1 / 37

Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions

Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions Arrhenius base: Substance that dissolves in water and provides OH - ions Examples: HCl  H + and Cl - Acid NaOH  Na + + OH - Base.

zhen
Download Presentation

Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H+ ions Arrhenius base: Substance that dissolves in water and provides OH- ions Examples: HCl  H+ and Cl-Acid NaOH  Na+ + OH- Base

  2. Bronsted Acid: Substance that donates proton to another substance Bronsted base: Substance that accepts proton from another substance Example: HCl + H2O  H3O+ + Cl- HCl acts as acid; H2O acts as base In the Reverse Reaction, H3O+ acts as an acid; Cl- acts as a base Note: (H3O+ = hydronium ion = H+ = proton)

  3. A monoprotic acid contains one acidic proton. HCl • A diprotic acid contains two acidic protons. H2SO4 • A triprotic acid contains three acidic protons. H3PO4 • A Brønsted–Lowry acid may be neutral or it may • carry a net positive or negative charge. HCl, H3O+, HSO4−

  4. Conjugate acid: Species formed after base accepts a proton • Conjugate base: Species remaining after an acid donates its proton • Conjugate acid-base pair: an acid and base on opposite sides of the equation that correspond to each other • Examples: HNO3 + H2O H3O+ + NO3- • acid base acid base • Conjugate pairs: HNO3 and NO3- • H2O and H3O+

  5. Example: HS- + H2O  H3O+ + S2- Conjugate pairs: HS- and S2- H2O and H3O+ Practice: HClO4 + H2O  H3O+ + ClO4- What are the conjugate pairs? HClO4 and ClO4- H2O and H3O+

  6. gain of H+ H2O H Br + Br− H3O+ + acid acid base base loss of H+ • HBr and Br− are a conjugate acid–base pair. • H2O and H3O+ are a conjugate acid–base pair. Note: The net charge must be the same on both sides of the equation.

  7. Water can act as both an acid and a base (amphiprotic)! HClO4 + H2O  H3O+ + ClO4 (base) NH3 + H2O OH- + NH4+ (acid) Strengths of Acids and Bases Strong acids/bases: dissociate completely when dissolved in solution Weak acids/bases: dissociate only partially when dissolved in solution

  8. Examples: Strong Acid: HCl  H+ + Cl- (100% dissociation) Strong Base: NaOH Na+ + OH (100% dissociation) Weak Acid: CH3COOH  H+ + CH3COO- (1.3% dissociation) Weak Base: NH3 + H+ NH4+

  9. Table 9.1

  10. Naming Acids Binary Acids: hydo + root of anion + ic + “acid” ex. HCl hydrochloric acid, HBr hydrobromic acid HI Polyatomic-based Acids: root of polyatomic ion + ic + “acid” ex. H2SO4 sulfuric acid, H3PO4 phosphoric acid H2CO3 HNO3 Hydroiodic acid carbonic acid nitric acid

  11. The Self-Ionization of Water H2O + H2O  H3O+ + OH- Pure water: [H3O+]=[OH- = 10-7 M (at 250C) Neutral Solution: Any solution in which the concentrations of H3O+ and OH- ions are equal (10-7 M) Acidic Solution: Solutions having a greater concentration ofH3O+ thanOH- ions ([H3O+] greater than 10-7 M) Example: A solution with [H3O+] = 10-5 M Basic Solution: solution having a greater concentration of OH- thanH3O+ions ([H3O+] less than 10-7 M) Example: A solution with [H3O+] = 10-12 M

  12. The pH Scale • pH is a measure of acidity • Scale ranges from 0-14 • pH = 7 Neutral • pH < 7 Acidic • pH > 7 Basic • pH represents the concentration of H+ ions in solution • Pure water: 1 x 10-7 moles H+ per liter and1 x 10-7 moles OH- per liter

  13. Solutions with equal concentrations of and ions are called Neutral Solutions with more than 1 x 10-7 moles H+ per liter are Acidic Solutions with less than 1 x 10-7 moles H+ per liter are Basic Note: [H+] x [OH-] = 10-14 (always!)

  14. pH Scale Summary • pH scale refers to amount of H+ ions in solution • pH 7 is neutral, less than 7 is acidic, greater than 7 is basic • Lower pH = more acidic = more H+ ions • Higher pH = more basic = less H+ ions

  15. Each pH unit represents a 10-fold change in H+ ion concentration! pH 4 has 10 times more H+ ions than pH 5 pH 9 has 10 times fewer H+ ions than pH 8 Mathematical equation for pH: pH is the negative log of the H3O+ concentration pH = -log [H3O+]

  16. pH = -log [H3O+] Any number can be expressed as 10 raised to some exponent: y = 10x Examples: 100 = 10 2 1000 = 103 0.10 = 10 -1 The log is that exponent! 100 = 10 2; Log of 100 =2 1000 = 103; Log of 1000 = 3 0.10 = 10 –1; Log of 0.10 = -1

