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Review of Everything

Review of Everything. 2 nd 9 weeks. Equilibrium. Chapter 13. Equations Sheet. Equilibrium. A state when two competing reactions are canceling each other out. H 3 O + + OH - H 2 O + H 2 O

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Review of Everything

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  1. Review of Everything 2nd 9 weeks

  2. Equilibrium Chapter 13

  3. Equations Sheet

  4. Equilibrium • A state when two competing reactions are canceling each other out. • H3O+ + OH- H2O + H2O • You reach equilibrium when the rate of forward reaction is equal to the rate of backwards reaction. • This does NOT mean the concentration of products and reactants are equal. • It does mean there is a stable ratio between products to reactants • The above reaction is at equilibrium when [H3O+] = [OH-] = 1x10-7mol/L. In 2 L of water there are 110 moles.

  5. Equilibrium expression and constant • K is calculated form EQUILIBRIUM CONCENTRATIONS! Its values may only be calculated experimentally. • aA+ bBcC+ dD • equilibrium constant = equilibrium expression • K = [C]c[D]d • [A]a[B]b • [ ] means concentration in M

  6. Determine the equilibrium expressions • For the following: • Br2(g) 2Br(g) • N2(g) + 3H2(g) 2NH3(g) • H2 (g) + Br2(g) 2 HBr(g) • HCN(aq) H+(aq) + CN-(aq)

  7. K values, K is always positive • Intermediate K. 0.1<K<10 • Significant concentration of all substances are present. • Very Large K. K >> 1 • The product concentration is very large with virtually no reactant concentration (the reaction has gone to completion). • Very Small K. K<<1 • The reactant concentration is very large with virtually no product concentration.

  8. Types of K • Keq= equilibrium constant • Kc= equilibrium constant in terms of concentration. • Kp= equilibrium concentration in terms of pressure. • Ka= acid dissociation constant • Kb= base dissociation constant • Kw= ion-product constant for water • Ksp= solubility product constant

  9. K problem • The following equilibrium concentrations were observed for the Haber process at 127oC. • [NH3] = 3.1 x 10-2 M, [N2] = 8.5 x 10-1 M, [H2] = 3.1 x10-3 M • Forward Reaction. • N2(g) + 3H2(g) 2NH3(g) • Reverse Reaction. • 2NH3(g) N2(g) + 3H2(g) • Multiply by Factor n. • 1/2N2(g) + 3/2H2(g) NH3(g) • K is unitless

  10. Summary • For forward reaction jA + kBlC + mD, • K = [C]l [D]m • [A]j [B]k • Forreverse reaction jA + kBlC + mD, • K’ = K-1= [A]j [B]k • [C]l [D]m • For reaction njA + nkBnlC + nmD • K’’ = Kn = [C]nl [D]nm • [A]nj[B]nk • For an overall reaction of two or more steps, • Koverall = K1 x K2 x K3 x ...

  11. Example N2(g) + 3H2(g) 2NH3(g)

  12. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. • The concentrations of pure liquids and solids are constant. 2KClO3(s) 2KCl(s) + 3O2(g)

  13. Reaction Quotient, Q • Q is calculated from INITIAL CONCENTRATIONS! • Q is useful in determining which direction a reaction must shift to establish equilibrium. • K vs. Q • K is calculated from equilibrium concentrations or pressures. • Q is calculated from initial concentrations or pressures.

  14. Reaction Quotient, Q • K= Q; The system is at equilibrium. No shift will occur. • K < Q; The system shifts to the left. • Consuming products and forming reactants, until equilibrium is achieved. • K > Q; The system shifts to the right. • Consuming reactants and forming products, to attain equilibrium.

