Equilibrium Acids and bases Titrations Solubility AP Review Big Idea 6 Equilibrium Acids and bases Titrations Solubility

Equilibrium Constant – Keq Most chemical reactions are reversible aA + bB cC + dD Keq = [C]c [D]d = products [A]a [B]b reactants Keq has no units Solids and pure liquids are not included in equilibrium

HC2H3O2(aq) H+(aq) + C2H3O2-(aq) [H+] [C2H3O2-] [HC2H3O2] Keq =

2H2S(g) + 3O2 (g) 2H2O (g) + 2SO2 (g) You can make an equilibrium expression for gases too. Keq = [H2O]2 [SO2]2 [H2S]2 [O2]3 KP = P2H2O P2 SO2 P2H2S P3 O2 You can use equilibrium expression if you have concentrations of the gases You can use equilibrium expression if you have pressures of the gases

Ksp Used for dissociation of a (very) slightly soluble solid Product is always a solid … so not part of equilibrium expression CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp = [Ca2+ ] [F-]2

Le Chatelier When stress is placed on equilibrium, equilibrium will shift to relieve the stress and re-establish equilibrium Note: heat is added need to know if exo or endothermic rxn. Exo = heat is a product Endo= heat is a reactant Pressure only effects gases. When pressure is increases = shift to side with LEAST # gas moles

Q Q is a particular point in the reaction Set up same as Keq We compare Q to Keq to see if this point in the reaction favors products (forward), favors reactants(reverse) or is at equilibrium If Q< Keq favors reactants(reverse) If Q> Keq favors products (forward) If Q= Keq at equilibrium

Keq and multistep problem If A + B C Keq = K1 and C D + E Keq = K2 Then A+ B D + E Keq = K1K2

Use ICE Box

NH3 H2O NH4+ OH- 0.0124M 0M 0M - 4.62x10-4M +4.62x10-4M +4.62x10-4M K = [NH4+] [OH-] ------------------ [NH3] K = [4.62x10-4] [4.62x10-4] ------------------ [.011936] K = 1.79x 10 -5

Common Ion Effect A solid is thrown in some water. The solid is only very slightly soluble in water so an equilibrium is established AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq) So using le Chatilier, what would happen if I add more Ag+ ? More Cl- ? Product side would be too high so ….Shift to the left Product side would be too high so ….Shift to the left

Common Ion Effect AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq) You can add more of the products, its called common ion effect. It does effect the way equilibrium will shift. If I add more [Cl- ]then the equil. will shift to the left and make more reactant while [Ag+ ] down

Common Ion Effect 1.6 x 10 -10 = [Ag+ ] [Cl-] AgCl(s) H2O Ag+ Cl- AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq) If Ksp = 1.6 x 10 -10, what is [Ag+ ] and [Cl-] ? Ksp = 1.6 x 10 -10 = [Ag+ ] [Cl-] 1.6 x 10 -10 = (x) (x) 1.6 x 10 -10 = x2 1.3 x 10 -5 = x = [Ag+ ] =[Cl-] So what happens if I add .1mol NaCl to 1L of AgCl solution? NaCl is completely soluble and will completely dissociate to Na+ and Cl- 0.1mol of Cl- is a significant amount of Cl- compared to [Cl-]=1.3 x 10 -5 1.6 x 10 -10 = [Ag+ ] [Cl-] 1.6 x 10 -10 = [Ag+ ] [.1M] [Ag+ ] = 1.6 x 10 -10 0.10 [Ag+ ] = 1.6 x 10 -9 [Ag+ ] has decreased drastically 0M 0M +x +x x x

Gibbs Free Energy and Equilibrium DG = -RT lnK (K is equilibrium constant) If DG = negative, K must be >1 so product are favored If DG = positive, K must be <1 so reactant are favored

Acids and bases Acids donate H+ Bases accept H+ (Ex. NaOH or NH3) pH = -log[H+] pOH = -log[OH-] pH = 7 neutral pH < 7 acid pH> 7 base

HCl HBr HI H2SO4 HNO3 HClO3 HClO4 They completely ionize. 100% Strong acids HCl HBr HI H2SO4 HNO3 HClO3 HClO4 They completely ionize. 100% Can find pH using pH = -log[H+] Strong bases LiOH NaOH KOH RbOH CsOH Ca(OH)2 Sr(OH)2 Ba(OH)2 and then 14 = pOH +pH