  17. We can also take the log of non-whole numbers, but we use our calculators for this. • Example: Find the log of 2.4 x 10-3 • Enter 2.4 x 10-3into calculator • Press the “log” key • 0.0024 “log” = -2.62 • Therefore, 10-2.62 = 2.4 x 10-3

  18. Calculating pH from [H3O+] • pH = -log [H3O+] • Enter [H3O+] into calculator • Press the “log” key • Change the sign • Example: [H3O+] = 1.0 x 10-7 M • pH = -log [H3O+] • pH = -log [1 x 10-7] = 7

  19. Example: [H3O+] = 1 x 10-11M • pH = -log [H3O+] • pH = -log [1 x 10-11] = 11 • Example: [H3O+] = 1 x 10-3 M • pH = -log [H3O+] • pH = -log [1 x 10-3] = 3

  20. Example: [H3O+] = 4.2 x 10-5 • pH = -log [H3O+] • Enter [H3O+] into calculator (4.2 x 10-5) • Press the “log” key (-4.3767507) • Change the sign (4.3767507) • pH = 4.3767507 = 4.4

  21. Example: [H3O+] = 8.1 x 10-9 • pH = -log [H3O+] • Enter [H3O+]into calculator (8.1 x 10-9) • Press the “log” key (-8.091515) • Change the sign (8.091515) • pH = 8.091515= 8.1

  22. Reactions Between Acids and Bases Neutralization: reaction between an acid and a base; always produces salt and water Example: Write a balanced equation for the reaction of hydrochloric acid with sodium hydroxide. HCl(aq) + NaOH(aq) H—OH(l) + NaCl(aq)

  23. Example: Write a balanced equation for the reaction of hydrochloric acid with magnesium hydroxide. 2 HCl + Mg(OH)2 2 H2O + MgCl 2

  24. Titration Titration:a technique used to determine the concentration of an acid or base in a solution • If we want to know the concentration of an acid • solution, a base of known concentration is • added slowly until the acid is neutralized. • When the acid is neutralized: # of moles of acid = # of moles of base • This is called the end point of the titration.

  25. Acid-Base Titration • Titration is a laboratory procedure used to determine the molarity of an acid. • In a titration, a base such as NaOH is added to a specific volume of an acid. Base (NaOH) Acid solution

  26. A few drops of an indicator is added to the acid in the flask. • The indicator changes color when the base (NaOH) has neutralized the acid.

  27. At the end point, the indicator has a permanent color. • The volume of the base used to reach the end point is measured. • The molarity of the acid is calculated using the neutralization equation for the reaction.

  28. Determining an unknown molarity from titration data requires three operations: mole–mole conversion factor [2] Moles of base Moles of acid M (mol/L) conversion factor M (mol/L) conversion factor [3] [1] Volume of base solution Molarity of acid solution

  29. HOW TO Determine the Molarity of an Acid Solution from Titration Example: What is the molarity of an HCl solution if 22.5 mL of a 0.100 M NaOH solution are needed to titrate a 25.0 mL sample of the acid? volume of base (NaOH) 22.5 mL volume of acid (HCl) 25.0 mL conc. of acid (HCl) ? conc. of base (NaOH) 0.100 M

  30. Determine the number of moles of base used to neutralize the acid. Step [1] M (mol/L) conversion factor Volume of base solution 1 L 1000 mL 0.100 mol NaOH 1 L x x = 22.5 mL NaOH mL–L conversion factor 0.00225 mol NaOH

  31. Determine the number of moles of acid that react from the balanced chemical equation. Step [2] HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) 1 mol HCl 1 mol NaOH x 0.00225 mol NaOH 0.00225 mol HCl = mole–mole conversion factor

  32. Determine the molarity of the acid from the number of moles and the known volume. Step [3] mol L 1000 mL 1 L 0.00225 mol HCl 25.0 mL solution M = = x mL–L conversion factor = 0.0900 M HCl Answer

  33. Buffers Buffer: a solution whose pH changes very little when acid or base is added. Most buffers are solutions composed of roughly equal amounts of • a weak acid • the salt of its conjugate base The buffer resists change in pH because • added base, −OH, reacts with the weak acid • added acid, H3O+, reacts with the conjugate base

  34. Buffers contain 2 compounds: • Compound with the ability to react with H+ ions • Compound with the ability to react with OH- ions • Example: HCO3- + H+ H2CO3 • If acids (H+) are added, react with HCO3- • H2CO3 + OH-  HCO3- + H2O • If OH- ions are added, react with H2CO3 • H2CO3 is unstable: H2CO3 H2O + CO2

  35. More Examples of a Buffer If an acid is added to the following buffer equilibrium, Adding more product… CH3COOH + H2O H3O+ + CH3COO− conjugate base weak acid …drives the reaction to the left. then the excess acid reacts with the conjugate base, so the overall pH does not change much.

  36. If a base is added to the following buffer equilibrium, Adding more reactant… H2O + CH3COO− CH3COOH + −OH conjugate base weak acid …drives the reaction to the right. then the excess base reacts with the weak acid, so the overall pH does not change much.

More Related