Equilibrium Acids and Bases A Dynamic Equilibrium equilibrium
Equilibrium, Acids and Bases A. Dynamic Equilibrium ¬ equilibrium theories and principles apply to a variety of phenomena in our world eg) blood gases in scuba diving CO 2 in carbonated beverages buffers in our blood ¬ reactions are often which means that reversible… not only are the products formed but the reactants can be reformed ¬we use the double arrow to show this relationship eg) A + B⇌C + D
¬the forward and reverse reaction will proceed at different rates…it depends on the concentration of the reactants and products u u u if we start with only the reactants A and B, the forward reaction will initially be the fastest as it is the only reaction possible as the products C and D are formed, the forward reaction will slow down and the reverse reaction will speed up at some point, the rates of forward and reverse reactions become equal
Dynamic Equilibrium forward reaction equilibrium Rate reverse reaction 0 Time
¬a system is said to be in a state of dynamic equilibrium when: 1. the of the forward and reverse reactions are rates equal observable (macroscopic) properties 2. the of the system, such as temperature, pressure, concentration, p. H are constant 3. the system is a system at closed constant temperature
B. Classes of Reaction Equilibria ¬there are three classes of chemical equilibria: 1. reactants favoured (percent rxn ) <50% A + B ⇌ C + D 2. products favoured (percent rxn ) >50% A + B ⇌ C + D
3. quantitative to the right (percent rxn ) >99% A + B ⇌ C + D or A + B C + D
C. The Equilibrium Constant ¬experiments have shown that under a given set of conditions (P and T) a specific quantitative relationship exists between the equilibrium concentrations of the reactants and products ¬one reaction that has been studied intensively is that between H 2(g) and I 2(g) (simple molecules and takes place in gas phase no solvent necessary!) H 2(g) + I 2(g) ⇌ 2 HI(g)
¬when different combinations of H 2(g), I 2(g), and HI(g) were mixed and the concentrations measured, it was discovered that equilibrium was reached in all cases different ¬even though the equilibrium [ ] are , the end was the same each time (within quotient experimental error)
¬this led to the empirical generalization known as the Law which says that there is a of Equilibrium between the concentrations of the constant ratio products and the concentrations of the reactants at equilibrium
¬this law can be expressed mathematically: For the reaction a. A + b. B ⇌ c. C + d. D The law is: Kc = [C]c [D]d [A]a [B]b Kc = equilibrium constant A, B = reactants C, D = products a, b, c, d = coefficients balancing where:
¬ is constant for a reaction at a given Kc …if you change the temperature, Kc also temperature changes ¬it is common to ignore the units for Kc and list it only as a numerical value (since depends on the powers of the various [ ] terms) ¬when determining Kc use only the species that are in gas or aqueous ***unless all states are the same, then use them all
¬the the value of K higher c, the greater the tendency for the reaction to favor the forward direction (the products) ¬if Kc is then the reaction is greater than 1, products favoured ¬if Kc is then the reaction is less than 1, reactants favoured ¬Kc indicates the and not the percent reaction rate of the reaction ¬catalysts will not affect the [ ] at equilibrium… they only increase the rate of the rxn
Example 1 Write the equilibrium law for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas. 2 NO(g) + O 2(g) ⇌ 2 NO 2(g) Kc = [NO 2(g)]2 [NO(g)]2[O 2(g)]
Example 2 Write the equilibrium law for the following reaction: Ca. CO 3(s) ⇌ Ca. O(s) + CO 2(g) Kc = [CO 2(g)] *** do not include solids in Kc
Example 3 Write the equilibrium law for the following reaction: 2 H 2 O(l) ⇌ 2 H 2(g) + O 2(g) Kc = [H 2(g)]2[O 2(g)] *** do not include liquids in Kc
Example 4 Phosphorus pentachloride gas can be decomposed into phosphorus trichloride gas and chlorine gas. a) Write the equilibrium law for this reaction. PCl 5(g) ⇌ PCl 3(g) + Cl 2(g) Kc = [PCl 3(g) ][Cl 2(g)] [PCl 5(g)]
b) If the [PCl 5(g)]eq = 4. 3 x 10 -4 mol/L, the [PCl 3(g) ]eq = 0. 014 mol/L and the [Cl 2(g)]eq = 0. 014 mol/L then calculate Kc. Kc = [PCl 3(g) ][Cl 2(g)] [PCl 5(g)] = (0. 014) (4. 3 x 10 -4) = 0. 46
Example 5 Find the [SO 3(g)] for the following reaction if Kc = 85. 0 at 25. 0 C. 2 SO 2(g) + 0. 500 mol/L Kc O 2(g) ⇌ 0. 500 mol/L = [SO 3(g) ]2____ [SO 2(g)]2[O 2(g)] 85. 0 = [SO 3(g) ]2 (0. 500)2(0. 500) [SO 3(g) ]2 = 10. 625 [SO 3(g) ] = 3. 26 mol/L 2 SO 3(g) ? ? ?
