Chapter 16: Aqueous Ionic Equilibrium

Buffers Solutions that resist changes in pH when an acid or base is added Act by neutralizing acid or base that is added to the buffered solution There is a limit to buffering capacity  eventually the pH changes Standard Buffer Solutions = – solution of a weak acid + solution of soluble salt containing the conjugate base anion

Making an Acid Buffer

How Acid Buffers Work: Addition of Base HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Buffers = Application of Le Châtelier’s Principle to weak acid equilibrium Buffer solutions contain significant amounts of the weak acid molecules, HA HA react with added A - to neutralize it HA(aq) + OH − (aq) → A − (aq) + H 2 O(l) – you can also think of the H 3 O + combining with the OH − to make H 2 O; the H 3 O + is then replaced by the shifting equilibrium

H2OH2O HA How Buffers Work HA  + H3O+H3O+ A−A− Added HO − new A − A−A−

How Acid Buffers Work: Addition of Acid HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) The buffer solution also contains significant amounts of the conjugate base anion, A − These ions combine with added acid to make more HA H + (aq) + A − (aq) → HA(aq) After the equilibrium shifts, the concentration of H 3 O + is kept constant

H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− A−A− Added H 3 O + new HA

Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding NaA, a salt containing the A − – A − = conjugate base of the acid (the common ion) – This shifts the equilibrium to the left The new pH is higher than the pH of the acid solution – Lower H 3 O + ion concentration

Common Ion Effect

Practice − What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? 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 HF + H 2 O  F  + H 3 O + [HA][A − ][H 3 O + ] initial ≈ 0 change equilibrium

[HA][A − ][H 3 O + ] initial change equilibrium Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? 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 +x+x+x+x xx 0.14  x x x HF + H 2 O  F  + H 3 O +

pK a for HF = 3.15K a for HF = 7.0 x 10 −4 Practice – What is the pH of a buffer that is 0.14 M and M KF? determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium x 0.14  x x

Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HF = 7.0 x 10 −4 the approximation is valid x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium x

Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? substitute x into the equilibrium concentration definitions and solve x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium E x x 0.14  x

Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium E−3

Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values are close enough [HA][A − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium E−3 K a for HF = 7.0 x 10 −4

Henderson-Hasselbalch Equation Calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson- Hasselbalch Equation The equation calculates the pH of a buffer from the pK a and initial concentrations of the weak acid and salt of the conjugate base – as long as the “x is small” approximation is valid

Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and M KF? find the pK a from the given K a 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 HF + H 2 O  F  + H 3 O +

When should you 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

How Much Does the pH of a Buffer Change When an Acid or Base Is Added? While buffers resist change in pH when acid or base is added to them, their pH does change Calculating the new pH after adding acid or base requires breaking the problem into two parts 1.a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A − to make more HA added base reacts with the HA to make more A − 2.an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ]

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before mols added– – mols after

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? fill in the table – tracking the changes in the number of moles for each component F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before mols added– – mols after

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction enter the initial number of moles for each F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before mols change mols after new molarity

F−F− H3O+H3O+ HF mols before mols change mols after new molarity Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities F − + H 3 O +  HF + H 2 O −

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? write the reaction for the acid with water construct an ICE table. assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation HF + H 2 O  F  + H 3 O + [HF][F − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium x

Basic Buffers B: (aq) + H 2 O (l)  H:B + (aq) + OH − (aq) Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl − H 2 O (l) + NH 3 (aq)  NH 4 + (aq) + OH − (aq)

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 Henderson-Hasselbalch Equation for Basic Buffers

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

Buffering Effectiveness A good buffer should be able to neutralize moderate amounts of added acid or base However, there is a limit to how much can be added before the pH changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the pH range the buffer can be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base

