Aqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

2012 Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

The Common-Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. 2012 Pearson Education, Inc.

The Common-Ion Effect The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. 2012 Pearson Education, Inc.

The Common-Ion Effect Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8 10 4. K a = [H 3 O + ] [F ] [HF] 2012 Pearson Education, Inc. = 6.8 10 4

The Common-Ion Effect HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M [H 3 O + ], M [F ], M Initially 0.20 0.10 0 Change At equilibrium 2012 Pearson Education, Inc.

The Common-Ion Effect 6.8 10 4 = (0.10) (x) (0.20) (0.20) (6.8 10 4 ) (0.10) = x 1.4 10 3 = x 2012 Pearson Education, Inc.

The Common-Ion Effect Therefore, [F ] = x = 1.4 10 3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4 10 3 = 0.10 M So, ph = log (0.10) ph = 1.00 2012 Pearson Education, Inc.

Common Ion Effect 10

Buffers Buffers are solutions of a weak conjugate acid base pair. They are particularly resistant to ph changes, even when strong acid or base is added. 2012 Pearson Education, Inc.

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water. 2012 Pearson Education, Inc.

Buffers Similarly, if acid is added, the F reacts with it to form HF and water. 2012 Pearson Education, Inc.

The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze Most brands of antifreeze contain ethylene glycol sweet taste initial effect drunkenness Metabolized in the liver to glycolic acid HOCH 2 COOH ethylene glycol (aka 1,2 ethandiol) glycolic acid (aka α-hydroxyethanoic acid) 14

Why Is Glycolic Acid Toxic? If present in high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO 3 in the blood, causing the blood ph to drop When the blood ph is low, its ability to carry O 2 is compromised acidosis HbH + (aq) + O 2 (g) HbO 2 (aq) + H + (aq) One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of 15 ethylene glycol

Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O K a = [H 3 O + ] [A ] [HA] H 3 O + + A 2012 Pearson Education, Inc.

Buffer Calculations Rearranging slightly, this becomes K a = [H 3 O + ] [A ] [HA] Taking the negative log of both side, we get log K a = log [H 3 O + ] + log [A ] [HA] base pk a ph 2012 Pearson Education, Inc. acid

Buffer Calculations So pk a = ph log [base] [acid] Rearranging, this becomes ph = pk a + log [base] [acid] This is the Henderson Hasselbalch equation. 2012 Pearson Education, Inc.

Henderson Hasselbalch Equation What is the ph of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4 10 4. 2012 Pearson Education, Inc.

Henderson Hasselbalch Equation ph = pk a + log [base] [acid] ph = log (1.4 10 4 ) + log (0.10) (0.12) ph = 3.85 + ( 0.08) ph = 3.77 2012 Pearson Education, Inc.

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations assuming the [H 3 O + ] from water is 0 HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change equilibrium 21

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change equilibrium x +x +x 0.100 x 0.100 + x x 22

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A ] eq = [A ] init, then solve for x K a for HC 2 H 3 O 2 = 1.8 x 10 5 [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change x +x +x equilibrium 0.100 x 0.100 +x x 23

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? K a for HC 2 H 3 O 2 = 1.8 x 10 5 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init x = 1.8 x 10 5 [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change x +x +x equilibrium 0.100 0.100 x the approximation is valid 24

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? substitute x into the equilibrium concentration definitions and solve x = 1.8 x 10 5 [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change x +x +x equilibrium 0.100 x 0.100 + x 1.8E-5 x 25

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? substitute [H 3 O + ] into the formula for ph and solve [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change x +x +x equilibrium 0.100 0.100 1.8E 5 26

Example: What is the ph of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a K a for HC 2 H 3 O 2 = 1.8 x 10 5 [HA] [A ] [H 3 O + ] initial 0.100 0.100 0 change x +x +x equilibrium 0.100 0.100 1.8E 5 the values match 27

Example: What is the ph of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2? assume the [HA] and [A ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch Equation check the x is small approximation HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 + H 3 O + K a for HC 7 H 5 O 2 = 6.5 x 10 5 28

