CALCULATING K - R. M. FABICON's BLOG

CALCULATING K - R. M. FABICON's BLOG

LECTURE 1: ACIDS AND BASES GENERAL PRINCIPLES OF CATALYSIS DEFINITIONS OF AN ACID Theory: Acid= When

Arrhenius increases H+ 1880s Brnsted proton donor

1923 Lowry proton donor 1923 Lewis

electron-pair acceptor 1923 Who Svante August Arrhenius (February 19, 1859 October 2, 1927) Swedish chemist; Nobel Prize in Chemistry, 1903

* Arrhenius equation (activation energy) * Greenhouse effect Johannes Nicolaus Brnsted (February 22, 1879-December 17, 1947) Danish physical chemist Thomas Martin Lowry (October 26, 1874November 2, 1936) English organic chemist

Gilbert Newton Lewis (October 23, 1875-March 23, 1946) American physical chemist ARRHENIUS DEFINITIONS Arrhenius acids and bases Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions (protons,

H+). Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. BRNSTEDLOWRY DEFINITION BrnstedLowry: must have both 1. an Acid: Proton donor

and 2. a Base: Proton acceptor Brnsted-Lowry acids and bases are always paired. The Brnsted-Lowry acid donates a proton, while the Brnsted-Lowry base accepts it.

Which is the acid and which is the base in each of these rxns? A BrnstedLowry acid must have a removable (acidic) proton. HCl, H2O, H2SO4 A BrnstedLowry base must have a pair of nonbonding electrons. NH3, H2O If it can be either

...it is amphiprotic. HCO3 HSO4 H2O

What Happens When an Acid Dissolves in Water? Water acts as a Brnsted Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

Movies CONJUGATE ACIDS AND BASES From the Latin word conjugare, meaning to join together. Reactions between acids and bases always yield their conjugate bases and acids. ACID AND BASE STRENGTH Strong acids are completely dissociated in water.

Their conjugate bases are quite weak. Weak acids only dissociate partially in water. Their conjugate bases are weak bases. ACID BASE STRENGTH Substances with negligible acidity

do not dissociate in water. Their conjugate bases are exceedingly strong. ACID BASE STRENGTH In any acid-base reaction, the equilibrium favors the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) H3O+(aq) + Cl(aq) H2O is a much stronger base than Cl, so the equilibrium

lies so far to the right K is not measured (K>>1). ACID BASE STRENGTH Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1). The stronger base wins the proton. HC2H3O2(aq) + H2O

H3O+(aq) + C2H3O2(aq) AUTOIONIZATION OF WATER As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This process is called autoionization.

EQUILIBRIUM CONSTANT FOR WATER The equilibrium expression for this process is Kc = [H3O+] [OH] This special equilibrium constant is referred to as the ion-product constant for water, Kw. At 25C, Kw = 1.0 10-14 pH pH is defined as the negative base-10 logarithm of the

hydronium ion concentration. pH = log [H3O+] pH In pure water, Kw = [H3O+] [OH] = 1.0 10-14 Because in pure water [H3O+] = [OH-], [H3O+] = (1.0 10-14)1/2 = 1.0 10-7 pH

Therefore, in pure water, pH = log [H3O+] = log (1.0 10-7) = 7.00 An acid has a higher [H3O+] than pure water, so its pH is <7 A base has a lower [H3O+] than pure water, so its pH is >7. pH These are the pH values for several common

substances. Other p Scales The p in pH tells us to take the negative log of the quantity (in this case, hydronium ions). Some similar examples are pOH log [OH-] pKw log Kw Watch This!

Because [H3O+] [OH] = Kw = 1.0 10-14, we know that log [H3O+] + log [OH] = log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00 If you know one, you know them all: [H+] [OH-]

pH pOH How Do We Measure pH? Litmus paper Red paper turns blue above ~pH = 8 Blue paper turns red below ~pH = 5 An indicator

Compound that changes color in solution. How Do We Measure pH? pH meters measure the voltage in the solution Strong Acids You will recall that the seven

strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H3O+] = [acid]. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal

(NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH)2, Sr(OH)2, and Ba(OH)2). Again, these substances dissociate completely in aqueous solution. [OH-] = [hydroxide added]. Dissociation Constants For a generalized acid dissociation, the equilibrium expression is

This equilibrium constant is called the aciddissociation constant, Ka. Dissociation Constants The greater the value of Ka, the stronger the acid. Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature.

