PRECIPITATION REACTIONS & Ksp

KEY IDEAS· Precipitation reactions are a result of a double replacement reaction where

one product is an insoluble salt.· Using equilibrium constants, we can predict if a precipitate (often written

"ppt") will form or not.

SOLUBILITY OF SALTS

AB (aq) + CD (aq) à AD(aq) + CB(s)

CaCl2(aq) + Na2CO3(aq) à CaCO3(s) + 2 NaCl(aq)

net ionic: Ca2+ + CO32- à CaCO3

REVIEW THE SOLUBILITY RULES!!!!!

THE SOLUBILITY PRODUCT

With any salt, even for salts you consider insoluble, some small amount dissolves:

AgBr(s) à Ag+ + Br-1

so we can write the Keq for this reaction: Keq = [Ag+][ Br-] (remember that solids and pure liquids DO NOT show up in equilibrium expressions!). Since this Keq expression deals with the amount of an insoluble salt that will dissolve (become soluble) and we multiply the two terms together, we refer to this as the solubility product constant, Ksp.

An experiment shows silver's concentration is 5.7 x 10-7 M @ 25OC, so:

Keq = Ksp = [5.7 x 10-7 ][5.7 x 10-7 ] = 3.2 x 10-13

Your textbook has tables of Ksp values.

DETERMINING Ksp FROM EXPERIMENTAL MEASUREMENTS

In practice, Ksp values are determined by careful laboratory measurements using various spectroscopic methods. Also, you will have to use STOICHIOMETRY in the problem to find all of the values. We can still use the ICE table (and in very rare cases the IRP table), but usually skip to the "E" part and almost always throw away the "- x" part!!

Example:

Lead (II) chloride dissolves to a slight extent in water according to the equation :

PbCl2(s) à Pb+2(aq) + 2 Cl-1(aq)

Calculate the Ksp if the lead ion concentration has been found to be

1.62 x 10-2 M.

PbCl2 à Pb+2 + 2 Cl-1

I ### 0 0

C - x + x + 2 x

E ### - x 1.62 x 10-2

If lead's concentration is 1.62 x 10-2 then chloride's concentration is twice that

amount: 2 x (1.62 x 10-2) = 3.24 x 10-2. So. . .

Ksp = [Pb2+][Cl-1]2 = (1.62 x 10-2)(3.24 x 10-2)2 = 1.70 x 10-5

Usually we skip right to the "E" line, and most people use an "S" instead of the "x" when using ICE tables with Ksp problems.

THE TWO MOST COMMON MISTAKES THAT STUDENTS MAKE ARE:

· FORGETTING TO PUT THE EXPONENT ON THE TERM IN THE Ksp EQUATION.

· FORGETTING TO DO THE STOICHIOMETRY TO GET THE CORRECT VALUES.

ESTIMATING SALT SOLUBILITY FROM Ksp

These type of problems are just an extension of standard equilibrium problems. Set up the ICE table, remove the "- S" if possible (5% rule -or- 100 times rule) and solve for the "S" value. Once you have the S value, convert to the units that you need for the problem.

Example: The Ksp for CaCO3 is 3.8 x 10-9 @ 25OC. Calculate the solubility of

calcium carbonate in pure water in (a) moles per liter & (b) grams per liter:

The relative solubilities can be deduced by comparing values of Ksp BUT BE CAREFUL! These comparisons can only be made for salts having the same ION:ION ratio.

Please don't forget solubility changes with temperature! Some substances become less soluble in cold water while some become more soluble! However, it is more common that substances will become more soluble in hot water.

PRECIPITATION OF INSOLUBLE SALTS

Metal-bearing ores often contain metals in the form of insoluble salts -- sometimes the ores contain several different metals. The problem is how to get one of the metals out and leave the others behind.

· Dissolve the metal salts to obtain the metal ion.· Concentrate the solution in some manner.· Select an ion to selectively precipitate the metal as an insoluble salt.

· To do this select a Ksp value for your salt that will allow it to precipitate and not any other salts (sometimes easier said than done).