  15. Change • Shift Right • 2 H2O⇌H3O+ + OH- • Shift -2y +y +y • Shift Left • 2 H2O⇌H3O+ + OH- • Shift +2y -y -y

  16. K vs Q Problem • For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: • N2(g) + 3 H2(g) 2 NH3(g) • Conc. (M) [NH3]o[N2]o [H2]o • Trial 1 1.0 x 10-31.0 x 10-52.0 x10-3 • Trial 2 2.0 x 10-41.5 x 10-3 3.54 x10-1 • Trial 3 1.0 x 10-45.0 x10-1 1.0 x10-3

  17. ICE tables • To do equilibrium problems set up an ICE table. • Write the balanced equation. Underneath it label three rows. • Initial Concentrations (pressure) • Change • Equilibrium concentration (pressure)

  18. Gas Problem • At a certain temperature a 1.00-L flask initially contained 0.298 molPCl3, 8.70 x 10-3molPCl5, and no Cl2. After the system had reached equilibrium, 2.00 x 10-3mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the following reaction • PCl5(g) PCl3(g) + Cl2(g) • Calculate the equilibrium concentrations of all species and the value of K.

  19. Answer • PCl5(g) PCl3(g) + Cl2(g) • 8.7 x10-3 M .298 M 0 • -x +x +x • 2.0x10-3 M • It has to shift right because chlorine increased • x must = 2.0x10-3 I C E .300 M .0067 M

  20. K • K = .3 (.002) / .0067 • = .0896

  21. Simplified Assumptions • *if you are using a solver function this is unnecessary. • You do have to understand the concept for a possible multiple choice • For some reactions the change will be very small compared to the initial amount. • You always have to check!

  22. Example • Gaseous NOCl decomposes to form the gases NO and Cl2. At 35o C, the K = 1.6 x10 -5. The initial conc. Of NOCl is .5 M. What is the equilibrium concentrations? • 2NOCl(g) 2NO(g) + Cl2(g) • The small K value means this will favor the reactant side so there wouldn’t be a large shift to the right.

  23. Problem • 2NOCl(g) 2NO(g) + Cl2(g) • I .50 0 0 • C -2x +2x + x • E .50-2x 2x x • 1.6 x10 -5= (2x)2 x / (.5-2x)2 • The algebra looks a little sticky on this problem • However, if the shift is really small then • .50 -2x  .5

  24. Solve now • 1.6 x10 -5 = (2x)2 x / (.5)2 • x = .01 • Of course this is ONLY ACCEPTABLE IF .50 -2x  .5 • .50 – 2 (.01) = .48 • If the change is less than 5% it is considered small enough to ignore • .48/.5 = 96% or a 4% change.

  25. Acid Equilibrium constant • For some acid “A” • HA(aq) + H2O(l) H3O+(aq) + A-(aq) • Ka= [H3O+] [ A-] • [HA]

  26. Acid Strength • Strong acids dissociate completely in water. • At equilibrium, Ka >> 1 because [HA] is approx. 0. • Weak acids are mostly undissociated. • At equilibrium, Ka<< 1 because [H3O+] and [ A-] are very small compared to the [HA]. • The smaller the Ka value, the weaker the acid.

  27. Ka values of some weak acids

  28. pH • pH = -log[H3O+] where neutral = 7.00 • acidic < 7.00 • [H3O+] = 10-pH basic >7.00 • pOH= -log[OH-] • pH + pOH = 14 • pK= -log K

  29. Sig Figs and pH • The number of decimal places in the log value, pH value, is equal to the number of significant figuresin the number that we took the logarithm of, concentration. • Calculate pH and pOH for each of the following solutions. • 2.7 x 10-3 M OH- • 3.4 x 10-5 M H3O+ • 1.54 x 10-10 M OH-

  30. pH problem • The pH of a sample of human blood was measured to be 7.41 at 25oC. • Calculate pOH, [H3O+], and [OH-] for the sample.

  31. Calculating pH of Strong Acids • Calculate the pH of 0.10 M HNO3 • Calculate the pH of 1.0 x 10-10 M HCl

  32. Calculating the pH of a weak acid • These require ICE tables • Again you could use simplified assumptions and test the 5% rule. • Solver functions on the calculator are completely legal to use on my test and the AP test.