D. Graphical Analysis ¬a graph of vs. can be used concentration time to see when equilibrium has been reached…as soon as the concentrations , you don’t change any more can read this time off the graph
Example 1 Consider this rxn: 2 SO 2(g) + O 2(g) ⇌ 2 SO 3(g) SO 2(g) 75 Concentration 50 (mol/L) 25 0 10 20 30 Time (s) At what time does equilibrium get reached and what is the value for Kc?
Kc = [SO 3]2 [SO 2]2[O 2] = (75)2 (50)2(25) = 0. 090 Equilibrium is reached at approximately 20 seconds.
E. Le Châtelier’s Principle ¬ states that when a Le Châtelier’s principle equilibrium chemical system at is disturbed by a the system change in property of the system, adjusts in a way that opposes the change ¬this takes place in a three-stage process 1. initial equilibrium state 2. shifting non-equilibrium state 3. new equilibrium state
¬a system can be affected by a change in concentration, temperature and or volume (pressure) 1. Concentration Changes increase ¬an in the [ ] of the products or reactants favours the other side of the equation ¬a in the [ ] of the products or reactants decrease favours the same side of the equation
eg) N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) ***Haber-Bosch process ↑ [N 2(g)] will shift the equilibrium to the products ↑ [NH 3(g)] will shift the equilibrium to the reactants [NH 3(g)] will shift the equilibrium to the products ¬changes in concentration have on the value no effect of Kc
2. Temperature Changes ¬energy is treated like a or reactant product eg) reactants + energy ⇌ products reactants ⇌ products + energy ¬if cooled, the equilibrium shifts so more heat is produced (same side) ¬if heated, the equilibrium shifts away from the heat so it cools down (opposite side)
¬a change in temperature is the only stress that will the value of K change c!!!!!!! ¬if the shift is towards the side, K product c will increase ¬if the shift is towards the side, K reactant c will decrease
3. Volume and Pressure Changes ¬with gases, volume and pressure are related (volume , ↑ pressure ) ¬the concentration of a gas is related to volume ↓ ↑ (pressure)…volume , concentration http: //michele. usc. edu/java/gassim. html ¬an caused by a in volume increase in [ ] drop causes a shift towards the side of the equation with fewer moles of gas eg) N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) 4 moles 2 moles will shift to NH 3(g)
¬if the number of moles are the on both sides of same the reaction, a change in volume (pressure) has no effect ¬changes in volume and pressure have on the no effect value of Kc
4. Colour Changes ¬in many equilibrium systems, the reactants will have a different colour than the products ¬predictions can be made about the equilibrium shift and the resulting change in colour
Example Use the following reaction to predict the equilibrium shift and resulting colour change when the stresses are applied 2 Cr. O 42 (aq) + 2 H 3 O+(aq) ⇌ 2 Cr 2 O 72 (aq) + 3 H 2 O(l) yellow orange a) a crystal of Na 2 Cr. O 4(s) is added products, orange b) a crystal of K 2 Cr 2 O 7(s) is added reactants, yellow c) a few drops of concentrated acid is added products, orange d) water is removed products, orange e) a few crystals of Na. OH(s) are added reactants, yellow
¬all of the changes that can happen to systems in equilibrium can be shown graphically: Example State what change to the equilibrium takes place at each of the labelled parts of the graph:
Manipulations of An Equilibrium System N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) + energy NH 3(g) N 2(g) Concentration (mol/L) H 2(g) A B Time (min) C D Stress Equilibrium Time A addition of H 2(g) B addition of inert gas, addition of catalyst C decrease in volume D increase in energy
F. ICE Tables ¬we can use a table set-up to calculate the equilibrium concentrations and/or Kc for any system ¬you must be able to calculate all equilibrium [ ] before you can use the equilibrium law
Example 1 Hydrogen iodide gas decomposes into hydrogen gas and iodine gas. If 2. 00 mol of HI(g) is place in a 1. 00 L container and allowed to come to equilibrium at 35 C, the final concentration of H 2(g) is 0. 214 mol/L. Find the value for Kc. 2 HI(g) Initial ⇌ 2. 00 mol/L Change – 0. 214 mol/L x 2/1 = – 0. 428 Equil. 1. 572 mol/L H 2(g) 0 + I 2(g) 0 +0. 214 mol/L x 1/1 0. 214 mol/L +0. 214 mol/L
Kc = [H 2(g)][I 2(g)] [HI(g)]2 = (0. 214) (1. 572)2 = 0. 0185
Example 2 In a 500 m. L stainless steel reaction vessel at 900 C, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction establishes an equilibrium with only partial conversion of reactants to products. Initially, 2. 00 mol of each reactant is placed in the vessel. Kc for this reaction is 4. 20 at 900 C. Calculate the concentration of each substance at equilibrium. CO(g) + H 2 O(g) I 2. 00/0. 5 L = 4. 00 mol/L C –x mol/L E 4. 00 x mol/L ⇌ CO 2(g) + H 2(g) 0 +x mol/L x 1/1 x mol/L
Kc = [CO 2(g)][H 2(g)] [CO(g)][H 2 O(g)] 4. 20 = (x)(x) (4. 00 x)( 4. 00 x) 4. 20 = x 2 (4. 00 x)2 2. 05 = x (4. 00 x) 2. 05(4. 00 x) = x 8. 05 2. 05 x = x 8. 05 = 3. 05 x x = 2. 69 mol/L ***Note, this is a perfect square so to solve for x simply square root both sides of the equation, then solve
[CO(g)] = 4. 00 – x = 4. 00 – 2. 69 = 1. 31 mol/L [H 2 O (g)] = 4. 00 – x = 4. 00 – 2. 69 = 1. 31 mol/L [CO 2(g)] = x = 2. 69 mol/L [H 2(g)] = x = 2. 69 mol/L
Example 3 Gaseous phosphorus pentachloride decomposes into gaseous phosphorus trichloride and chlorine gas at a temperature where Keq = 1. 00 10 3. Suppose 2. 00 mol of PCl 5(g) in a 2. 00 L vessel is allowed to come to equilibrium. Calculate the equilibrium [ ] of each species. PCl 5(g) ⇌ I 2. 00 mol/2. 00 L = 1. 00 mol/L C –x mol/L x 1/1 E 1. 00 – x mol/L PCl 3(g) 0 mol/L +x mol/L + Cl 2(g) 0 mol/L +x mol/L x 1/1 x mol/L
Kc = [PCl 3(g)][Cl 2(g)] [PCl 5(g)] 1. 00 10 -3 = (x)(x) (1. 00 - x) ***at this point, you would have to use the quadratic formula to solve for x ¬when the concentrations are greater than the 1000 X equilibrium constant, we can make an approximation that greatly simplifies our calculations ¬if Kc is very small, the equilibrium doesn’t lie very far to the right and x is a very small number
***in this example 1. 00 – x can be assumed to be 1. 00 since x is really small, so… 1. 00 10 -3 = (x)(x) (1. 00 ) x 2 = 1. 00 10 -3 x 1. 00 x = 0. 0316 ***now you can calculate the [ ]eq for each species …substitute x into the equilibrium values in the ICE table [PCl 5(g)]eq = 1. 00 mol/L – 0. 0316 mol/L = 0. 967 mol/L [PCl 3(g)]eq = 0 + 0. 0316 mol/L = 0. 0316 mol/L [Cl 2(g)]eq = 0 + 0. 0316 mol/L = 0. 0316 mol/L
Example 4 Gaseous NOCl decomposes to form gaseous NO and Cl 2. At 35 C the equilibrium constant is 1. 6 10 -5. Calculate the equilibrium [ ] of each species when 1. 0 mol of NOCl is placed in a 2. 0 L covered flask. 2 NOCl(g) I C ⇌ 1. 0 mol/2. 0 L = 0. 50 mol/L –x mol/L x 2/1 = – 2 x mol/L E 0. 50 – 2 x mol/L 2 NO(g) + Cl 2(g) 0 mol/L +x mol/L x 2/1 = +2 x mol/L +x mol/L 2 x mol/L