We know that Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature. To calculate Ka, we need all equilibrium concentrations. We can find [H3O+], which is the same as [HCOO], from the pH.

Calculating Ka from the pH pH = log [H3O+] 2.38 = log [H3O+] 10-2.38 = 10log [H3O+] = [H3O+] 4.2 10-3 = [H3O+] = [HCOO] Calculating Ka from pH In table form: [HCOOH], M

[H3O+], M [HCOO], M Initially 0.10 0

0 Change 4.2 10-3 +4.2 10-3 +4.2 10-3

0.10 4.2 10-3 = 0.0958 = 0.10 4.2 10-3 4.2 10 - 3 At Equilibrium Calculating Ka from pH

Ka = [4.2 10-3] [4.2 10-3] [0.10] = 1.8 10-4 Calculating Percent Ionization In the example:

[A-]eq = [H3O+]eq = 4.2 10-3 M [A-]eq + [HCOOH]eq = [HCOOH]initial = 0.10 M Calculating Percent Ionization Percent Ionization = 4.2 10-3 100 0.10

= 4.2% Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25C. Ka for acetic acid at 25C is 1.8 10-5. Is acetic acid more or less ionized than formic acid (Ka=1.8 x 10-4)?

Calculating pH from Ka The equilibrium constant expression is: Calculating pH from Ka Use the ICE table: Initial Change Equilibrium

[C2H3O2], M [H3O+], M [C2H3O2], M 0.30 0

0 x +x +x 0.30 x

x x Calculating pH from Ka Use the ICE table: Initial Change Equilibrium

[C2H3O2], M [H3O+], M [C2H3O2], M 0.30 0

0 x +x +x 0.30 x

x x Simplify: how big is x relative to 0.30? Calculating pH from Ka Use the ICE table:

Initial Change Equilibrium [C2H3O2], M [H3O+], M [C2H3O2], M

0.30 0 0 x +x

+x 0.30 x 0.30 x x Simplify: how big is x relative to 0.30?

Calculating pH from Ka Now, (1.8 10-5) (0.30) = x2 5.4 10-6 = x2 2.3 10-3 = x Check: is approximation ok? CALCULATING Ka from pH The pH of a 0.01M hypochlorous acid (HClO) is 4.76.

Calculate its Ka. Polyprotic Acids Have more than one acidic proton. If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. Weak Bases Bases react with water to produce hydroxide ion.

Weak Bases The equilibrium constant expression for this reaction is where Kb is the base-dissociation constant. Weak Bases Kb can be used to find [OH] and, through it, pH.

pH of Basic Solutions What is the pH of a 0.15 M solution of NH3? Kb = [NH4+] [OH] [NH3] = 1.8 10-5

pH of Basic Solutions Tabulate the data. [NH3], M Initial Equilibrium 0.15 0.15 - x 0.15 Simplify: how big is x relative to 0.15?

[NH4+], M [OH], M 0 0 x

x pH of Basic Solutions 1.8 10-5 = (x)2 (0.15) (1.8 10-5) (0.15) = x2

2.7 10-6 = x2 1.6 10-3 = x2 Check: is approximation ok? pH of Basic Solutions Therefore, [OH] = 1.6 10-3 M pOH = log (1.6 10-3) pOH = 2.80 pH = 14.00 2.80

pH = 11.20 SAMPLE PROBLEM A 0.01M solution of caffeine, a weak organic base, has a pH of 11.3. Calculate its dissociation constant. Ka and dissociation constant of a conjugate base HCN CN-

+ + H2O H2O KaKb = Kw CNHCN

+ + H3O+ OH- SALT HYDROLYSIS Salts of strong acids and bases Salts of strong base and a weak acid

Salt of strong acid and a weak base. Ka and Kb are linked: Combined reaction = ? Ka and Kb are linked: Combined reaction = ?

