Aqueous Ionic Equilibrium

Buffers buffers are solutions that resist changes in pH when an acid or base is added they act by neutralizing the added acid or base but just like everything else, there is a limit to what they can do, eventually the pH changes many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion

Demo of a pH Buffer Aqueous Solution pH Δ pH Bottled Water (30 mL) 5.6 0.7 Bottled Water + 1 drop of 0.1M HNO3 4.9 Deionized Water (30 mL) 6.0 2.2 Deionized Water + 1 drop of 0.1M HNO3 3.8 pH Buffer 7.8 0.0 pH Buffer + 1 drop of 0.1M HNO3 What should the pH be when we add 1 drop of 0.1M HNO3 to 30mL of deionized water ? Assume 1 drop = 0.05mL When this 1 drop is dissolved in 30 mL of H2O

Why are pH buffers important? Life on Earth is water-based Human 48-75% water Plants as high as 95% water H+ and OH- are chemically and structurally reactive in cells The functioning of the cell is very pH dependent Blood plasma pH = 7.4 pH < 6.9 fatal (acidosis) pH > 7.9 fatal (alkalosis) Fish die if the pH of the water goes below 5 or above 9

Making an Acid Buffer

How Acid Buffers Work HA(aq) + H2O(l) A−(aq) + H3O+(aq) buffers work by applying Le Châtelier’s Principle to weak acid equilibrium Adding OH- to the buffer: the OH- reacts with the H3O+ to make 2H2O(l) so it removes H3O+ so drives the equilibrium to the right effectively doing the following Effectively 𝐻𝐴 (𝑎𝑞) + 𝑂𝐻 (𝑎𝑞) − → 𝐴 𝑎𝑞 − + 𝐻 2 𝑂 (𝑙) By Le Châtelier it drives the equilibrium to the right causing more HA to make A- and maintaining [H3O+] Adding H3O+ to the buffer: By Le Châtelier it drives the equilibrium to the left (because we are adding more product) causing A- to react with the H3O+ to make more HA maintaining [H3O+] Effectively 𝐴 𝑎𝑞 − + 𝐻 3 𝑂 (𝑎𝑞) + ⇢ 𝐻𝐴 (𝑎𝑞) + 𝐻 2 𝑂 (𝑙)

How Buffers Work H2O new HA HA HA A− A− H3O+ + Added H3O+

How Buffers Work H2O new A− A− HA A− HA H3O+ + Added HO−

Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left this causes the pH to be higher than the pH of the acid solution lowering the H3O+ ion concentration

Common Ion Effect

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 HC2H3O2 + H2O C2H3O2- + H3O+ Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 [HA] [A-] [H3O+] initial 0.100 ≈ 0 change equilibrium

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 HC2H3O2 + H2O C2H3O2- + H3O+ 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 [HA] [A-] [H3O+] initial 0.100 change equilibrium -x +x +x x 0.100 -x 0.100 + x

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium x 0.100 -x 0.100 +x

the approximation is valid What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium x x = 1.8 x 10-5 the approximation is valid

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 substitute x into the equilibrium concentration definitions and solve [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 1.8E-5 0.100 -x 0.100 + x x x = 1.8 x 10-5

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 substitute [H3O+] into the formula for pH and solve [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 1.8E-5

What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 1.8E-5 the values match

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Your turn!

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O F- + H3O+ Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change equilibrium

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O F- + H3O+ 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 [HA] [A-] [H3O+] initial 0.14 0.071 change equilibrium -x +x +x 0.14 -x 0.071 + x x

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.012 0.100 x 0.14 x 0.071 +x

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? [HA] [A2-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium x Ka for HF = 7.0 x 10-4 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init x = 1.4 x 10-3 the approximation is valid

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? [HA] [A2-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.072 1.4E-3 substitute x into the equilibrium concentration definitions and solve 0.14 x 0.071 + x x x = 1.4 x 10-3

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? substitute [H3O+] into the formula for pH and solve [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.072 1.4E-3

What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0 What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.072 1.4E-3 the values are close enough

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

Deriving the Henderson-Hasselbalch Equation

What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0 What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2?

What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0 What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2? HC7H5O2 + H2O C7H5O2- + H3O+ 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

What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0 What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2? HC7H5O2 + H2O C7H5O2- + H3O+ [HA] [A-] [H3O+] initial 0.050 0.150 ≈ 0 change -x +x equilibrium 0.050-x 0.150-x x

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA]o and [A-]o into the equation assuming they are equilibrium values), makes it an approximation For this reason 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: the initial concentrations of acid and salt are not very dilute the Ka is fairly small for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka

How Much Does the pH of a Buffer Change When an Acid or Base Is Added? though buffers do resist change in pH when acid or base are added to them, their pH does change calculating the new pH after adding acid or base requires breaking the problem into 2 parts 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 𝐴 𝑎𝑞 − + 𝐻 3 𝑂 (𝑎𝑞) + ⇢ 𝐻𝐴 (𝑎𝑞) + 𝐻 2 𝑂 (𝑙) added base reacts with the HA to make more A− 𝐻𝐴 (𝑎𝑞) + 𝑂𝐻 (𝑎𝑞) − ⇢ 𝐴 𝑎𝑞 − + 𝐻 2 𝑂 (𝑙) an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0 What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2- + H2O 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 it with A−. Construct a stoichiometry table for the reaction HA A- OH− mols Before 0.100 mols added - 0.010 mols After