Kc = [NO(g)]2[Cl 2(g)] [NOCl(g)]2 1. 6 10 -5 = (2 x)2(x) (0. 50 - 2 x)2 ***using approximation, 0. 50 – 2 x = 0. 50 1. 6 10 -5 = (4 x 2)(x) (0. 50 )2 4 x 3 = 1. 6 10 -5 x 0. 502 x 3 = 4. 0 10 -6 / 4 x = 0. 010 mol/L [NOCl(g)]eq = 0. 50 mol/L – (2)(0. 010) mol/L = 0. 48 mol/L [NO(g)]eq = 0 + (2)(0. 010 mol/L) = 0. 020 mol/L [Cl 2(g)]eq = 0 + 0. 010 mol/L = 0. 010 mol/L
G. Ionization of Water ¬the equilibrium of water can be written as follows: H 2 O(l) + H 2 O(l) H 3 O+(aq) + OH-(aq) ¬the equilibrium law is: Kc = [H 3 O+(aq) ][ OH-(aq)] ¬the equilibrium constant for water is designated as Kw ¬at 25 C, neutral water has [H+(aq)] = [OH-(aq)] = 1. 0 10 -7 mol/L Kw = (1. 0 10 -7) = 1. 0 10 -14 (on pg 3 of Data Booklet)
¬Kw is constant and therefore can be used to determine [H 3 O+(aq) ] [ OH-(aq)] the or the eg) if [H 3 O+(aq)] = 1. 0 10 4 mol/L then [OH (aq)] = 1. 0 10 14 = 1. 0 10 10 mol/L 1. 0 10 4 if [H 3 O+(aq)] = [OH (aq)], then solution is neutral if [H 3 O+(aq)] > [OH (aq)], then solution is acidic if [H 3 O+(aq)] < [OH (aq)], then solution is basic
Try These: 1. [H 3 O+(aq)] = 1. 0 10 9 mol/L basic [OH (aq)] = 1. 0 10 5 mol/L 2. [H 3 O+(aq)] = 1. 0 10 1 mol/L [OH (aq)] = 1. 0 10 13 mol/L 3. [H 3 O+(aq)] = 1. 0 10 12 mol/L [OH (aq)] = 1. 0 10 2 mol/L acidic basic 4. [H 3 O+(aq)] = 6. 3 10 9 mol/L [OH (aq)] = 1. 6 10 6 mol/L basic 5. [H 3 O+(aq)] = 8. 1 10 3 mol/L [OH (aq)] = 1. 2 10 12 mol/L acidic 6. [H 3 O+(aq)] = 3. 6 10 8 mol/L [OH (aq)] = 2. 8 10 7 mol/L basic
H. Review of p. H and p. OH ¬ the number of digits following the decimal place in p. H value sig digs the is equal to the number of in the [H 3 O+(aq)] p. H = log[H 3 O+(aq)] p. OH = log[OH-(aq)] [H 3 O+(aq)] = 10 -p. H [OH-(aq)] = 10 -p. OH p. H + p. OH = 14
Example 1 Find the p. H of a solution where the [H 3 O+(aq)] = 4. 7 10 -11 mol/L. p. H = log[H 3 O+(aq)] = log(4. 7 10 -11) = 10. 33
Example 2 Find the p. H of a solution where the [OH-(aq)] = 2. 4 10 -3 mol/L. p. OH = log[OH (aq)] = log(2. 4 10 3 ) = 2. 619… p. H = 14 – p. OH = 14 – 2. 619… = 11. 38
Example 3 Calculate the [H 3 O+(aq)] if the p. H of the solution is 5. 25. [H 3 O+(aq)] = 10 -p. H = 10 -5. 25 = 5. 6 10 -6 mol/L
Example 4 Calculate the p. H of a solution where 10. 3 g of Ca(OH)2(s) is dissolved in 500 m. L of water. Ca(OH)2(s) Ca 2+(aq) + 2 OH-(aq) m = 10. 3 g v = 0. 500 L M = 74. 10 g/mol n = 0. 139…mol 2/1 n = m M = 0. 278…mol = 10. 3 g 74. 10 g/mol C = n V = 0. 139…mol = 0. 278…mol 0. 500 L = 0. 556…mol/L
Example 4 (continued) p. OH = log[OH (aq)] = log(0. 556… ) = 0. 254… p. H = 14 – p. OH = 14 – 0. 254… = 13. 745
I. Brønsted-Lowry Definition of Acids & Bases (1923) ¬ this theory looks at the role of the acid or base ¬ an acid is a chemical species (anion, cation or molecule) that loses a proton ¬ a base is a chemical species that gains a proton ¬ like in electrochemistry where e are transferred…now we transfer H+
H + HCl(aq) + H 2 O(l) ⇌ Cl-(aq) + H 3 O+(aq) H + NH 3(aq) + H 2 O(l) ⇌ NH 4+(aq) + OH-(aq) ¬ water does not have to be involved! H + HCl(g) + NH 3(g) ⇌ NH 4 Cl(s)