Ka and Kb Ka and Kb are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other. SAMPLE PROBLEM Calculate the pH of 0.10 M NH4Cl solution. Kb(NH3)=1.8 x 10-5

Calculate the % hydrolysis of a 0.36M CH3COONa. Ka=1.75 x 10-5 ACID BASE TITRATIONS A. Strong Acid Strong Base Titration. NaOH + HCl H2O + NaCl B. Weak acid with a strong Base. Polyprotic acids

H3PO4 + H2O H2PO4- + H3O+ H2PO4- = 7.5 x 10 -3 + H2O

HPO42- + H3O+ HPO42- = 6.2 x 10-8 H2O

PO43- + H3O+ + = 4.8 x 10-13

Polyprotic Acids Have more than one acidic proton. If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. SAMPLE PROBLEM What is the pH of 0.025 M H2S solution? K1= 5.7 x 10-8 K2 = 1.2 x 10-15 SAMPLE PROBLEM

What is the pH of 0.012 M Na2CO3 solution? K1= 4.2 x 10-7 K2 = 4.8 x 10-11 SAMPLE PROBLEM A 50 ml of 0.05M formic acid solution (Ka = 1.77 x 10-4) is titrated with 0.05 M NaOH solution. What is the pH at equivalence point? PRACTICE EXERCISES 1. Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution?

(b) What is the acid-dissociation constant, Ka, for niacin? 2. What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? What percentage of the bases are ionized? 3. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Ka for HF is 6.8 x10-4. Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH and the conjugate acid:

X(aq) + H2O(l) HX(aq) + OH(aq) Reactions of Cations with Water Cations with acidic protons (like NH4+) lower the pH of a solution by releasing H+. Most metal cations (like Al3+) that are

hydrated in solution also lower the pH of the solution; they act by associating with H2O and making it release H+. Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and

the water more acidic. Greater charge and smaller size make a cation more acidic. Effect of Cations and Anions 1. An anion that is the conjugate base of a strong acid will not affect the pH. 2. An anion that is the conjugate base of a weak acid will increase the pH. 3. A cation that is the conjugate acid of a

weak base will decrease the pH. Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the pH. 5. Other metal ions will cause a decrease in pH. 6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base,

the affect on pH depends on the Ka and Kb values. What effect on pH? Why? An anion that is the conjugate base of a strong acid does not affect pH.

= very weak base An anion that is the conjugate base of a weak acid increases pH. = strong base A cation that is the conjugate acid of a weak base decreases pH.

= strong acid Cations of the strong Arrhenius bases (Na+, Ca2+) do not affect pH. = very weak acid Other metal ions cause a decrease in pH. Weak acid + weak base

(not really acidic at all) = moderate bases (cations) Depends on Ka and Kb Factors Affecting Acid Strength

The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. Acidity increases from left to right across a row and from top to bottom down a group. Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y

is, the more acidic the acid. Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens. Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids. A compound with no Hs can be a Lewis acid. Lewis Bases Lewis bases are defined as electron-pair donors.

Anything that is a BrnstedLowry base is also a Lewis base. (BL bases also have a lone pair.) Lewis bases can interact with things other than protons. The Common-Ion Effect Consider a solution of acetic acid: HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) If acetate ion is added to the solution, Le Chtelier says

the equilibrium will shift to the left. 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. 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.

Ka for HF is 6.8 104. Ka = [H3O+] [F] [HF] = 6.8 10-4 The Common-Ion Effect

HF(aq) + H2O(l) H3O+(aq) + F(aq) Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M. Initially Change At Equilibrium

[HF], M [H3O+], M [F], M 0.20 0.10

0 x 0.20 x 0.20 +x 0.10 + x 0.10 +x x

The Common-Ion Effect (0.10) (x) (0.20) 6.8 10 = 4 (0.20) (6.8 104) (0.10)

=x 1.4 103 = x The Common-Ion Effect Therefore, [F] = x = 1.4 103 [H3O+] = 0.10 + x = 0.10 + 1.4 103 = 0.10 M So, pH = log (0.10) pH = 1.00

Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added. 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. Buffers If acid is added, the F reacts to form HF and water. Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H2O H3O+ + A Ka = [H3O+] [A] [HA]

Buffer Calculations Rearranging slightly, this becomes Ka = [H3O+] [A] [HA] Taking the negative log of both side, we get

log Ka = pKa pH log [H3O ] + log + base [A ]

[HA] acid Buffer Calculations So pKa = pH log [base]

[acid] Rearranging, this becomes pH = pKa + log [base] [acid] This is the HendersonHasselbalch equation.

HendersonHasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 104. HendersonHasselbalch Equation pH = pKa + log [base]

[acid] pH = log (1.4 10 ) + log 4 pH = 3.85 + (0.08) pH = 3.77 (0.10) (0.12)

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 pKa close to the desired pH. 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. 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 HendersonHasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. a) Calculate the pH of this solution after 0.020 mol of NaOH is added. Ka = 1.8 x 10-5 b) calculate the pH after 0.020 mole HCl is added. Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2

pH = pKa = log (1.8 105) = 4.74 Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH(aq) C2H3O2(aq) + H2O(l) HC2H3O2 C 2H 3O 2 OH

Before reaction 0.300 mol 0.300 mol 0.020 mol After reaction

0.280 mol 0.320 mol 0.000 mol Calculating pH Changes in Buffers Now use the HendersonHasselbalch equation to calculate the new pH:

pH = 4.74 + log pH = 4.74 + 0.06 pH = 4.80 (0.320) (0. 200) Titration A known concentration of

base (or acid) is slowly added to a solution of acid (or base). 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.

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. Titration of a Strong Acid with a Strong Base Just before and after the

equivalence point, the pH increases rapidly. 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. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off. 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. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations. Titration of a Weak Acid with a Strong Base

At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time. 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. Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation. Solubility Equilibria Solubility Rules

Salts are generally more soluble in HOT water (Gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li3PO4, Na2SO4 etc... Ammonium salts are very soluble in water. NH4Br, (NH4)2CO3 etc Salts containing the nitrate ion, NO3-, are very soluble in water. Most salts of Cl-, Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+. soluble salts: FeCl2, AlBr3, MgI2 etc...

insoluble salts: AgCl, PbBr2 etc... Dissolving a salt... A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The dissolving of a salt is an example of equilibrium. The cations and anions are attracted to each other in the salt. They are also attracted to the water

molecules. The water molecules will start to pull out some of the ions from the salt crystal. At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are pulled back in to the salt.

(precipitation) Eventually the rate of dissociation is equal to the rate of precipitation. The solution is now saturated. It has reached equilibrium. Solubility Equilibrium: Dissociation = Precipitation Na+ and Cl - ions surrounded by

water molecules In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. NaCl Crystal

Dissolving NaCl in water The rate at which the salt is dissolving into solution equals the rate of precipitation. Dissolving silver sulfate, Ag2SO4, in water When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists:

Ag2SO4 (s) 2 Ag+ (aq) + SO42- (aq) Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag+]2[SO42-] Notice that the Ag2SO4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a solubility product constant - Ksp.

Writing solubility product expressions... For each salt below, write a balanced equation showing its dissociation in water. Then write the Ksp expression for the salt. Iron (III) hydroxide, Fe(OH)3 Nickel sulfide, NiS Silver chromate, Ag2CrO4 Zinc carbonate, ZnCO3 Calcium fluoride, CaF2

Some Ksp Values Note: These are experimentally determined, and may be slightly different on a different Ksp table. Calculating Ksp of Silver Chromate A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x 10-4 M. What is the Ksp for Ag2CrO4?

Ag2CrO4 (s) 2 Ag+ (aq) + CrO42- (aq) ---- 1.3 x 10-4 M ---- Calculating the Ksp of silver sulfate The solubility of silver sulfate is 0.014 mol/L. This means that 0.0144 mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution.

Calculate the value of the equilibrium constant, Ksp for this salt. Ag2SO4 (s) 2 Ag+ (aq) + SO42- (aq) --- --- Calculating solubility, given Ksp The Ksp of NiCO3 is 1.4 x 10-7 at 25C. Calculate its molar solubility.