What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0 What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2- + H2O Fill in the table – tracking the changes in the number of moles for each component HA A- OH− mols Before 0.100 ≈ 0 mols added - 0.010 mols After 0.090 0.110

What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0 What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O C2H3O2- + H3O+ Substitute into the Henderson-Hasselbalch Equation Check the “x is small” approximation pKa for HC2H3O2 = 4.745

Compare the effect on pH of adding 0 Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water? Pure Water Buffer HC2H3O2 + H2O C2H3O2- + H3O+ pKa for HC2H3O2 = 4.745

Basic Buffers B:(aq) + H2O(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− H2O(l) + NH3 (aq) NH4+(aq) + OH−(aq)

What is the pH of a buffer that is 0. 50 M NH3 (pKb = 4. 75) and 0 What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? NH3 + H2O NH4+ + OH− find the pKa of the conjugate acid (NH4+) from the given Kb 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

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 (2) the absolute concentrations of acid and base

Buffering Effectiveness and Buffering Range a buffer will be most effective when the [base]:[acid] = 1 equal concentrations of acid and base a buffer will be most effective when the [acid] and the [base] are large The Buffer is effective when 0.1 < [base]/[acid] < 10 𝑝𝐻 𝑚𝑖𝑛 = 𝑝𝐾 𝑎 +log⁡(0.1), 𝑝𝐻 𝑚𝑎𝑥 = 𝑝𝐾 𝑎 +log⁡(10), The buffering range is therefore 𝑝𝐻 𝑟𝑎𝑛𝑔𝑒 = 𝑝𝐾 𝑎 ±1

Effect of Relative Amounts of Acid and Conjugate Base a buffer is most effective with equal concentrations of acid and base pH change after adding 0.010M NaOH Buffer 2 0.18 mol HA & 0.020 mol A- Initial pH = 4.05 Buffer 1 0.100 mol HA & 0.100 mol A- Initial pH = 5.00 pKa (HA) = 5.00 HA + OH− A- + H2O after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25 HA A- OH− mols Before 0.100 mols added - 0.010 mols After 0.090 0.110 ≈ 0 HA A- OH− mols Before 0.18 0.020 mols added - 0.010 mols After 0.17 0.030 ≈ 0

Effect of Absolute Concentrations of Acid and Conjugate Base a buffer is most effective when the concentrations of acid and base are largest pH change after adding 0.010M NaOH Buffer 1 0.50 mol HA & 0.50 mol A- Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A- Initial pH = 5.00 pKa (HA) = 5.00 HA + OH− A- + H2O after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18 HA A- OH− mols Before 0.50 0.500 mols added - 0.010 mols After 0.49 0.51 ≈ 0 HA A- OH− mols Before 0.050 mols added - 0.010 mols After 0.040 0.060 ≈ 0

Example from Homework You have two acetic acid buffers (pKa = 4.475) and to these buffers you add 0.010 mole of NaOH Buffer 1 0.50 mol HC2H3O2 & 0.50 mol C2H3O2 - Buffer 2 0.050 mol HC2H3O2 & 0.050 mol C2H3O2 – Calculate the pH before adding the NaOH to each buffer  Calculate the pH after the NaOH to each buffer Calculate the percentage change in the pH of each buffer Which is the best buffer?

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, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54

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, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.

What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25? Formic Acid, HCHO2, pKa = 3.74 to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2

Example from Homework You have four weak acids HA Chlorous Acid, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54   Which would be the best acid for making a buffer at pH = 7.0 ? What ratio will you need [A-]/[HA] to be to obtain the correct pH?

Buffering Capacity buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness 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]

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

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 H3O+ = moles of OH−, the titration has reached its equivalence point

Titration Curve a plot of pH vs. amount of added titrant the inflection point of the curve is the equivalence point of the titration prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH 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

Monitoring pH During a Titration

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 [H3O+] using a probe that specifically measures just H3O+ 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

Titration Curves for Strong and Weak Acids with a Strong Base

Titration of a Polyprotic Acid if Ka1 >> Ka2, there will be two equivalence points in the titration the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH pKa2 pKa1

Titration Curve of a Weak Base with a Strong Acid

Monitoring a Titration with an Indicator for most titrations, the titration curve shows a very large change in pH for very small additions of base 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 pKa of HInd ≈ pH at equivalence point pKa = 4.2 Methyl Orange pKa = 9.7 Phenolphthalein

HInd(aq) + H2O(l) Ind-(aq) + H3O+(aq) Indicators many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution HInd(aq) + H2O(l) Ind-(aq) + H3O+(aq) the color of the solution depends on the relative concentrations of 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

Titration

Indicators: Phenolphthalein Used to detect pH = 8 good for titrations of weak acids with a strong base

Indicators: Methyl Red Used to detect pH = 5 good for titrations of weak bases with a strong acid