¬a Brønsted-Lowry acid doesn’t necessarily have to produce an acidic solution…it depends on what accepts the proton ¬an acid/base reaction is a chemical reaction in which a proton (H+) is transferred from an acid to a base forming a new acid and a new base ¬this theory explains how some chemical species can be used to neutralize both acids and bases eg) HCO 3 -(aq) + H 3 O+(aq) ⇌ H 2 O(l) + H 2 CO 3(aq) HCO 3 -(aq) + OH-(aq) ⇌ H 2 O(l) + CO 32 -(aq)
¬a substance that appears to act as a Brønsted-Lowry acid in some rxns and a Brønsted-Lowry base in other rxns is said to be amphiprotic or amphoteric eg) H 2 O, HCO 3 , HSO 4 , HOOCCOO etc
J. Conjugate Acids and Bases ¬ a pair of substances that differ only by a proton is acid called a conjugate acid-base pair …the is on one side of the reaction and the is on the other base ¬ in general, the reaction can be shown as follows: HA(aq) + H 2 O(l) ⇌ H 3 O+(aq) + A-(aq) acid base conjugate acid conjugate base ¬ the an acid, the its conjugate base stronger weaker ¬ the an acid, the its conjugate base weaker stronger
K. Strengths of Acids and Bases ¬ two different acids (or bases) can have the same [ ] but have different strengths eg) 1 M CH 3 COOH(aq) and 1 M HCl(aq) will react in the same way but not to the same degree ¬ the stronger the acid, the electricity it conducts, more lower the the p. H and the it reacts with faster other substances
1. Strong Acids ¬ acids that ionize in water to form quantitatively H 3 O+(aq) ¬ percent rxn = 100% ¬ the bigger the Ka (Kc for acids) the more the products are favoured ¬ top 6 acids on the table (pg 8 -9 in Data Book) have a very large Ka …note the H 3 O+ is the strongest acid on the chart (leveling effect)…all strong acids react to form H 3 O+(aq) so it is the strongest
[SA] = [H 3 O+(aq)] ¬when calculating p. H, the so use p. H = -log[H 3 O+(aq)] Example What is the p. H of a 0. 500 mol/L solution of HNO 3(aq)? [H 3 O+(aq)] = [HNO 3(aq)] = 0. 500 mol/L p. H = -log[H 3 O+(aq)] = -log(0. 500 mol/L) = 0. 301
2. Weak Acids ¬ a weak acid is one that only partially ionizes in water to form H+(aq) ions ¬ most ionize <50% ¬ Ka value is small (<1) Ka value ¬ to calculate p. H, you need to use the …you cannot use just the because it is not [WA] 100% dissociated
¬the Ka law is an and is devised equilibrium law the same way we did Kc eg) HA(aq) + H 2 O(l) ⇌ H 3 O+(aq) + A-(aq) Ka = [H 3 O+(aq)][A-(aq)] [HA(aq)] ¬you will be required to figure out the [H 3 O+(aq)] before you can calculate the p. H
¬you have the and the value but you don’t Ka [WA] have the [A-(aq)] [H 3 O+(aq)]: [A-(aq)] ¬since the mole ratio for is , 1: 1 dissociation they have the same [ ] (this is a !) Ka = [H 3 O+(aq)]2 [HA(aq)] ¬now you can solve for x to get the [H 3 O+(aq)] = (Ka)([WA])
Example 1 What is the p. H of a 0. 10 mol/L acetic acid solution? ***check in DB…weak acid!!!!! CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O+(aq) + CH 3 COO-(aq) [H 3 O+(aq)] = (Ka)([WA]) = (1. 8 x 10 -5 mol/L )(0. 10 mol/L) = 1. 34… 10 -3 mol/L p. H = -log[H 3 O+(aq)] = -log(1. 34… 10 -3 mol/L) = 2. 87
Example 2 What is the p. H of a 1. 0 mol/L acetic acid solution? CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O+(aq) + CH 3 COO-(aq) [H 3 O+(aq)] = (Ka)([WA]) = (1. 8 x 10 -5 mol/L )(1. 0 mol/L) = 4. 24… 10 -3 mol/L p. H = -log[H 3 O+(aq)] = -log(4. 24… 10 -3 mol/L) = 2. 37