  33. Problem The hypochlorite ion (ClO-) is a strong oxidizing agent found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-, for example) and forms the weakly acidic hypochlorousacid (HOCl, Ka = 3.5 x 10-8). Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid.

  34. Answer • HOCl + H2O H3O+ + ClO- • I .10 M 0 0 • C -x +x +x • E .1-x x x • Ka = [H3O+] [ClO-] 3.5x10-8 = x2 • [HOCl] .1-x • x = 5.9 x10-5 • The x value is [H3O+] so pH = 4.23

  35. Percent Dissociation • For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

  36. Base dissociation constant • For some base B • B + H2OBH+ + OH- (aq) • The Base Dissociation Constant (Kb) • Kb = [BH+] [OH-] • [B]

  37. The pH of Strong Bases. • This works the same as the pH of a strong acid. • Calculate the pH of a 5.00 x 10-2 M NaOH solution.

  38. pH of weak bases • Calculate the pH for a 15.0 M solution of NH3 (Kb = 1.8 x 10-5).

  39. Polyprotic Acids • A polyproticacid has more than one ionizable proton (H+). • e.g. H2SO4, H2CO3, H3AsO4 • Each successive Ka value get smaller (Ka1 > Ka2 > Ka3). Therefore, the first dissociation step makes the most significant contribution to the equilibrium concentration of [H3O+].

  40. Sulfuric acid is unique in being a strong acid in its first dissociationstep and a weak acid in its second step.

  41. LeChâtelier’s Principle • For 1.0 M or solutions of sulfuric acid, the large concentration of H3O+ from the first dissociation step represses the second step, which can be neglected as a contributor of H3O+ ions. For dilute solutions of sulfuric acid, the second step does make a significant contribution, ICE tables must be used to obtain thetotal H3O+ concentration.

  42. Problem • Calculate the pH of .50 M H2SO4 • H2SO4  HSO4- + H+Ka is very large • HSO4- ⇌SO42- + H+Ka= 1.2x10-2 • The main difference from sulfuric acid from other problems will be the first step. • 0.50 M H2SO4 means .50 M H+ and .5 M HSO4- is formed. • Now to the second dissociation…

  43. Cont. • HSO4-  SO42- + H+ • I .50 M .50 M • C -x +x +x • E .50 –x x .50 + x • Ka = x(.5+x) / (.5-x) = 1.2 x10-2 • x = .011462 • [HSO4-]= .49 M [SO42- ]= .011 M • [H+] = .51 M pH = .29

  44. Salts

  45. Salts and pH • The salts of strong acids or base will be neutral, excluding H2SO4. • To be a strong acid or base, the conjugate base or acid must have no affinity for protons or hydroxide. • That is to say it won’t run the reverse reaction under any conditions.

  46. Salts of weak acids and bases • Salts of weak acids and bases will run the reverse reactions. • Salts of weak acids are weak bases • HF ⇌ H+ + F- • F- + H2O⇌ HF + OH- • Salts of weak bases are weak acids • NH3⇌ NH4+ + OH- • NH4+⇌NH3 + H+

  47. pH of salt • Would potassium chlorite be acidic or basic? • KClO2⇌ K++ ClO2- • Potassium is neutral (KOH = strong base), chlorite is basic(HClO2 = weak acid) • Therefore it is basic

  48. What about salts of a weak acid and a weak bases… • If you have NH4F • You have to look at the Ka value of NH4+ and compare it to the Kb value of F- to see which weak acid/base is stronger. • The larger the k value the stronger the acid/base.

  49. Ka and Kb • The Ka and Kb of a weak acid and its conjugate base or vice versa are related and easily calculated from one another • For a weak acid and its conjugate base or a weak base and its conjugate acid • Kax Kb = Kw

  50. Problem • Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3is 1.8 x 10-5. • Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 x 10-4.

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