To help with this, the Reaction Quotient, Q, is very important. Types of solutions:· Unsaturated

· not enough solute is dissolved· can have more solute dissolved· the maximum amount of solute is not dissolved

· Saturated· under normal conditions, the maximum amount of solute is dissolved · no more solute will dissolve

· Supersaturated· temporarily has more than the maximum amount of solute

Ksp AND THE REACTION QUOTIENT, Q

1. Q < Ksp, the system is not at equilibrium. (unsaturated)2. Q = Ksp, the system is at equilibrium. (saturated)3. Q > Ksp, the system is not at equilibrium. (supersaturated)

Precipitates form when the solution is supersaturated!!!

Think about the equation and the expression for Q. Try to see how Q will have to move to become equal to Ksp. With some knowledge of the reaction quotient, we can decide (a) whether a ppt will form AND (b) what concentrations of ions are required to begin the ppt of an insoluble salt.

SOLUBILITY AND THE COMMON ION EFFECT

Experimentation shows that the solubility of any salt is always less in the presence of a common ion.

LeChatelier's Principle

Be reasonable and use approximations when you can!!

SOLUBILITY, ION SEPARATIONS, AND QUALITATIVE ANALYSIS

1. Introduce you to some basic chemistry of various ions2. Illustrate how the principles of chemistry equilibria can be applied.

Problem: Separate the following metal ions: silver lead cadmium nickel

1. From solubility rules, lead and silver chloride will ppt. so add dilute HCl.2. Nickel and cadmium will stay in solution.3. Separate PbCl2 and AgCl by filtration

· Lead chloride will dissolve in HOT water.· Filter while HOT and those two will be separated.

4. Cadmium and nickel are more subtle. · Use their Ksp's with sulfide ion. Which metal precipitates first?

REMEMBER THE PROBLEM WITH THE LABELS FALLEN OFF.

SIMULTANEOUS EQUILIBRIA

When another reagent is added to a saturated solution of an insoluble salt, the salt may be converted to another, even less soluble salt.lead chloride and lead chromate:

PbCl2 , white solid Ksp = 1.7 x 10-5

PbCrO4 , yellow solid Ksp = 1.8 x 10-14

If you add a few drops of potassium chromate to lead chloride and shake the mixture, the lead chloride is changed to lead chromate.

PbCl2(s) + CrO4-2 (aq) à PbCrO4 (s) + 2 Cl-1(aq) Knet = 9.4 x 108 =

K1K2

simply the sum of the two reactions:

PbCl2 (s) à Pb+2 (aq) + 2 Cl-1(aq) K1 = Ksp = 1.7 x 10-5

Pb+2 (aq) + CrO4-2 (aq) à PbCrO4 (s) K2 = 1 = 1

Ksp 1.8 x 10-14

This large constant indicates the reaction should proceed from the left to the right.

SOLUBILITY AND COMPLEX IONS

When a metal ion (a Lewis acid) reacts with a Lewis base, a complex ion can form. Such ions are often insoluble in water, their formation can be used to dissolve otherwise insoluble salts. If sufficient aqueous ammonia is added to

silver chloride, the latter can be dissolved in the form of [Ag(NH3)2]+

AgCl (s) à Ag+ (aq) + Cl-1 (aq) Ksp = 1.8 x 10-10

Ag+ (aq) + 2 NH3 (aq) à [Ag(NH3)2]+ (aq) Kform = 1.6 x 107

SUM:

K = Ksp x Kform = 2.0 x 10-3 = {[Ag(NH3)2+}[Cl-1]

[NH3]2

The equilibrium constant for dissolving silver chloride in ammonia is not large; however, if the concentration of ammonia is sufficiently high, the complex ion and chloride ion must also be high, and silver chloride will dissolve.

To get the expression write the overall equation and then write the Keq expression.

ACID-BASE AND PPT. EQUILIBRIA OF PRACTICAL SIGNIFICANCE SOLUBILITY OF SALTS IN WATER AND ACIDS

a) the solubility of PbS in water:

PbS (s) à Pb+2 + S-2 Ksp = 8.4 x 10-28

b) the hydrolysis of the S-2 ion in water

S-2 + H2O à HS-1 +OH-1 Kb = 0.077

Overall process:

PbS + H2O à Pb+2 + HS-1 + OH-1 Ktotal = Ksp x Kb = 6.5 x 10-29

It may not seem like much but it can increase the environmental lead concentration by a factor of about 10,000 over the solubility of PbS calculated from simply Ksp!

Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by the Ksp. This means salts of sulfate, phosphate, acetate, carbonate, and cyanide, as well as sulfide can be affected. If a strong acid is added to water-insoluble salts such as ZnS or CaCO3, then hydroxide ions from the anion hydrolysis are removed by the

formation of water. This shifts the anion hydrolysis further to the right; the weak acid is formed and the salt dissolves.

Carbonates and many metal sulfides along with metal hydroxides are generally soluble in strong acids.

The only exceptions are sulfides of mercury, copper, cadmium and a few others.

Insoluble inorganic salts containing anions derived from weak acids tend to be soluble in solutions of strong acids.

Salts are not soluble in strong acids if the anion is the conjugate base of a strong acid!!

PREPARING SALTS BY ACID-BASE AND PPT. REACTIONSWant to prepare Nickel (II) nitrate:

· Convert NiCl2 into an insoluble material using an anion that is the conjugate

base of a weak acid. Carbonate is a good choice. · Filter, wash, add a strong acid HNO3 (just until the ppt. dissolves) since you

want to make a nitrate salt. Water and carbon dioxide will also be formed.· Evaporate to dryness and you get a hexahydrate of nickel (II) nitrate.

To make an inorganic salt, it is best to begin with a hydroxide soluble or insoluble or a carbonate and add a strong acid containing the desired anion. All

reactions have a strong driving force: the formation of water or a gas such as carbon dioxide that is poorly soluble in water.

SEPARATION OF IONS BY SELECTIVE PRECIPITATION

Chloride vs. phosphate ions in solution:

· Add silver nitrate to ppt. silver chloride or silver phosphate. Silver chloride will not dissolve with the addition of nitric acid but silver phosphate will.

Metal Sulfides:

· Carefully add sulfide ion, mercury, copper and cadmium will ppt. leaving zinc, cobalt and manganese in solutions. Lowering the pH will also cause sulfide ppts. Raising pH causes the sulfide ppts. to dissolve.

AP QUESTION 1985 A

At 25°C the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.6 x 10-7. The solubility

product constant for strontium fluoride, SrF2, at the same temperature is 7.9 x 10-10.

(a) What is the molar solubility of SrSO4 in pure water at 25°C?

(b) What is the molar solubility of SrF2 in pure water at 25°C?

(c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 litre of a well-stirred solution containing

0.020 mole F- and 0.10 mole SO42- at 25°C. You may assume that the added Sr(NO3)2 solution

does not materially affect the total volume of the system.)1. Which salt precipitates first?

2. What is the concentration of strontium ion, Sr2+, in the solution when the firstprecipitate begins to form?

(d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that stage,

what percent of the anion of the first precipitate remains in solution?

AP QUESTION 1971Solve the following problem

AgBr(s) à Ag+(aq) + Br-(aq) Ksp = 3.3 x 10-13

Ag+(aq) + 2 NH3(aq) à Ag(NH3)2+(aq) K = 1.7 x 10+7

(a) How many grams of silver bromide, AgBr, can be dissolved in 50 milliliters of water?(b) How many grams of silver bromide can be dissolved in 50 milliliters of 10 molar ammonia solution?

AP QUESTION 1972 D(a) How many moles of Ba(IO3)2 is contained in 1.0 liter of a saturated solution of this salt

at 25OC. Ksp of Ba(IO3)2 = 6.5 x 10-10

(b) When 0.100 liter of 0.060 molar Ba(NO3)2 and 0.150 liter of 0.12 molar KIO3 are mixed

at 25OC, how many milligrams of barium ion remains in each milliliter of the solution? Assume that the volumes are additive.

AP QUESTION 1977 DThe solubility of Zn(OH)2 is not the same in the following solutions as it is in pure water. In each case state

whether the solubility is greater or less than that in water and briefly account for the change in solubility.

(a) 1–molar HCl(b) 1–molar Zn(NO3)2(c) 1–molar NaOH(d) 1–molar NH3

AP QUESTION 1980 DAccount for the differences in solubility described in each of the following experimental observations:

(a) BaCO3, BaSO3, and BaSO4 are only slightly soluble in water, but the first two dissolve in HCl

solution whereas BaSO4 does not.

(b) CuS cannot be dissolved by warm dilute HCl but it does dissolve in warm dilute HNO3.