Example 3 A 0. 25 mol/L solution of carbonic acid has a p. H of 3. 48. Calculate Ka. H 2 CO 3(aq) + H 2 O(l) ⇌ H 3 O+(aq) + HCO 3 -(aq) [H 3 O+(aq)] = 10 -p. H = 10 -3. 48 = 3. 31… 10 -4 mol/L Ka = [H 3 O +(aq)]2 [H 2 CO 3 (aq)] = (3. 31…x 10 -4 mol/L)2 0. 25 mol/L = 4. 4 x 10 -7 mol/L
¬the can be written as a % reaction (% ionization) % above the in a chemical reaction: ⇌ 1. 3% eg) CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O+(aq) + CH 3 COO-(aq) Ka = 1. 8 x 10 -5 ¬the % reaction be calculated using [H 3 O+] and [WA] % ionization = [H 3 O+(aq)] 100 [WA(aq)]
Example 1 Calculate the % ionization for a 0. 500 mol/L solution of hydrosulphuric acid if the [H+(aq)] is 5. 0 10 -4 mol/L. % ionization = [H 3 O +(aq)] 100 [WA(aq)] = 5. 0 10 -4 mol/L 100 0. 500 mol/L = 0. 10 %
Example 2 The p. H of a 0. 10 mol/L solution of methanoic acid is 2. 38. Calculate the % ionization. [H 3 O+] = 10 -p. H = 10 -2. 38 = 0. 00416… mol/L % ionization = [H 3 O +(aq)] 100 [WA(aq)] = (0. 00416…mol/L) x (100) 0. 10 mol/L = 4. 2 %
3. Strong Bases ¬ according to Arrhenius, bases are substances that increase the of a solution hydroxide [ ] ¬ all are strong bases ionic hydroxides ¬ percent rxn = 100% ¬ strength depends on … # of hydroxide ions Ba(OH) Na. OH (aq) is a stronger base than at 2(aq) the same [ ] because it produces 2 OH-(aq)
[OH-(aq)] = x[BH(aq)] ¬ where x is the number of ions (think about the dissociation hydroxide equation!) Example Calculate the p. H of a 0. 0600 mol/L solution of Ca(OH)2(aq). [OH-(aq)] = x[BH(aq)] = x[Ca(OH)2(aq)] = 2(0. 0600 mol/L) = 0. 120 mol/L p. OH = log(0. 120) = 0. 9208… p. H = 14 – 0. 9208… = 13. 079
4. Weak Bases ¬ do not in water…just like dissociate completely weak acids Kb ¬ is the dissociation constant or equilibrium constant for bases B + H 2 O(l) ⇌ BH+(aq) + OH-(aq) Kb = [BH+(aq)][OH-(aq)] [WB(aq)]
¬ you can calculate Kb using for the Ka base ¬ Ka Kb = KW = 1. 00 10 14 eg) Calculate Kb for SO 42 (aq). Kb = KW Ka = 1. 00 10 14 1. 0 10 2 = 1. 0 10 12 mol/L
¬ once you have Kb, you can then solve for [OH (aq)] using the equilibrium law (just like with weak acids!) Kb = [BH+(aq)][OH-(aq)] [WB(aq)] Kb = [OH-(aq)]2 [WB(aq)] ¬ now you can solve for [OH (aq)] [OH-(aq)] = (Kb)([WB])
Example Find the p. H of a 15. 0 mol/L NH 3(aq) solution. NH 3(aq) + H 2 O(l) ⇌ NH 4+(aq) + OH-(aq) Ka = 5. 6 10 -10 mol/L Kb = Kw Ka = 1. 00 10 -14 5. 6 10 -10 = 1. 78… 10 -5 mol/L
[OH-(aq)] = (Kb)([NH 3(aq)]) [OH-(aq)] = (1. 78… 10 -5)(15. 0) [OH-(aq)] = 1. 63… x 10 -2 mol/L p. OH = -log[OH-(aq)] = -log(1. 63…x 10 -2 mol/L) = 1. 78… p. H = 14 – p. OH = 14 – 1. 78… = 12. 214
L. Predicting Acid-Base Equilibria ¬ acids are listed in order of strength on decreasing the left side and bases are listed in order of strength on the right side increasing ¬ when predicting reactions, the substance with the greatest attraction for protons (the strongest base) will react with the substance that gives up its proton most easily (strongest acid) one proton ¬ we will assume that only is transferred per reaction
¬ to predict the acid-base reaction, follow the following steps: Steps 1. List all species (ions, atoms, molecules) initially present. Note: strong acids ionize into H 3 O+ and the anion weak acids are NOT dissociated dissociate ionic compounds don’t forget to include water 2. Identify all possible acids and bases.