HAA−A− OH − mols before mols added ── mols after ≈ 0 Effect of Relative Amounts of Acid and Conjugate Base Buffer mol HA & mol A − Initial pH = 5.00 Buffer mol HA & mol A − Initial pH = 4.05 pK a (HA) = 5.00 after adding mol NaOH pH = 5.09 HA + OH −  A  + H 2 O HAA−A− OH − mols before mols added ── mols after ≈ 0 after adding mol NaOH pH = 4.25 A buffer is most effective with equal concentrations of acid and base

HAA−A− OH − mols before mols added ── mols after ≈ 0 HAA−A− OH − mols before mols added ── mols after ≈ 0 Effect of Absolute Concentrations of Acid and Conjugate Base Buffer mol HA & 0.50 mol A − Initial pH = 5.00 Buffer mol HA & mol A − Initial pH = 5.00 pK a (HA) = 5.00 after adding mol NaOH pH = 5.02 HA + OH −  A  + H 2 O after adding mol NaOH pH = 5.18 A buffer is most effective when the concentrations of acid and base are largest

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

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

Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pHHighest pH Therefore, the effective pH range of a buffer is pK a ± 1 When choosing an acid to make a buffer, choose one whose is pK a closest to the pH of the buffer

Practice – What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 to make a buffer with pH 3.75, you would use times as much NaC 7 H 5 O 2 as HC 7 H 5 O 2

Buffering Capacity Buffering capacity = 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]

Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration 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

Titration

Titration Curve: pH vs Amount of Titrant Added Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH The inflection point of the curve is the equivalence point of the titration – The pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH

Titration Curve: Unknown Strong Base Added to Strong Acid

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

HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new HClNaClNaOH mols before2.50E-305.0E-4 mols change mols end molarity, new HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new 5.0 x 10 −4 mole NaOH added Titration of 25 mL of M HCl with M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10 −3 Before equivalence point added 5.0 mL NaOH 42

Titration of 25 mL of M HCl with 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

HClNaClNaOH mols before2.50E-302.5E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new x 10 −3 mole NaOH added Titration of 25 mL of M HCl with M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10 −3 At equivalence point added 25.0 mL NaOH

HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new HClNaClNaOH mols before2.50E-303.0E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new Titration of 25 mL of M HCl with M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10 −3 After equivalence point added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added

added 5.0 mL NaOH mol HCl pH = 1.18 added 10.0 mL NaOH mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 added 30.0 mL NaOH mol NaOH pH = added 40.0 mL NaOH mol NaOH pH = added 50.0 mL NaOH mol NaOH pH = added 35.0 mL NaOH mol NaOH pH = added 15.0 mL NaOH mol HCl pH = 1.60 added 20.0 mL NaOH mol HCl pH = 1.95 Adding M NaOH to M HCl 25.0 mL M HCl mol HCl pH = 1.00

HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new 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 = L x 0.25 mol/L=1.25 x 10 −2 Before equivalence point added 10.0 mL NaOH

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 = L x 0.25 mol/L=1.25 x 10 −2 At equivalence point: moles of NaOH = 1.25 x 10 −2

HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E E-2−1.25E-2 mols end 01.25E molarity, new HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E E-2−1.25E-2 mols end 01.25E molarity, new Practice – Calculate the pH of the solution that results when 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 = L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added mL NaOH

HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E E-2−1.25E-2 mols end 01.25E molarity, new HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E E-2−1.25E-2 mols end 01.25E molarity, new Practice – Calculate the pH of the solution that results when 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 = L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added mL NaOH

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

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

Titration of 25 mL of M HCHO 2 with M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial pH [HCHO 2 ][CHO 2 − ][H 3 O + ] initial ≈ 0 change −x−x+x+x+x+x equilibrium − xxx K a = 1.8 x 10 −4

Titration of 25 mL of M HCHO 2 with M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = L x mol/L = 2.50 x 10 −3 Before equivalence added 5.0 mL NaOH HAA−A− OH − mols before2.50E-300 mols added ––5.0E-4 mols after 2.00E-35.0E-4≈ 0

