Chemical Equilibrium
THE NATURE OF CHEMICAL EQUILIBRIUM
Reversible Reactions In theory, every reaction can continue in two directions, forward and reverse Reversible reaction! chemical reaction in which the products can react to re-form the reactants
Mercury (II) oxide Decomposes when heated 2HgO(s)! 2Hg(l) + O 2 (g) Mercury and oxygen combine to form mercury (II) oxide when heated gently 2Hg(l) + O 2 (g)! 2HgO(s)
Suppose HgO is heated in closed container where mercury and oxygen can t escape Once decomposition begins, mercury and oxygen released can recombine to form HgO again Both reactions happen at same time
Rate of composition will eventually equal rate of decomposition At equilibrium, mercury and oxygen will combine at same rate that mercury (II) oxide decomposes Amounts of mercury(ii) oxide, mercury and oxygen are expected to remain constant as long as conditions continue
At this point, a state of dynamic equilibrium has been reached between two reactions No net change in composition of system
A reversible reaction is in chemical equilibrium when the rate of its forward reaction equals the rate of its reverse reaction and the concentrations of its products and reactants remain unchanged
Equation for chemical equilibrium written using double arrows 2HgO(s) 2Hg(l) + O 2 (g)
Equilibrium, a Dynamic State In some cases, the forward reaction is almost done before the rate of the reverse reaction is high enough for equilibrium
Here, products of forward reaction are favored (at equilibrium, there is higher concentration of products than reactants) HBr(aq) + H 2 O(l) (aq) H 3 O + (aq) + Br
Other cases, reverse reaction is favored More reactants than products H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 (aq) Sometimes forward and reverse happen almost same amount before equilibrium H 2 SO 3 (aq) + H 2 O(l) H 3 O + (aq) + HSO 3 (aq)
The Equilibrium Expression Let s say 2 substances, A and B, react to make products C and D C and D react to make A and B na + mb xc + yd
At first, concentrations of C and D are zero and concentrations of A and B are max Over time, rate of forward reaction decreases as A and B are used up Rate of reverse increases as C and D are formed
When two reaction rates become equal, equilibrium is created Individual concentrations of A, B, C, and D don t change if conditions stay the same
After EQ, concentrations of products and reactants remain constant, so the ratio of their concentrations should also stay the same
Ratio of product [C] x x [D] y to product [A] n x [B] m for this reaction has definite value at given temp Definite value is EQ constant of the reaction and is given letter K
Following equation describes equilibrium constant for theoretical equilibrium system [C] = concentration of C in moles/liter
Concentrations of products in chemical reaction appear in numerator of equilibrium ratio Each concentration to the power of the coefficient These are products of forward reaction
Concentrations of reactants found in denominator (risen to power of coefficient) These are reactants of forward reaction K independent of original concentrations, but dependent on temperature
The Equilibrium Constant Value of K for particular equilibrium system is found experimentally Chemist analyzes equilibrium mixture and determines concentrations of all substances Value of K for given reaction at given temp shows degree to which reactants converted to products
If K = 1, products of concentrations raised to appropriate power in numerator and denominator have the same value Therefore, at equilibrium, there are equal concentrations of reactants and products
If K is low, forward reactions happens before equilibrium, and reactants are favored If K is high, the original reactants are largely converted to products
Only concentrations of substances that can actually change are included in K This means PURE solids and liquids are left out because their concentrations cannot change
Chemical-equilibrium Expression In general, the equilibrium constant, K,! is the ratio of the mathematical product of the concentrations of the products at equilibrium to the mathematical products of the concentrations of reactants
Each concentration raised to power equal to coefficient Chemical-equilibrium expression! the equation for K
The H 2, I 2, HI Equilibrium System Consider reaction between H2 and I2 vapor in sealed flask at elevated temp Rate of reaction followed by observing rate at which the violet color of iodine vapor lessens
Suppose hydrogen gas is present in excess and reaction continues until all iodine is used up Color (from iodine) gradually disappears because the product, HI and hydrogen gas are both colorless
In reality, color fades to constant intensity but does not disappear completely The reaction is reversible HI decomposes to form hydrogen and iodine
Rate of reverse reaction increases as concentration of HI increases Rate of forward reaction decreases accordingly
Concentrations of hydrogen and iodine decrease as they are used up As rates of opposing reactions become equal, equilibrium is established
Constant color indicates that equilibrium exists Net chemical equation: H 2 (g) + I 2 (g) 2HI(g)
H 2 (g) + I 2 (g) 2HI(g) From this chemical equation we can write chemical-equilibrium expression
Sample Problem An equilibrium mixture of N, O, and 2 2 NO gases at 1500 K is determined to consist of 6.4 x 10 3 mol/l of N, 1.7 x 2 10 3 mol/l of O, and 1.1 x 10 5 mol/l of 2 NO. What is the equilibrium constant for the system at this temperature?