3. Identify the and strongest acid (SA) strongest base (SB) …like redox rxns the SA and the is top left SB is bottom right. 4. To write the reaction, transfer one proton from the acid to the base to predict the conjugate acid and conjugate base.
5. Predict the position of the equilibrium. Note: if acid is above base, then >50% (favours products) ⇌ if base is above acid, then <50% (favours reactants) ⇌
Example 1 Predict the acid-base reaction that occurs when sodium hydroxide is mixed with vinegar. List: OH-(aq) + Na+(aq) OH-(aq) CH 3 COOH(aq) H 2 O(l) A B A/B SB SA CH 3 COOH(aq) ⇌ H 2 O(l) + CH 3 COO-(aq)
Example 2 Predict the acid-base reaction when ammonia is mixed with HCl(aq). List: NH 3(aq H 3 O+(aq) A B SB SA NH 3(aq) + H 3 O+(aq) ⇌ Cl-(aq) B H 2 O(l) A/B H 2 O(l) + NH 4+ (aq)
M. Monoprotic vs. Polyprotic Acids and Bases ¬ an acid capable of donating only one proton is called monoprotic eg) HCl(aq), HNO 3(aq), HOCl(aq) etc. ¬if an acid can transfer more than one proton, it is polyprotic diprotic called ( if 2 protons, if 3 protons) triprotic
eg) Label each of the following acids as monoprotic or polyprotic: 1. H 2 SO 4(aq) polyprotic 2. HOOCCOOH(aq) polyprotic 3. HCOOH(aq) monoprotic 4. CH 3 COOH(aq) monoprotic 5. H 2 PO 4 -(aq) polyprotic 6. NH 4+(aq) monoprotic
¬ a base capable of accepting only one proton is called a monoprotic base ¬ a base that can accept more than one proton is called a polyprotic base ( or ) diprotic triprotic + to form PO 43 -(aq) eg) can accept up to 3 H 2 and H PO H 3 PO 4(aq) HPO , 2 4 (aq), 4 (aq) respectively
eg) Label each of the following as monoprotic or polyprotic acids, monoprotic or polyprotic bases: 1. HSO 4 -(aq) 2. H 2 PO 4 -(aq) 3. HPO 42 -(aq) 4. HCO 3 -(aq) 5. H 2 O(l) monoprotic acid; monoprotic base polyprotic acid; monoprotic base monoprotic acid; polyprotic base monoprotic acid; monoprotic base
¬ reactions involving polyprotic acids or polyprotic bases substances involve the same principles of reaction prediction ¬only is transferred at a time and one proton always from strongest acid to strongest base
Example 1 Potassium hydroxide is continuously added to oxalic acid until no more reaction occurs. List: K+(aq) OH-(aq) HOOCCOOH(aq) H 2 O(l) HOOCCOO-(aq) OOCCOO B SB A SA A/B SA (aq) 2 - B OH-(aq) + HOOCCOOH(aq)⇌ H 2 O(l) + HOOCCOO-(aq) OH-(aq) + HOOCCOO-(aq) ⇌ H 2 O(l) + OOCCOO 2 -(aq)
¬ Net Reaction: Add all reactions together (only if all quantitative), cancelling out any species that occur in the same quantity on both the reactant and product sides and summing any species that occur more than once on the same side OH-(aq) + HOOCCOOH(aq) H 2 O(l) + HOOCCOO-(aq) OH-(aq) + HOOCCOO-(aq) H 2 O(l) + OOCCOO 2 -(aq) 2 OH-(aq)+ HOOCCOOH(aq) 2 H 2 O(l) + OOCCOO 2 -(aq)
Example 2 Sodium hydrogen phosphate is titrated with hydroiodic acid. If we assume all steps are quantitative, give the net reaction. List: Na+(aq) HPO 42 -(aq) I-(aq) H 3 O+(aq) H 2 O(l) H 2 PO 4 -(aq) H 3 PO 4(aq) A/B B SB A SA A/B SB A H 3 O+(aq) + HPO 42 -(aq) H 2 O(l) + H 2 PO 4 -(aq) H 3 O+(aq) + H 2 PO 4 -(aq) H 2 O(l) + H 3 PO 4(aq) 2 H 3 O+(aq) + HPO 42 -(aq) 2 H 2 O(l)+ H 3 PO 4(aq)
O. Titrations ¬ titrations are used to determine the p. H of the of acid-base reactions endpoint ¬ the information from the titration can be plotted on a graph, buffer regions can be analyzed and stoichiometric calculations can be performed