Acid-Base Indicators

Titrating Weak Acid with a Strong Base: Calculating the pH using Ka the initial pH is that of the weak acid solution calculate like a weak acid equilibrium problem before the equivalence point, the solution becomes a buffer calculate mol HAinit and mol A−init using reaction stoichiometry calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init half-neutralization pH = pKa

Titrating Weak Acid with a Strong Base: Calculating the pH using Ka 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 [A−]init = mol A−/total liters calculate like a weak base equilibrium problem beyond equivalence point, the OH is in excess [OH−] = mol MOH xs/total liters [H3O+][OH−]=1 x 10-14

Example from homework A 10.0 mL sample of 0.300 M HNO2 is titrated with 0.200 M KOH. Calculate the pH before any base is added Calculate the volume of base needed to reach the equivalence point Calculate the volume of base needed to reach the half equivalence point Calculate the pH at the half equivalence point  Calculate the pH at the equivalence point Calculate the pH when 20 mL of base is added

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, Ksp for an ionic solid MnXm, the dissociation reaction is: MnXm(s) nMm+(aq) + mXn−(aq) the solubility product would be Ksp = [Mm+]n[Xn−]m for example, the dissociation reaction for PbCl2 is PbCl2(s) Pb2+(aq) + 2 Cl−(aq) and its equilibrium constant is Ksp = [Pb2+][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 MnXm(s) nMm+(aq) + mXn−(aq)

Calculate the molar solubility of PbCl2 in pure water at 25°C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid PbCl2(s) Pb2+(aq) + 2 Cl−(aq) Ksp = [Pb2+][Cl−]2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S

Calculate the molar solubility of PbCl2 in pure water at 25°C Substitute into the Ksp expression Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Pb2+][Cl−]2 Ksp = (S)(2S)2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S

Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

PbBr2(s) Pb2+(aq) + 2 Br−(aq) Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid PbBr2(s) Pb2+(aq) + 2 Br−(aq) Ksp = [Pb2+][Br−]2 [Pb2+] [Br−] Initial Change +(1.05 x 10-2) +2(1.05 x 10-2) Equilibrium (1.05 x 10-2) (2.10 x 10-2)

Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10-2)(2.10 x 10-2)2 Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M Substitute into the Ksp expression plug into the equation and solve Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10-2)(2.10 x 10-2)2 [Pb2+] [Br−] Initial Change +(1.05 x 10-2) +2(1.05 x 10-2) Equilibrium (1.05 x 10-2) (2.10 x 10-2)

Ksp and Relative Solubility molar solubility is related to Ksp but you cannot always compare solubilities of compounds by comparing their Ksps in order to compare Ksps, 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 PbCl2 decreases the solubility of PbCl2 PbCl2(s) Pb2+(aq) + 2 Cl−(aq) addition of Cl− shifts the equilibrium to the left

Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid CaF2(s) Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S

Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C Substitute into the Ksp expression assume S is small Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Ca2+][F−]2 Ksp = (S)(0.100 + 2S)2 Ksp = (S)(0.100)2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S

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) Mn+(aq) + nOH−(aq) for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32− (aq) HCO3− (aq) + H2O(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 Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation Q < Ksp, the solution is unsaturated, no precipitation Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions

precipitation occurs if Q > Ksp 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 Ksp values are significantly different

What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059M]) from seawater assuming Ksp=2.06x10-13)? precipitating may just occur when Q = Ksp

What is the [Mg2+] when Ca2+ (with [0 What is the [Mg2+] when Ca2+ (with [0.011M]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M

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

Ag+(aq) + 2 H2O(l)  [Ag(H2O)2+](aq) Complex Ion Formation transition metals tend to be good Lewis acids they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag+(aq) + 2 H2O(l)  [Ag(H2O)2+](aq) ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2+ the attached ions or molecules are called ligands e.g., H2O

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(H2O)2+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) + 2 H2O(l) generally H2O is not included, since its complex ion is always present in aqueous solution Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)

Ag+(aq) + 2 NH3(aq) Ag(NH3)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 NH3(aq) Ag(NH3)2+(aq) the equilibrium constant for the formation reaction is called the formation constant, Kf

Formation Constants

Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq) 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Write the formation reaction and Kf expression. Look up Kf value Determine the concentration of ions in the diluted solutions Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)

Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq) 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq) [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq) 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq) Substitute in and solve for x confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid

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) Ksp = 1.77 x 10-10 Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Kf = 1.7 x 107 adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+

Solubility of Amphoteric Metal Hydroxides many metal hydroxides are insoluble eg Fe(OH)3, Al(OH)3, Co(OH)2 all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH− Fe(OH)3(s) [Fe(OH)2+](aq) + OH-(aq) H3O+(aq)+OH-(aq)  2H2O(l) Amphoteric metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+

Al3+ Al3+ is hydrated in water to form an acidic solution Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq) addition of OH− drives the equilibrium to the right and continues to remove H from the molecules Al(H2O)5(OH)2+(aq) + OH−(aq) Al(H2O)4(OH)2+(aq) + H2O (l) Al(H2O)4(OH)2+(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O (l)