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Test 4 Review

Test 4 Review. Advanced Chemistry. Equilibrium. m A + n B g s P + r Q K eq = [P] s [Q] r [A] m [B] n K sp = [A + ][B - ] for dissolving of a solid AB g A + + B -

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Test 4 Review

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  1. Test 4 Review Advanced Chemistry

  2. Equilibrium mA + nB gsP + rQ Keq = [P]s[Q]r [A]m[B]n Ksp = [A+][B-] for dissolving of a solid AB g A+ + B- Ka = [H+][A-] for the dissociation of an acid [HA] Ka is small for a weak acid, large for a strong acid. Never include solids or pure liquids in a Keq.

  3. Solubility Product Constant AB(s) D A+(aq) + B-(aq) Ksp = [A+][B-] At 25°C the solubility product constant for strontium sulfate, SrSO4, is 7.6 x 10-7. What is the concentration of Sr2+ at 25°C? [Sr2+][SO42-] = 7.6 x 10-7 [Sr2+]=[SO42-] [Sr2+] = 8.7 x 10-4M

  4. The solubility product constant for strontium fluoride is 7.9 x 10-10. What is the molar solubility of SrF2 at 25°C? SrF2 D Sr+2 + 2F- [Sr2+][F-]2=7.9 x 10-10 [Sr2+]= x [F-]= 2x (x)(2x)2=7.9 x 10-10 x = 5.8 x 10-4M

  5. Strong bases: hydroxides of groups 1 & 2 (except Be) • Strong acids: HCl, HBr, HI HClO4, H2SO4, HNO3

  6. Ionization constant, Ka, for a weak acid HA D H+ + A- Ka = [H+][A-] [HA] Kb = Ka ∙Kb= Kw = 1.00 x 10-14

  7. What is the [H+] in 0.100M formic acid? Ka for formic acid is 1.77 x 10-4 HCOOH D H+ + COOH- Since this is a weak acid, [HCOOH] ͌ 0.100M Ka = [H+][COOH-]= 1.77 x 10-4 [HCOOH] Let x = [H+] = [COOH-] x2= 1.77 x 10-4 0.100 X = 4.21 x 10-3M

  8. Percent ionization [amount ionized] [original acid] What is the percent ionization of [H+] from the previous problem? [H+] = 4.21 x 10-3M, [HCOOH] = 0.100M 4.21 x 10-3= 4.21% 0.100

  9. hydrolysis • The reaction of a salt with water to form an acidic or basic solution. • Example: FeCl3 D Fe3+(aq) + Cl-(aq) Fe3+ + 3OH-D Fe(OH)3 H2O D H+ + OH- shifts right, creating more H+

  10. [H+] ∙ [OH-] = 10-14 pH + pOH = 14.0 pH = -log[H3O+] [H3O+] = antilog(-pH) Find the pH of a solution with [H3O+] of 9.85 x 10-8M. pH = -log (9.85 x 10-8) = 7.01 What is the [H3O+] in a solution with pH 7.01? [H3O+] = antilog(-7.01)= 9.85 x 10-8M

  11. What is the pH of the following solutions?0.1M HCl0.1M NaOH0.1M H2CO3 [H+] = 10-1, pH = 1 [OH-]=10-1, pOH=1, pH=13 Ka = 4.4 x 10-7 = x2 0.1 X = 2.1x10-4 pH = 3.68

  12. Titration • Standard solution- one whose concentration is known • Endpoint- the point at which equivalent amounts of reactants are present. • M∙V = moles • MaVa=MbVb

  13. Titration curves

  14. Calculating pH of a buffer • What is the pH of a buffer that is 0.12M lactic acid (HC3H5O3) and 0.10M sodium lactate? • Ka = 1.4 x 10-4  -x +x +x .12-x x .10+x

  15. Henderson-Hasselbalch Conjugate base of the acid • pH = pKa + log [base] [acid] = -log(1.4x10-4) + log .10 .12 = 3.85 + (-.079) = 3.77

  16. Common Ion (Buffers) Calculate the pH of a 0.100M solution of formic acid and 0.020M sodium formate. Ka= 1.77 x 10-4 HCOOH g H+ + COOH- 0.100 x 0.020 Ka = [H+][COOH-]= .020x [HCOOH] 0.100 X = 8.85 x 10-4M pH = 3.05

  17. Redox reactions involve a change in oxidation number Oxidation Loss of electrons Reduction Gain electrons 3Cu2+ + 2Fe g 3Cu + 2Fe3+copper gains 2 electrons iron loses 3 electrons Reducing agent- is oxidized (iron) Oxidizing agent- is reduced (copper) (reduced) (oxidized)

  18. Steps for balancing redox reactions using the half-reaction method Write ionic equation for half reactions Balance chemically Balance non- O and H atoms Add H2O to balance O’s Add H+ to balance H’s (in a basic solution add OH-) Balance electrically- add e-’s to the more + side Check for balanced charges on both sides Combine half reactions and cancel common items Add spectator ions and balance

  19. Activity Series lithium potassium magnesium aluminum zinc iron nickel lead HYDROGREN copper silver platinum gold More active Oxidizes easily The metal must be above (more active than) the ion for it to be a spontaneous reaction. Reduces easily Less active

  20. Voltaic Cell Anode attracts anions where oxidation occurs Cathode attracts cations where reduction occurs Salt bridge connects the two half cells contains a strong electrolyte

  21. Zn Zn+ Cu2+ Cu in shorthand Two half cells connected by a salt bridge

  22. Reduction half reactions F2 is the strongest oxidizing agent Li is the strongest reducing agent

  23. Reduction Potentials (E) If E is positive, the reaction is spontaneous. If E is negative, the reverse reaction is spontaneous. Eo is the standard electrode potential all ions are 1M and gases are 1 atm The net Eois the sum of the Eo of the half reactions The stronger oxidizing agent reduces. Reverse the sign of the substance oxidized.

  24. What is the voltage produced from the reaction of Zn metal with Cu2+ ions? Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s) Zn2+ + 2e-g Zn -0.7628 Cu2+ + 2e-g Cu 0.3402 -(-0.7628) + 0.3402 = 1.103 volts Will happen spontaneously Zn2+ + Ni(s) g Zn(s) + Ni2+ -(-0.23) + (-0.7628) = -.53 Will not occur spontaneously

  25. Faraday’s Law Coulombs = amperes x seconds 1 C = 1amp·1sec 96,485 coulombs = 1 mole e-

  26. What mass of copper will be deposited by a current of 7.89 amps flowing for 1200 seconds? Cu2+ + 2e-g Cu at the cathode 7.89A x 1200s x 1C x 1 mol e-= .0981 mol e- A·s 96,485C .0981 mol e- x 1 mol Cux63.5g Cu = 3.1g Cu 2 mol e- 1 mole Cu

  27. Nernst Equation Eo Voltage under standard conditions (1M solutions at 25°C and 101.3kPa) At non-standard conditions, use Nernst equation E = E° - 0.05916 log [products] n [reactants] n = no. of electrons transferred Coefficients in front of reactants or products are used as powers of their concentrations.

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