1. p. H Curves ¬ a p. H curve is a graph showing the continuous change of p. H during an acid-base reaction ¬ graph of vs p. H (y-axis) volume of titrant added to sample (x-axis) ¬ the endpoint is the point (usually shown by a change in indicator colour) when the reaction has gone to completion ¬ the equivalence point is the of titrant volume required for the reaction to go to completion
¬ they contain a relatively flat region called the buffer region ¬ all p. H curves have 4 major features: v the initial p. H of the curve must be the p. H of the sample v the co-ordinate of the equivalence point must be correct in terms of p. H and volume v the “over-titration” must be asymptotic with the p. H of the titrant v number of equivalence points must match the number of quantitative reactions occurring
¬ titrant selection: v if the sample is an acid, titrant should be a strong base such as Na. OH(aq) or KOH(aq) v if the sample is a base, titrant should be HCl(aq)
¬ you need to be able to interpret p. H curves: 1. Strong Monoprotic with Strong Monoprotic 7 v p. H of at the equivalence point SA titrated with SB SB titrated with SA 14 14 p. H EP 7 0 0 volume
2. Weak Monoprotic with Strong Monoprotic v if weak acid, then p. H of at equivalence point >7 v if weak base, then p. H of at equivalence point <7 v bottom “flat” region is not as flat as with strong acid/strong base WA titrated with SB WB titrated with SA 14 p. H 14 EP 7 0 p. H volume EP 7 0 volume
3. Polyprotic with Strong Monoprotic v more than 1 equivalence point WA(poly) titrated with SB WB(poly) titrated with SA 14 14 EP 2 EP 1 7 p. H EP 2 p. HEP 1 0 7 volume 0 volume
2. Indicators ¬an indicator is a substance that changes colour when it reacts with an acid or base and are usually weak acids themselves ¬they exist in one of two conjugate forms that are reversible and distinctly different in color HIn(aq) + H 2 O(l) ⇌ acid eg) litmus red base In-(aq) + H 3 O+(aq) conjugate base blue conjugate acid
¬ recall that to show the equivalence point of an acidbase titration, choose an indicator: 1. whose includes the colour change range equivalence point of the titration 2. that will react right after the sample reacts …this means the indicator is a weaker acid or base than the sample
3. Buffers ¬ buffers are chemicals that, when added to water, protect the solution from large p. H changes when acids or bases are added to them ¬they are used to and calibrate p. H meters contro l the rate of p. H sensitive reactions (eg. in the blood) ¬ typical buffers are solutions containing relatively conjugate pairs large amounts of such as a weak acid and the salt of the conjugate base eg) H 2 CO 3 and Na. HCO 3
¬can be selected by using …this tells p. Ka = –log. Ka you the p. H at which the buffer is most useful eg) Choose a buffer that would be useful for each of the following solutions: 1. p. H of 7. 0 H 2 S – HS- p. Ka = -log(8. 9 × 10 -8) = 7. 1 2. p. H of 4. 5 CH 3 COOH – CH 3 COOp. Ka = -log(1. 8 × 10 -5) = 4. 7 3. p. H of 10. 0 HCO 3 - – CO 32 p. Ka = -log(4. 7 × 10 -11) = 10. 3
acid ¬the in the conjugate pair of the buffer protects against any added base ¬the in the conjugate pair of the buffer protects base acid against any added ¬buffers can be by the addition of overwhelmed too much acid or base
Example Using an acetic acid – sodium acetate buffer system, show what happens when: a) a small amount of HCl(aq) is added CH 3 COOH(aq) Na+(aq) CH 3 COO-(aq) H 3 O+(aq) Cl-(aq) H 2 O(l) A – B SB CH 3 COO-(aq) + H 3 O+(aq) ⇌ A SA B A/B CH 3 COOH(aq) + H 2 O(l)
b) a small amount of Na. OH(aq) is added CH 3 COOH(aq) Na+(aq) CH 3 COO-(aq) OH-(aq) A SA – B CH 3 COOH(aq)+ OH-(aq) ⇌ B SB H 2 O(l) A/B CH 3 COO-(aq) + H 2 O(l)
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