NiCO3 (s) Ni2+ (aq) + CO32- (aq) --- --- Other ways to express solubility... We just saw that the solubility of nickel (II) carbonate is 3.7 x 10 -4 mol/ L. What mass of NiCO3 is needed to prepare 500 mL of saturated solution? 3.7 x 10 4 mol NiCO3

0.500 L 118.72 g x x 0.022 g 1L 1 mol NiCO3 0.022 g of NiCO3 will dissolve to make 500 mL solution.

Calculate the solubility of MgF2 in water. What mass will dissolve in 2.0 L of water? Ksp = 7.4 x 10-11 MgF2 (s) Mg2+ (aq) + 2 F- (aq) Solubility and pH Calculate the pH of a saturated solution of silver hydroxide, AgOH. Ksp = 2.0 x 10-8. AgOH (s) Ag+ (aq) + OH- (aq)

The Common Ion Effect on Solubility The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to the solubility if we dissolve the MgF2 in a solution of NaF, instead of pure water? Calculate the solubility of MgF2 in a solution of 0.080 M NaF. MgF2 (s) Mg2+ (aq) + 2 F- (aq) Explaining the Common Ion Effect

The presence of a common ion in a solution will lower the solubility of a salt. LeChateliers Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower! Ksp and Solubility Generally, it is fair to say that salts with very small solubility product constants (Ksp) are only sparingly soluble in water.

When comparing the solubilities of two salts, however, you can sometimes simply compare the relative sizes of their Ksp values. This works if the salts have the same number of ions! For example CuI has Ksp = 5.0 x 10-12 and CaSO4 has Ksp = 6.1 x 10-5. Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can say that calcium sulfate is more soluble. But be careful... Salt

Solubility (mol/L) Ksp -45 -23 CuS

8.5 x 10 Ag2S 1.6 x 10-49 3.4 x 10-17 Bi2S3

-73 -15 1.1 x 10 9.2 x 10 1.0 x 10

Mixing Solutions - Will a Precipitate Form? If 15 mL of 0.024-M lead nitrate is mixed with 30 mL of 0.030-M potassium chromate - will a precipitate form? Pb(NO3)2 (aq) + K2CrO4 (aq) PbCrO4 (s) + 2 KNO3 (aq) Pb(NO3)2 (aq) + K2CrO4 (aq) PbCrO4 (s) + 2 KNO3 (aq) Step 1: Is a sparingly soluble salt formed? We can see that a double replacement reaction can occur and produce PbCrO4. Since this salt has a very small Ksp, it may precipitate from the mixture. The solubility equilibrium is:

PbCrO4 (s) Pb2+ (aq) + CrO42- (aq) Ksp = 2 x 10-16 = [Pb2+][CrO42-] If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS. This will happen only if Qsp > Ksp in our mixture. Step 2: Find the concentrations of the ions that form the sparingly soluble salt. Since we are mixing two solutions in this example, the concentrations of the Pb2+ and CrO42- will be diluted. We have to do a dilution

calculation! Dilution: C1V1 = C2V2 [Pb2+] = [CrO42-] = C1V1 (0.024 M)(15 mL) 0.0080 M Pb 2 V2 (45 mL)

C1V1 (0.030 M)(20 mL) 0.020 M CrO 4 2V2 (45 mL) Step 3: Calculate Qsp for the mixture. Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M) Qsp = 1.6 x 10-4 Step 4: Compare Qsp to Ksp. Since Qsp >> Ksp, a precipitate will form when

solutions are mixed! Note: If Qsp = Ksp, the mixture is saturated If Qsp < Ksp, the solution is unsaturated Either way, no ppte will form! the two A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is

mixed with 0.200 L of 0.060 M NaF? Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water: BaSO4(s) Ba2+(aq) + SO42(aq) Solubility Products

The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42] where the equilibrium constant, Ksp, is called the solubility product. Solubility Products Ksp 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).

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. BaSO4(s) Ba2+(aq) + SO42(aq)

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.

Factors Affecting Solubility Complex Ions Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. Factors Affecting Solubility Complex Ions The formation of these complex ions increases the

solubility of these salts. 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

Al3+, Zn2+, and Sn2+. Will a Precipitate Form? In a solution, If Q = Ksp, the system is at equilibrium and the solution is saturated. If Q < Ksp, more solid will dissolve until Q = Ksp. If Q > Ksp, the salt will precipitate until Q = Ksp. Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

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