Practice What is the ph of a buffer that is 0.14 M HF (pk a = 3.15) and 0.071 M KF? 29

Practice What is the ph of a buffer that is 0.14 M HF (pk a = 3.15) and 0.071 M KF? find the pk a from the given K a HF + H 2 O F + H 3 O + assume the [HA] and [A ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the x is small approximation 30

Henderson-Hasselbalch Equation for Basic Buffers The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H 2 O H:B + + OH To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction H:B + + H 2 O B: + H 3 O + this does not affect the concentrations, just the way we are looking at the reaction 31

Relationship between pk a and pk b Just as there is a relationship between the K a of a weak acid and K b of its conjugate base, there is also a relationship between the pk a of a weak acid and the pk b of its conjugate base K a K b = K w = 1.0 x 10 14 log(k a K b ) = log(k w ) = 14 log(k a ) + log(k b ) = 14 pk a + pk b = 14 32

Example: What is the ph of a buffer that is 0.50 M NH 3 (pk b = 4.75) and 0.20 M NH 4 Cl? find the pk a of the conjugate acid (NH 4+ ) from the given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the x is small approximation NH 3 + H 2 O NH 4 + + OH 33

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? The Henderson-Hasselbalch equation is generally good enough when the x is small approximation is applicable Generally, the x is small approximation will work when both of the following are true: a) the initial concentrations of acid and salt are not very dilute b) the K a is fairly small For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of K a 34

ph Range The ph range is the range of ph values over which a buffer system works effectively. It is best to choose an acid with a pk a close to the desired ph. 2012 Pearson Education, Inc.

Example: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with ph 4.25? Chlorous acid, HClO 2 pk a = 1.95 Nitrous acid, HNO 2 pk a = 3.34 Formic acid, HCHO 2 pk a = 3.74 Hypochlorous acid, HClO pk a = 7.54 The pk a of HCHO 2 is closest to the desired ph of the buffer, so it would give the most effective buffering range 36

Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer 37

Effectiveness of Buffers A buffer will be most effective when the [base]:[acid] = 1 equal concentrations of acid and base A buffer will be effective when 0.1 < [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large 38

Buffering Capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in ph The buffering capacity increases with increasing absolute concentration of the buffer components As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] 39

When Strong Acids or Bases Are Added to a Buffer When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction. 2012 Pearson Education, Inc.

Addition of Strong Acid or Base to a Buffer 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson Hasselbalch equation to determine the new ph of the solution. 2012 Pearson Education, Inc.

Calculating ph Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is 4.74. Calculate the ph of this solution after 0.020 mol of NaOH is added. 2012 Pearson Education, Inc.

Calculating ph Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 ph = pk a = log (1.8 10 5 ) = 4.74 2012 Pearson Education, Inc.

Calculating ph Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) HC 2 H 3 O 2 C 2 H 3 O 2 OH Before reaction 0.300 mol 0.300 mol 0.020 mol After reaction 2012 Pearson Education, Inc.

Calculating ph Changes in Buffers Now use the Henderson Hasselbalch equation to calculate the new ph: ph = 4.74 + log (0.320) (0.200) ph = 4.74 + 0.06 ph = 4.80 2012 Pearson Education, Inc.

Titration In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base). 2012 Pearson Education, Inc.

Titration In an acid-base titration, a solution of unknown concentration is slowly added to a solution of known concentration(titrant) from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint an indicator is a chemical that changes color when the ph changes When the moles of H 3 O + = moles of OH, the titration has reached its equivalence point 48

Titration A ph meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. 2012 Pearson Education, Inc.

Titration 50

Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the ph goes up slowly. 2012 Pearson Education, Inc.

Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the ph increases rapidly. 2012 Pearson Education, Inc.

Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. 2012 Pearson Education, Inc.

Titration of a Strong Acid with a Strong Base As more base is added, the increase in ph again levels off. 2012 Pearson Education, Inc.