(c) AgCl, Hg2Cl2 and PbCl2 are only slightly soluble in water, but AgCl does dissolve in ammonia

solution whereas the other two do not.(d) Fe(OH)3 and Al(OH)3 are only slightly soluble in water, but Al(OH)3 dissolves in concentrated

NaOH whereas Fe(OH)3 does not.

AP QUESTION 1990 A

The solubility of iron(II) hydroxide, Fe(OH)2, is 1.43 x 10-3 grams per litre at 25°C.

(a) Write a balanced equation for the solubility equilibrium.

(b) Write the expression for the solubility product constant, Ksp, and calculate its value.

(c) Calculate the pH of a saturated solution of Fe(OH)2 at 25°C.

(d) A 50.0 millilitre sample of 3.00 x 10-3 molar FeSO4 solution is added to 50.0 millilitres of

4.00 x 10-6 molar NaOH solution. Does a precipitate of Fe(OH)2 form? Explain and show

calculations to support your answer.

15-1) How many milliliters of 0.100 M NaOH are required to react completely with 0.976 g of the weak, monoprotic acid benzoic acid (C6H5COOH)? What is the pH of the solution after the reaction?

15-2) Assume you have a 0.30 M solution of formic acid (HCOOH) and add enough sodium formate (NaHCOO) to make the solution 0.10 M in the salt. Calculate the pH of the formic acid solution before and after adding sodium formate.

15-3) Calculate the pH change that occurs when 1.0 mL of 1.0 M HCl is added to (a) 1.0 L of pure water and (b) 1.0 L if acetic acid/sodium acetate buffer with [CH3COOH] = 0.700 M and [CH3COO--] = 0.600 M.

15-4) Suppose you wish to prepare a buffer solution to maintain the pH at 4.30. A list of possible acids (and their conjugate bases) is

ACID CONJUGATE BASE Ka

HSO4 SO42-- 1.2 x 10-2

CH3COOH CH3COO-- 1.8 x 10-5

HCN CN-- 4.0 x 10-10

Which combination should be selected and what should the ratio of acid to conjugate base be?

15-5) Assume you titrate 20.0 mL of 0.11 M NH3 with 0.10 M HCl.(a) What is the pH of the NH3 solution before the titration begins?(b) What is the pH of the equivalence point?(c) What is the pH at the midpoint of the titration?(d) Which indicator in figure 15.8 on page 756 would be best to detect the equivalence point?(e) Calculate the pH of the solution after adding 5.00, 11.0, 15.0, 20.0, 22.0, and 25.0 mL of the acid. Combine this information with that from (a) through (c) and plot the titration curve.

15-6) How many grams of ammonium chloride, NH4Cl, would have to be added to exactly500 mL of 0.10 M NH3 solution to have a pH of 9.0?

15-7) If a buffer solution is made of 12.2 g of benzoic acid (C6H5COOH) and 7.20 g of sodium benzoate (C6H5COONa) in exactly 250 mL of solution, what is the pH of the buffer? If the solution is diluted to exactly 500 mL with pure water, what is the new pH of the solution?

15-8) Which of the following combinations would be the best to buffer the pH at approximately 9?(a) acetic acid/sodium acetate, [CH3COOH/Na(CH3COO)](b) HCl/NaCl(c) ammonia/ammonium chloride, [NH3/NH4Cl]

15-9) Estimate the solubility of silver cyanide in (a) moles per liter and (b) grams per liter in pure water at 25 degrees Celsius.

AgCN(s) Ag+(aq) + CN—(aq)

15-10) Zn(OH)2 is a relatively insoluble base. A saturated solution has a pH of 8.65. Calculate the Ksp for Zn(OH)2.

15-11) If you mix 50 mL of 0.0012 M BaCl2 with 25 mL of 1.0 x 10-6 M of H2SO4, will a precipitate of BaSO4 form? The equilibrium process involved is

BaSO4(s) Ba2+(aq) + SO42—(aq).

15-12) The cations Co2+, Mn2+, and Ni2+ can all be precipitated as very insoluble sulfides, CoS, MnS, and NiS. If you add a soluble sulfide containing salt (say (NH4)2S) to a solution containing all these ions, each with a concentration of 0.10 M, in what order are the sulfides precipitated?