HAA−A− OH − mols before2.50E-300 mols added ––2.50E-3 mols after 02.50E-3≈ 0 [HCHO 2 ][CHO 2 − ][OH − ] initial ≈ 0 change +x+x−x−x+x+x equilibrium x5.00E-2-xx Titration of 25 mL of M HCHO 2 with M NaOH added 25.0 mL NaOH CHO 2 − (aq) + H 2 O (l)  HCHO 2(aq) + OH − (aq) K b = 5.6 x 10 −11 [OH − ] = 1.7 x 10 −6 M HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = L x mol/L = 2.50 x 10 −3 At equivalence

HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new HAA−A− NaOH mols before2.50E-303.0E-3 mols change mols end molarity, new Titration of 25 mL of M HCHO 2 with M NaOH 56 added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial mol of HCHO 2 = L x mol/L = 2.50 x 10 −3 After equivalence

added 35.0 mL NaOH mol NaOH xs pH = initial HCHO 2 solution mol HCHO 2 pH = 2.37 added 5.0 mL NaOH mol HCHO 2 pH = 3.14 added 10.0 mL NaOH mol HCHO 2 pH = 3.56 added 25.0 mL NaOH equivalence point mol CHO 2 − [CHO 2 − ] init = M [OH − ] eq = 1.7 x 10 −6 pH = 8.23 added 30.0 mL NaOH mol NaOH xs pH = added 20.0 mL NaOH mol HCHO 2 pH = 4.34 added 15.0 mL NaOH mol HCHO 2 pH = 3.92 added 12.5 mL NaOH mol HCHO 2 pH = 3.74 = pK a half-neutralization Adding NaOH to HCHO 2 added 40.0 mL NaOH mol NaOH xs pH = added 50.0 mL NaOH mol NaOH xs pH = 12.52

Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution – calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 Before the equivalence point, the solution becomes a buffer – calculate mol HA init and mol A − init using reaction stoichiometry – calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a

Titrating Weak Acid with a Strong Base At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established – mol A − = original mole HA calculate the volume of added base as you did in Example 4.8 – [A − ] init = mol A − /total liters – calculate like a weak base equilibrium problem e.g., Beyond equivalence point, the OH is in excess – [OH − ] = mol MOH xs/total liters – [H 3 O + ][OH − ]=1 x 10 −14

Titration Curve of a Weak Base with a Strong Acid

Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5

Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5

Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = L x mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new NH 4 + (aq) + H 2 O (l)  NH 4 + (aq) + H 2 O (l) pK b = 4.75 pK a = − 4.75 = 9.25

Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = L x mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new

Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = L x mol/L = 2.50 x 10 −3 At equivalence mol NH 3 = mol HCl = 2.50 x 10 −3 added 25.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new

Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) at equivalence [NH 4 Cl] = M [NH 3 ][NH 4 + ][H 3 O + ] initial ≈ 0 change +x+x−x−x+x+x equilibrium x0.050−xx NH 4 + (aq) + H 2 O (l)  NH 3(aq) + H 3 O + (aq)

Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = L x mol/L = 2.50 x 10 −3 After equivalence: after adding 30.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new when you mix a strong acid, HCl, with a weak acid, NH 4 +, you only need to consider the strong acid

Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration – the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of M H 2 SO 3 with M NaOH

Monitoring pH During a Titration The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ] – using a probe that specifically measures just H 3 O + The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

Monitoring pH During a Titration

Indicators Many dyes change color depending on the pH of the solution These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  :HInd – when Ind  :HInd ≈ 1, the color will be mix of the colors of Ind  and HInd – when Ind  :HInd > 10, the color will be mix of the colors of Ind  – when Ind  :HInd < 0.1, the color will be mix of the colors of HInd

Phenolphthalein

Methyl Red

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

Acid-Base Indicators

Solubility Equilibria All ionic compounds dissolve in water to some degree – however, many compounds have such low solubility in water that we classify them as insoluble We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

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

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)

