11

Summary 8 Provide an example of: An acidic compound A basic compound A neutral compound Add a minimum of 3 acids and 3 bases to your pH scale

Self ionization of pure water H2O + H2O  H3O+ + OH- = Equilibrium system For pure water: [H3O+] = [OH-] = 1x10-7M pH = -log[H3O+] = 7 pH 7 = neutral [H3O+] > [OH-] = pH < 7= acidic solution [H3O+] < [OH-] = pH > 7 = basic solution

Summary 9 Identify the following solutions as acidic, basic, or neutral: pH 3 pH 5 pH 10 pH 7

Acids ‘produce’ H3O+ in solution pH measurement: e.x. HCl + H2O  H3O+ + Cl- pH measurement: [H3O+] = 0.1M = 1x10-1M pH = -log [H3O+] = -log [1x10-1] pH = 1 High [H3O+] = ‘Low’ pH reading = acidic solution

Summary 10 pH measures the concentration of ___________________ ions in solution. A solution with pH = 2: has a high / low concentration of these ions is acidic / basic

Bases ‘decrease’ H3O+ in solution pH measurement: e.x. NaOH + H3O+  Na+ + 2H2O pH measurement: [H3O+] = 1x10-14M pH = -log [H3O+] = -log[1x10-14] pH = 14 Low [H3O+] = ‘High’ pH reading = basic solution (Alkaline)

Summary 11 What is the hydronium ion concentration for a solution with pH = 12? Is the solution acidic or basic?

pH = -log[H3O+] 5f: calculate pH Example: [H3O+] = 4.5 x10-5M pH should be between 4 - 5 pH = -log[4.5x10-5] = 4.35

Summary 13 Estimate then calculate the pH for a solution with: [H3O+] = 7 x 10-2 M [H3O+] = 2.2 x 10-9 M

pOH = -log[OH-] pOH + pH = 14 Relationship between [OH-] and [H3O+] Example: [OH-] = 1x10-2M pOH = -log[1x10-2] = 2 pH = 14 - 2 = 12

Summary 14 Complete the following table: [H3O+] [OH-] pH pOH 10-5M 1

5e: acid/base definitions Arrhenius: acids are hydrogen containing compounds that ionize to yield H+ ions in aqueous solution. e.x. HCl(aq)  H+(aq) + Cl-(aq) Bases ionize to yield OH- ions in aqueous solutions. e.x. NaOH(aq)  Na+(aq) + OH-(aq) Problems: NH3 is not a base according to Arrhenius

Summary 15 Explain why NH3 is not a base according to the Arrhenius definition of a base.

Base = hydrogen acceptor BrØnsted-Lowry: Acid = hydrogen donor Base = hydrogen acceptor e.x. NH3(aq) +H2O  NH4+ + OH-

An acid accepts a pair of electrons (accepts a negative charge) Lewis: An acid accepts a pair of electrons (accepts a negative charge) Acid = proton donor = electron acceptor A base donates a pair of electrons Base = proton acceptor = electron donor e.x. HCl + H2O  H3O+ + Cl-

Electron acceptor

Electron donor

Summary 16 Explain the difference between the Bronsted-Lowry definition and the Lewis definition of acids and bases

5g: buffers Buffer = a solution whose pH does not change (much) when acid or base are added. A buffer is a solution made from an acid and its conjugate base

There are 2 buffer systems in human blood: H2PO4-/HPO42- & H2CO3/HCO3- Example: human blood must have a pH between 7.35 – 7.45 (or we will die) There are 2 buffer systems in human blood: H2PO4-/HPO42- & H2CO3/HCO3- B- + H2PO4-  HPO42- + HB B- + H2CO3  HCO3- + HB HB = acid (H+ donor) B- = base (H+ acceptor)

Summary 17 Write a general equation showing the buffering ability of HSO4- and its conjugate base. (use B- as the base that reacts with HSO4-)