Download
chapter 14 acids and bases n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 14 Acids and Bases PowerPoint Presentation
Download Presentation
Chapter 14 Acids and Bases

Chapter 14 Acids and Bases

96 Vues Download Presentation
Télécharger la présentation

Chapter 14 Acids and Bases

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chapter 14Acids and Bases • An acid is a proton(H+) donor • A base is a proton acceptor HA(aq) + H2O(l)  H3O+(aq) + A-(aq) acid base conjugate conjugate acid base Conjugate base- everything that remains of an acid molecule after a proton is lost Conjugate acid- proton is transferred to base

  2. Acid dissociation constant What is the equilibrium expression for the general reaction? Ka= [H3O+] [A-]/[HA] or [H+] [A-]/[HA] Ka is called the acid dissociation constant

  3. Acid Strength Strong acid- equilibrium lies far to the right Ka is large HCl, H2SO4, HNO4, HClO4 Weak acid- equilibrium lies far to the left Ka is small Phosphoric acid, Nitrous acid, Acetic acid

  4. Water as an acid and a base • Amphoteric- acts as either an acid or a base • Autoionization of water-transfer of a proton from 1 H20 molecule to another to produce hydroxide ion and the hydronium ion H2O + H2O  H3O+ + OH- or 2H20(l)  H3O+ (aq) + OH-(aq) This leads to the Ion-product constant,Kw

  5. Kw con. Kw= [H3O+] [OH-] or [H+] [OH-] At 25° C, [H+] = [OH-]= 1.0x10-7M Kw= (1.0x10-7) (1.0x10-7)= 1.0x10-14 At 25° C no matter what it contains, the product of [H+] and [OH-] 1.0x10-14

  6. 3 Possible situations • Neutral solution [H+] = [OH-] • Acid solution [H+]  [OH-] • Basic solution [H+] [OH-] Calculate [H+] or [OH-] at 25°C. State if neutral, acidic, or basic. • 1.0x10-5M OH- • 1.0x10-7M OH- • 10.0 M H+

  7. pH Scale pH= -log[H+] Example: [H+]=1.0x10-7M pH= 7.00 Other Log Scales pOH= -log[OH-] pK= -logK

  8. Example14.5 Calculate pH and pOH for the following at 25°C. • 1.0x10-3 M OH • 1.0 M H+

  9. Converting between p and concentration • pX= -log[X]  (pH or pOH) = -log[H+ or OH-] • [X] = 10-pX  [H+ or OH-] = 10-pX

  10. Calculate “p” or [ ] of the following: • Calculate [Cl-] if pCl = 7.32 • Calculate pAg if [Ag+] = .034 M The pH of a sample of blood equals 7.41 at 25°C. Calculate the pOH, [H+], and [OH-]

  11. Calculating pH of Strong Acid Solutions What are some examples of strong acids? HCl, HNO3, H2SO4, and HClO4 Calculate the pH and [OH-] of a 5 x10-3 M HClO4 solution. Calculate the pH of 1.0 x10-10 M HCl.

  12. Calculating the pH of a Weak Acid Calculate the pH of a .100M aqueous solution of hypochlorous acid(HOCl) where Ka= 3.5x10-8. • List the major species. HOCl H2O

  13. Con. • Write the balanced equations for the reactions producing H+. HOCl(aq) H+(aq) + OCl-(aq) H2O(l)  H+(aq) + OH-(aq) • Compare equilibrium constants to determine the dominate H+ Ka= 3.5x10-8 Kw= 1.0x10-14

  14. Con. • Write the equilibrium expression Ka= [H+] [OCl-]/ [HOCl] • Do an ICEbox

  15. Con. • Substitute in Ka 3.5x10-8 = [x] [x] / [.100-x] 3.5x10-8 = x2/ .100 x= 5.9 x 10-5

  16. Con. • Validate the approximation by 5% - compare x to [HA]0 x100% 5.95 x10-5/.100 x100= .059% • Calculate [H+] and pH [H+]= 5.95x10-5 M pH= -log[5.95x10-5] = 4.225

  17. pH of a Weak Acid Mixtures • Calculate the pH of a solution that contains 1.00 M HCN(Ka=6.2x10-10) and 5.00 M HNO2 (Ka= 4.0x10-4). HCN(aq)H+(aq) + CN-(aq) Ka= 6.2x10-10 HNO2(aq)  H+(aq) + NO2-(aq) Ka= 4.0 x10-4 H2O(l)  H+(aq) + OH-(aq) Kw= 1.0x10-14

  18. Con. Ka= [H+] [NO2-]/ [HNO2] HNO2 H+ NO2- I 5.00 0 0 C -x +x +x E 5.00 –x x x

  19. Con. 4.0 x10-4 = [x] [x]/ 5.00-x approximate so 4.0 x10-4 = x2/5.00 X=4.5 x10-2 Compare: 4.5 x10-2 /5.00 x100 = .90% [H+]= 4.5 x10-2 M pH= -log[4.5 x10-2 ] = 1.35

  20. Bases • Proton acceptors • Examples of strong bases: • NaOH, KOH, all groups in 1A&IIA