Titration of 25 ml of 0.100 M HCl with 0.100 M NaOH Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes Before Equivalence (excess acid) After Equivalence (excess base) Equivalence Point equal moles of HCl and NaOH ph = 7.00 55

Titration of 25 ml of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Initial ph = log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 Before equivalence point added 5.0 ml NaOH 5.0 x 10 4 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 0 5.0E-4 mols change 5.0E-4 +5.0E-4 5.0E-4 mols end 2.00E-3 5.0E-4 0 molarity, new 0.0667 0.017 0 Tro: Chemistry: A Molecular Approach, 2/e 56

Titration of 25 ml of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (aq) To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10 3 moles At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over Because the NaCl is a neutral salt, the ph at equivalence = 7.00 57

Titration of 25 ml of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Initial ph = log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 At equivalence point added 25.0 ml NaOH 2.5 x 10 3 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 0 2.5E-3 mols change 2.5E-3 +2.5E-3 2.5E-3 mols end 0 2.5E-3 0 molarity, new 0 0.050 0 58

Titration of 25 ml of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Initial ph = log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 After equivalence point 0.100 mol NaOH L of added NaOH 1 L = moles added NaOH added 30.0 ml NaOH 3.0 x 10 3 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 0 3.0E-3 mols change 2.5E-3 +2.5E-3 2.5E-3 mols end 0 2.5E-3 5.0E-4 molarity, new 0 0.045 0.0091

Adding 0.100 M NaOH to 0.100 M HCl added 25.0 ml 35.0 30.0 10.0 25.0 0.100 ml M NaOH HCl 0.00100 0.00050 0.00150 0.00200 0.00250 equivalence mol point NaOH HCl ph = 12.22 11.96 1.37 1.18 1.00 7.00 added 40.0 15.0 ml NaOH 0.00150 0.00100 mol NaOH HCl ph = 12.36 1.60 added 50.0 20.0 ml NaOH 0.00250 0.00050 mol NaOH HCl ph = 12.52 1.95 Tro: Chemistry: A Molecular Approach, 2/e 60

Practice Calculate the ph of the solution that results when 10.0 ml of 0.15 M NaOH is added to 50.0 ml of 0.25 M HNO 3 61

Practice Calculate the ph of the solution that results when 10.0 ml of 0.15 M NaOH is added to 50.0 ml of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H 2 O(l) Initial ph = log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/l=1.25 x 10 2 Before equivalence point added 10.0 ml NaOH HNO 3 NaNO 3 NaOH mols before 1.25E-2 0 1.5E-3 mols change 1.5E-3 +1.5E-3 1.5E-3 mols end 1.1E-3 1.5E-3 0 molarity, new 0.018 0.025 0 62

Practice Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 ml of 0.25 M HNO 3 to reach equivalence 63

Practice Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 ml of 0.25 M HNO 3 to reach equivalence HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H 2 O(l) Initial ph = log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/l=1.25 x 10 2 At equivalence point: moles of NaOH = 1.25 x 10 2 64

Practice Calculate the ph of the solution that results when 100.0 ml of 0.15 M NaOH is added to 50.0 ml of 0.25 M HNO 3 65

Practice Calculate the ph of the solution that results when 100.0 ml of 0.15 M NaOH is added to 50.0 ml of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H 2 O(l) Initial ph = log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/l=1.25 x 10 2 After equivalence point added 100.0 ml NaOH HNO 3 NaNO 3 NaOH mols before 1.25E-2 0 1.5E-2 mols change 1.25E-2 +1.25E-2 1.25E-2 mols end 0 1.25E-2 0.0025 molarity, new 0 0.0833 0.017

Practice Calculate the ph of the solution that results when 100.0 ml of 0.15 M NaOH is added to 50.0 ml of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H 2 O(l) Initial ph = log(0.250) = 0.60 initial mol of HNO 3 = 0.0500 L x 0.25 mol/l=1.25 x 10 2 After equivalence point HNO 3 NaNO 3 NaOH mols before 1.25E-2 0 1.5E-2 mols change added 100.0 ml NaOH 1.25E-2 +1.25E-2 1.25E-2 mols end 0 1.25E-2 0.0025 molarity, new 0 0.0833 0.017 67

Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 68

Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the ph when it is formed. At the equivalence point the ph is >7. Phenolphthalein is commonly used as an indicator in these titrations. 2012 Pearson Education, Inc.

Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial ph is determined using the K a of the weak acid The ph in the excess acid region is determined as you would determine the ph of a buffer The ph at the equivalence point is determined using the K b of the conjugate base of the weak acid The ph after equivalence is dominated by the excess strong base the basicity from the conjugate base anion is negligible 70

Titration of 25 ml of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) Initial ph [HCHO 2 ] [CHO 2 ] [H 3 O + ] K a = 1.8 x 10 4 initial 0.100 0.000 0 change x +x +x equilibrium 0.100 x x x 71

Titration of 25 ml of 0.100 M HCHO 2 with 0.100 M NaOH (K a = 1.8 x 10-4 ) HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 Before equivalence HCHO2 CHO OH mols before 2.50E-3 0 0 mols added 5.0E-4 +5.0E4 5.0E-4 mols after 2.00E-3 5.0E-4 0 added 5.0 ml NaOH 72

Titration of 25 ml of 0.100 M HCHO 2 with 0.100 M NaOH (K a = 1.8 x 10-4 ) HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 At equivalence [HCHO 2 ] [CHO 2 ] [OH ] HA A OH initial 0 0.0500 0 mols before 2.50E-3 0 0 mols change added +x x +x 2.50E-3 mols equilibrium after x 0 5.00E-2-x 2.50E-3 x 0 CHO 2 (aq) + H 2 O (l) HCHO 2(aq) + OH (aq) added 25.0 ml NaOH K b = 5.6 x 10 11 [OH ] = 1.7 x 10 6 M

Titration of 25 ml of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/l = 2.50 x 10 3 After equivalence HA A NaOH mols before 2.50E-3 0 3.0E-3 mols change 2.5E-3 +2.5E-3 2.5E-3 mols end 0 2.5E-3 5.0E-4 molarity, new 0 0.045 0.0091 74 added 30.0 ml NaOH 3.0 x 10 3 mole NaOH added

Adding NaOH to HCHO 2 added 10.0 25.0 30.0 ml ml NaOH initial HCHO 2 solution 0.00150 equivalence 0.00050 0.00200 added 0.00250 35.0 mol mol ml point HCHO NaOH HCHO NaOH 2 xs 0.00250 2 0.00100 3.14 3.56 mol ph 2.37 mol CHO ph = 11.96 NaOH 2 xs [CHO ph = 12.22 2 ] init = 0.0500 M [OH ] eq = 1.7 x 10 6 ph added = 8.23 12.5 ml NaOH added 40.0 ml NaOH 0.00125 mol HCHO 0.00150 mol NaOH 2 xs ph 3.74 = pk ph = 12.36 a half-neutralization added 50.0 15.0 ml NaOH 0.00250 0.00100 mol NaOH HCHO 2 xs ph = 12.52 3.92 added 20.0 ml NaOH 0.00050 mol HCHO 2 ph = 4.34 75

Titration of a Weak Acid with a Strong Base With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle. 2012 Pearson Education, Inc.

Titration of a Weak Base with a Strong Acid The ph at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea. Methyl red is the indicator of choice. 2012 Pearson Education, Inc.

Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. 2012 Pearson Education, Inc.

Phenolphthalein 79

Methyl Red H C H C H C H C C C C CH C H C H (CH 3 ) 2 N N N N H H 3 O+ OH- NaOOC C C H H C H C H C H C (CH 3 ) 2 N C C N N N C CH C H C H NaOOC C C H 80

Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in ph for very small additions of titrant near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in ph pk a of HInd ph at equivalence point 81

Acid-Base Indicators 82

Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) 2012 Pearson Education, Inc.

Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2 ] where the equilibrium constant, K sp, is called the solubility product. 2012 Pearson Education, Inc.

Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s) nm m+ (aq) + mx n (aq) The solubility product would be K sp = [M m+ ] n [X n ] m For example, the dissociation reaction for PbCl 2 is PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) And its equilibrium constant is K sp = [Pb 2+ ][Cl ] 2 85

Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction M n X m (s) nm m+ (aq) + mx n (aq) 87

Example: Calculate the molar solubility of PbCl 2 in pure water at 25 C (K sp = 1.17 x 10-5 ) write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K sp = [Pb 2+ ][Cl ] 2 [Pb 2+ ] [Cl ] Initial 0 0 Change +S +2S Equilibrium S 2S 88

Example: Calculate the molar solubility of PbCl 2 in pure water at 25 C substitute into the K sp expression find the value of K sp from Table 16.2, plug into the equation, and solve for S K sp = [Pb 2+ ][Cl ] 2 K sp = (S)(2S) 2 [Pb 2+ ] [Cl ] Initial 0 0 Change +S +2S Equilibrium S 2S 89

Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 2 M 90

Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 2 M write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) K sp = [Pb 2+ ][Br ] 2 [Pb 2+ ] [Br ] initial 0 0 change +(1.05 x 10 2 ) +2(1.05 x 10 2 ) equilibrium (1.05 x 10 2 ) (2.10 x 10 2 ) 91

Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 2 M substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br ] 2 K sp = (1.05 x 10 2 )(2.10 x 10 2 ) 2 [Pb 2+ ] [Br ] initial 0 0 change +(1.05 x 10 2 ) +2(1.05 x 10 2 ) equilibrium (1.05 x 10 2 ) (2.10 x 10 2 ) 92

Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/l) or 100 ml (g/ml) of solution, or in mol/l (M). 2012 Pearson Education, Inc.

Factors Affecting Solubility The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) addition of Cl shifts the equilibrium to the left 2012 Pearson Education, Inc.

Example: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25 C (K sp = 1.46 x 10-10 ) write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid CaF 2 (s) Ca 2+ (aq) + 2 F (aq) K sp = [Ca 2+ ][F ] 2 [Ca 2+ ] [F ] initial 0 0.100 change +S +2S equilibrium S 0.100 + 2S 95

Example: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25 C substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S Tro: Chemistry: A Molecular Approach, 2/e K sp = [Ca 2+ ][F ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2 [Ca 2+ ] [F ] initial 0 0.100 change +S +2S equilibrium S 0.100 + 2S 96

Practice Determine the concentration of Ag + ions in seawater that has a [Cl ] of 0.55 M K sp of AgCl = 1.77 x 10 10 Tro: Chemistry: A Molecular Approach, 2/e 97

Practice Determine the concentration of Ag + ions in seawater that has a [Cl ] of 0.55 M write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid AgCl(s) Ag + (aq) + Cl (aq) K sp = [Ag + ][Cl ] [Ag + ] [Cl ] initial 0 0.55 change +S +S equilibrium S 0.55 + S 98

Practice Determine the concentration of Ag + ions in seawater that has a [Cl ] of 0.55 M (K sp =1.77 x10-10 ) substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S K sp = [Ag + ][Cl ] K sp = (S)(0.55 + S) K sp = (S)(0.55) [Ag + ] [Cl ] Initial 0 0.55 Change +S +S Equilibrium S 0.55 + S 99

Factors Affecting Solubility ph If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. 2012 Pearson Education, Inc.

Factors Affecting Solubility Complex Ions The formation of these complex ions increases the solubility of these salts. 2012 Pearson Education, Inc.

Factors Affecting Solubility Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al 3+, Zn 2+, and Sn 2+. 2012 Pearson Education, Inc.

Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur Q = K sp, the solution is saturated, no precipitation Q < K sp, the solution is unsaturated, no precipitation Q > K sp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > K sp will not precipitate unless disturbed these are called supersaturated solutions 103

precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added 104

Selective Precipitation A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different 105

Will a Precipitate Form? In a solution, If Q = K sp, the system is at equilibrium and the solution is saturated, no precipitation. If Q < K sp, more solid can dissolve until Q = K sp, no precipitation. If Q > K sp, the salt will precipitate until Q = K sp, solution will precipitate. 2012 Pearson Education, Inc.

Example: Will a precipitate form when we mix Pb(NO 3 ) 2(aq) with NaBr (aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any insoluble product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Pb(NO 3 ) 2(aq) + 2 NaBr (aq) PbBr 2(s) + 2 NaNO 3(aq) Pb(NO 3 ) 2 = 0.0150 M Pb 2+ = 0.0150 M, NO 3 = 2(0.0150 M) NaBr = 0.0350 M Na + = 0.0350 M, Br = 0.0350 M K sp of PbBr 2 = 4.67 x 10 6 PbBr 2(s) Pb 2+ (aq) + 2 Br (aq) Q < K sp, so no precipitation 107

Practice Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? K sp of Ca(OH) 2 = 4.68 x 10 6 Tro: Chemistry: A Molecular Approach, 2/e 108

Practice Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any insoluble product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Ca(NO 3 ) 2(aq) + 2 NaOH (aq) Ca(OH) 2(s) + 2 NaNO 3(aq) Ca(NO 3 ) 2 = 0.0175 M Ca 2+ = 0.0175 M, NO 3 = 2(0.0175 M) NaOH = 0.0175 M Na + = 0.0175 M, OH = 0.0175 M K sp of Ca(OH) 2 = 4.68 x 10 6 Ca(OH) 2(s) Ca 2+ (aq) + 2 OH (aq) Q > K sp, so precipitation 109

Example: What is the minimum [OH ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? Mg(OH) 2(s) Mg 2+ (aq) + 2 OH (aq) precipitating may just occur when Q = K sp 110

Practice What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq)? K sp of Ca(OH) 2 = 4.68 x 10 6 111

Practice What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq)? Ca(NO 3 ) 2(aq) + 2 NaOH (aq) Ca(OH) 2(s) + 2 NaNO 3(aq) Ca(OH) 2(s) Ca 2+ (aq) + 2 OH (aq) precipitating may just occur when Q = K sp [Ca(NO 3 ) 2 ] = [Ca 2+ ] = 0.0153 M 112

Example: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? Ca(OH) 2(s) Ca 2+ (aq) + 2 OH (aq) precipitating may just occur when Q = K sp 113

Example: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH ] = 1.9 x 10 6 M precipitating Ca 2+ begins when [OH ] = 2.06 x 10 2 M Mg(OH) 2(s) Mg 2+ (aq) + 2 OH (aq) when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 10 M 114

Practice A solution is made by mixing Pb(NO 3 ) 2(aq) with AgNO 3(aq) so both compounds have a concentration of 0.0010 M. NaCl (s) is added to precipitate out both AgCl (s) and PbCl 2(aq). What is the [Ag + ] concentration when the Pb 2+ just begins to precipitate? 115

Practice What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? AgNO 3(aq) + NaCl (aq) AgCl (s) + NaNO 3(aq) AgCl (s) Ag + (aq) + Cl (aq) precipitating may just occur when Q = K sp 116

Practice What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? Pb(NO 3 ) 2(aq) + 2 NaCl (aq) PbCl 2(s) + 2 NaNO 3(aq) PbCl 2(s) Pb 2+ (aq) + 2 Cl (aq) precipitating may just occur when Q = K sp 117

Practice What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate precipitating Ag + begins when [Cl ] = 1.77 x 10 7 M precipitating Pb 2+ begins when [Cl ] = 1.08 x 10 1 M AgCl (s) Ag + (aq) + Cl (aq) when Pb 2+ just begins to precipitate out, the [Ag + ] has dropped from 0.0010 M to 1.6 x 10 9 M 118