This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1/15/2010 Announcements: Friday, January 15, 2010
Homework WebAssign #1- Due Friday, January 15 at 5:00 PM WebAssign #2- Due Wednesday, January 20 at 8:00 AM Recitation Next Week Week 2- “Equilibrium-Weak and Strong Acids” ** Monday recitation sectionsPlease attend another session next week! ** Lab Next Week Lab Familiarization and Lab Check-In ** Monday lab sections will check in week 3.** Suggested Textbook Questions Continue Chapter 18 problems. Acid-Base Clarification
All Arrhenius acids and bases are Brønsted-Lowry acids and bases. Acid definitions are the same. Forms H+ (H3O+) in water = H+ donor. HF + H2O ⇄ H3O+ + BrArrhenius base forms OH- in water and OH- is a H+ acceptor = Brønsted-Lowry base. NaOH → Na+ + OHOH- + H+ ⇄ H2O Acid/Base Strength
The stronger the acid, the easier it is to remove a proton. HCl is strong acid. HCl + H2O Cl– + H3O+ HCl dissociates completely! Vinegar is a weak acid. HC2H3O2 + H2O C2H3O2– + H3O+ Ka=1.8x10-5 Only some HC2H3O2 dissociates! Ka, the acid dissociation constant, is a measure of the strength of the acid. Ka strength of the acid Ka>>1 January 15, 2010 – CT#1 Arrange the following acids by increasing acid strength. HCN HCN HCN HCN Ka = 6.2x10-10 HCN HCN HClO2 HCIKa = 1.1x10-2 HCN HCN HBrO HCNKa = 2.3x10-9 A. HCN < HClO2 < HBrO B. HClO2 < HCN < HBrO C. HBrO < HCN < HClO2 D. HClO2 < HBrO < HCN E. HCN < HBrO < HClO22 HBrO < HClO ↑Ka→↑acid strength (See Appendix C for Ka values) Kw
Kw is the Ka for water. H2O(l) + H2O(l) OH–(aq) + H3O+(aq) Kw = 1 x 10–14 at 25ºC Kw = [H3O+][OH–] = 1 x 10–14 In pure water at 25ºC: [H3O+] = [OH–] = 10–7 M If [H3O+] = 10–4 M, [H3O+][OH–] = 10–14 (10-4)[OH–] = 10–14 [OH–] = 10–10 M January 15, 2010 - CT#2 At 25ºC, if [OH-] = 10-9 M, what is the [H3O+]? 10- M [H3O+][OH-] = 10-14 A. 10-55M [H3O+](10-9) = 10-14 B. 10-7 M [H3O+] = 10-14+9 = 10-5 M C. 10-9 M D. 10-14 M E. Not enough information 1 1/15/2010 Acidic and Basic Solutions
In pure water at 25ºC: [H3O+] = [OH–] = 10–7 M H3O+ and OH- are formed in a 1:1 ratio. At 25ºC, [H3O+][OH–] = 10-14 so if one , the other must . When [H3 from another source [OH-] and solution is acidic! O+ ] When [OH–] from another source [H3O+] and solution is basic! pH and pOH pH = -log[H3O+] [H3O+] = 10–7 M, pH = 7 Neutral [H3O+] = 10– 4 M, pH = 4 Acidic [H3O+] = 10–10 M, pH = 10 Basic pOH = -log[OH-] Recall Kw = [H3O+][OH–] = 10–14 -log([H3O+][OH–]) = -log(10–14) -log[H3O+]+ -log[OH–] = 14 pH + pOH = 14
acidic basic pH Scale: 1----------7----------14
neutral Calculating the [H3O+] Concentration
Strong acids Reaction goes to completion (Ka ∞) Strong Acids
Strong acids (and strong bases) dissociate completely. What is the [H3O+] in a 0.25 M HCl(aq) solution? HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) Weak acids Reaction is not extensive and we must use Ka & ICE diagram to calculate the [H3O+] concentration. initial: 0.25 M : -0.25 M equil.: 0M All of it dissociates! 0M +0.25 M 0.25 M 0M +0.25 M 0.25 M [H3O+] = 0.25 M pH = 0.60 Very acidic!
pH does not have to fall between 1 & 14! Weak Acids
Weak acids (and bases bases) do not dissociate completely. Use Ka and ICE diagram to calculate [H3O+]. What is the [H3O+] in a 0.25 M HC2H3O2(aq) solution? Weak Acids
+ Ka = [H3O ][C2H3O2 ] = (x)(x) = 1.8 x 10-5 [HC2H3O2] (0.25 – x) 0.25 Make an assumption to avoid quadratic equation. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) initial: : equil.: Must 0M calculate -x how much + x 0.25 M - x dissociates! x 0.25 M Ka = [H3O+][C2H3O2-] [HC2H3O2] = (x)(x) (0.25 – x) 0M +x x x2 = 4.5 x 10-6 [H3O+] = x = 2.1 x 10-3 M pH = 2.68 Check assumption: (2.1 x 10-3)/0.25 x 100 = 0.84% << 5% = 1.8 x 10-5 2 1/15/2010 Significant Figures with Logs
1.0 x 10-9 → 2 sig figs -log(1.0 x 10-9) = 9.00 1.00 x 10-9 → 3 sig figs -log(1.00 x 10-9) = 9.000 1.00 x 10-11 → 3 sig figs -log(1.00 x 10-11) = 11.000 Numbers before the decimal point come from the exponent. These digits are NOT significant. Numbers after the decimal point are significant! 3 ...
View Full Document