Weak acids Weak bases HA H+(aq) + A- (aq) B HA+(aq) + OH- (aq) Only a small amount ionizes Need to set up equilibrium HA H+(aq) + A- (aq) Ka = [H+] [A- ] [HA ] To find pH -ice box! Only a small amount ionizes Need to set up equilibrium B HA+(aq) + OH- (aq) Kb = [HA+] [OH- ] [B ] To find pH -ice box

Find pH of .2M HC2H3O2, with Ka = 1.8 x 10-5 HC2H3O2 H+ + C2H3O2- E .2M 0M 0M -x +x +x .2-x x x 1.8 X 10-5 = x2 .2 x = [H+] = 1.9 x 10 -5 pH = -log (1.9 x 10 -5) pH = 2.7

Buffers and Salt solutions

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 NaF(aq) Na+ + F- step 1: identify major species Na+, F-, H2O Step 2: write dominate reaction F-(aq) + H2O (l) HF(aq) + OH-(aq) expect something basic If from weak acid then stronger conjugate base) (from strong base) (from weak acid)

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 F-(aq) + H2O (l) HF(aq) + OH-(aq) Since F- is a stronger conjugate base we need Kb Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 Step 3: calculate Kb, using Ka and Kw Ka x Kb = Kw Acid Conj base Kb=Kw/Ka (for HF) Kb = 1.0x10-14/7.2x10-4 Kb= 1.4 x 10 -11

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 Step 4 make icebox F- + H2O (l) HF + OH- 0.30 -x +x +x 0.30-x x x Use short cut 5% X is significantly smaller than .3 so x is negligible and .3M-x is about equal to .3M Kb = 1.4x10-11 = (x)(x) 0.30 x= 2.0x10-6= [OH-]

x= 2.0x10-6= [OH-] pOH = -log [OH-] = -log(2.0x10-6) = 5.69 14=pH +pOH 14 – pOH = pH 14- 5.69 = 8.31 = pH

Buffer…. Weak acid + its conjugate base Or Weak base + its conjugate acid

Which of following pairs make a buffer? A. NH3, NH4Cl B. RbOH, HBr C. KOH, HF D. NaC2H3O2, HCl

Calculate the pH of a buffer solution that contains 0 Calculate the pH of a buffer solution that contains 0.15M of acetic acid and 0.20mol of sodium acetate in 1L. (Ka =1.8 x 10-5) Solve by icebox and then solve by Henderson-Hasselbach HC2H3O2 H+ + C2H3O2- .15M .2M -x +x +x .15-x +x .2 +x 5% siort cut K= x(.2) ____ .15 1.8x10 -5= x(.2) ____ .15 x= 1.35x10 -5 and x also = [H+] pH =-log[H+] pH =-log(1.35x10 -5) pH = 4.87

Yes- Henderson-Hasselbalch equation Is there an easier way to calculate pH of a buffered system? Yes- Henderson-Hasselbalch equation pH = pKa + log[A-]/[HA] pH =pKa + log[base]/[acid] Is useful for calculating pH of solutions when the HA and A- are known pH =pKa + log[base]/[acid] look back at our problem Calculate the pH of a buffer solution that contains 0.15M of acetic acid and 0.20mol of sodium acetate in 1L. (Ka =1.8 x 10-5) pH = -log(1.8x10 -5) + log[.2]/[.15] pH = 4.745 + .124 pH = 4.87 same as by ice box method just faster 

Equivalence pt. During strong acid/strong base titration is always pH7 Equivalence pt. happens during titration. It is when moles H+ = moles OH- Equivalence pt. During strong acid/strong base titration is always pH7 7 7 Volume Base added Volume Acid added Titration curve of strong acid with strong base Titration curve of strong base with strong acid Type 1: strong acid-strong base titration

Weak acid- Strong base HF + OH- H2O + F- (A) Maximum buffering where [HA] / [A-]= 1 (B) Equivalence point (above pH = 7) and [HA] =0 (C) Basic (D) ½ way point and pH =pKa No calculation needed C A B D Definition of buffer = weak acid + its salt (conj. base) Or weak base + its salt (conj.acid)

Weak Base- Strong Acid NH3 + HCl NH4+ +Cl- Equivalence pt