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 [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) PbBr 2 (s)  Pb 2+ (aq) + 2 Br − (aq) K sp = [Pb 2+ ][Br − ] 2

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 − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 )

K sp and Relative Solubility Molar solubility is related to K sp But you cannot always compare solubilities of compounds by comparing their K sp s To compare K sp s, the compounds must have the same dissociation stoichiometry

The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) addition of Cl − shifts the equilibrium to the left

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 [Ag + ][Cl − ] initial00.55 change+S+S+S+S equilibriumS S AgCl(s)  Ag + (aq) + Cl − (aq) K sp = [Ag + ][Cl − ]

Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M 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 [Ag + ][Cl − ] Initial00.55 Change+S+S+S+S EquilibriumS S K sp = [Ag + ][Cl − ] K sp = (S)( S) K sp = (S)(0.55)

The Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide – and the lower the pH, the higher the solubility – higher pH = increased [OH − ] M(OH) n (s)  M n+ (aq) + nOH − (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l)

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

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

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

Practice – Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 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) K sp of Ca(OH) 2 = 4.68 x 10 –6 Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Ca(NO 3 ) 2 = M  Ca 2+ = M, NO 3 − = 2( M) NaOH = M  Na + = M, OH − = M Q > K sp, so precipitation

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

Practice – A solution is made by mixing Pb(NO 3 ) 2(aq) with AgNO 3(aq) so both compounds have a concentration of 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?

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

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

Practice – What is the [Ag + ] concentration when the Pb 2+ ( 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 when Pb 2+ just begins to precipitate out, the [Ag + ] has dropped from M to 1.6 x 10 −9 M AgCl (s)  Ag + (aq) + Cl − (aq)

Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme – wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out

Qualitative Analysis

Group 1 Group one cations are Ag +, Pb 2+ and Hg 2 2+ All these cations form compounds with Cl − that are insoluble in water – as long as the concentration is large enough – PbCl 2 may be borderline molar solubility of PbCl 2 = 1.43 x 10 −2 M Precipitated by the addition of HCl

Group 2 Group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+ All these cations form compounds with HS − and S 2− that are insoluble in water at low pH Precipitated by the addition of H 2 S in HCl

Group 3 Group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides All these cations form compounds with S 2− that are insoluble in water at high pH Precipitated by the addition of H 2 S in NaOH

Group 4 Group four cations are Mg 2+, Ca 2+, Ba 2+ All these cations form compounds with PO 4 3− that are insoluble in water at high pH Precipitated by the addition of (NH 4 ) 2 HPO 4

Group 5 Group five cations are Na +, K +, NH 4 + All these cations form compounds that are soluble in water – they do not precipitate Identified by the color of their flame

Complex Ion Formation Transition metals tend to be good Lewis acids They often bond to one or more H 2 O molecules to form a hydrated ion – H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l)  Ag(H 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions – e.g., Ag(H 2 O) 2 + The attached ions or molecules are called ligands – e.g., H 2 O

Complex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) – generally H 2 O is not included, because its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq)

Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f

Practice – What is [HgI 4 2− ] when 125 mL of M KI is reacted with 75 mL of M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq)

Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq)

Practice – What is [HgI 4 2− ] when 125 mL of M KI is reacted with 75 mL of M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Create an ICE table. Because K f is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium. [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) I − is the limiting reagent

Practice – What is [HgI 4 2− ] when 125 mL of M KI is reacted with 75 mL of M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) substitute in and solve for x confirm the “x is small” approximation 2 x 10 −8 << 1.6 x 10 −4, so the approximation is valid Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) [HgI 4 2− ] = 1.6 x 10 −4 [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4

The Effect of Complex Ion Formation on Solubility The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s)  Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 −10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 Adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +

Solubility of Amphoteric Metal Hydroxides Many metal hydroxides are insoluble All metal hydroxides become more soluble in acidic solution – shifting the equilibrium to the right by removing OH − Some metal hydroxides also become more soluble in basic solution – acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric Some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+

Al 3+ Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq)  Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)