1. Analyze Given: [N2] = 6.4 10 3 mol/l [O2] = 1.7 10 3 mol/l [NO] = 1.1 10 5 mol/l Unknown: K
2. Plan The balanced chemical equation is N 2 (g) + O 2 (g) 2NO(g) The chemical equilibrium expression is
= 1.1 x 10-5 3. Compute
Practice Problems At equilibrium a mixture of N 2, H 2, and NH 3 gas at 500 C is determined to consist of 0.602 mol/l of N 2, 0.420 mol/l of H 2, and 0.113 mol/l of NH 3.What is the equilibrium constant for the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) at this temperature? 0.286
The reaction AB 2 C(g) B 2 (g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB 2 C, 0.035 mol of B 2, and 0.059 mol of AC were detected. What is the equilibrium constant at this temperature for this system? 4.9 10 3
At equilibrium a 1.0 L vessel contains 20.0 mol of H 2, 18.0 mol of CO 2, 12.0 mol of H 2 O, and 5.9 mol of CO at 427 C. What is the value of K at this temperature for the following reaction? CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) 0.20
A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600 C. At that temperature, the concentration of SO 2 is found to be 1.50 mol/l, the concentration of O 2 is 1.25 mol/l, and the concentration of SO 3 is 3.50 mol/l. Using the balanced chemical equation, calculate the equilibrium constant for this system. 4.36
Shifting Equilibrium
In chemical equilibrium, forward and reverse reactions happen at same rate Any change alters the rates and disturbs equilibrium
System will try to find a new equilibrium state By shifting equilibrium in desired direction, chemists can improve the yield of product they want
Predicting the Direction of Shift Le Chatelier s principle provides a way of predicting the influence of stress factors on equilibrium systems
Le Chatelier s principle! if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that relieves the stress
Principle true for all dynamic equilibria Changes in pressure, concentration, and temperature demonstrate the application of Le Chatelier s principle to chemical equilibrium
Changes in Pressure Change in pressure only affects equilibrium systems where gases are involved For changes to affect system, the total number of moles of gas on left side must be different from total moles on right side CaCO 3 (s) CaO(s) + CO 2 (g) 1 mol!1 mol + 1 mol
CaCO 3 (s) CaO(s) + CO 2 (g) High pressure favors reverse reaction b/c fewer CO 2 molecules are produced Lower pressure favors increased production of CO 2
Haber process For synthesis of ammonia N 2 (g) + 3H 2 (g) 2NH 3 (g) Consider an increase in pressure which causes increase in concentration of all species System reduces number of molecules (total pressure) by shifting equilibrium to right
N 2 (g) + 3H 2 (g) 2NH 3 (g) For each four molecules of reactants there are two molecules of products By making more NH 3, and using up N 2 and H 2 the system reduces the number of molecules
This decreases pressure New equilibrium pressure higher than before, but not as high as original stress
Ammonia made in Haber process is continuously removed by condensation to liquid This removes most of the product from gas phase where reaction happens
Result: decrease in partial pressure of NH 3 gas Same as decrease in product concentration Shifts equilibrium to the right
Changes in Concentration Increase in concentration of reactant is a stress on equilibrium system Causes increase in collision frequency, and increase in reaction rate Consider the following reaction: A + B C + D
A + B C + D Increase in [A] shift equilibrium to right Both A and B used up faster, more of C and D are formed Equilibrium reestablished with lower [B]
A + B C + D Increase in [B] drives reaction to right Increase in [C] or [D] increases rate of reverse reaction (shift left) Decrease in [C] or [D] equilibrium shifts right
Changes in concentration have no effect on the value of the equilibrium constant This is b/c these changes have effect on numerator and denominator of chemical equilibrium expression So, all concentrations give same value or ratio for equilibrium constant when equilibrium is reestablished
Many chemical reactions involve heterogeneous reaction where reactants and products are in different phases Concentrations of solids and liquids not changed by adding or removing it
This is b/c concentration is densitydependent Density of solids/liquids is constant, regardless of amount present A pure substance in a condensed phase (solid, liquid) can be eliminated from the expression for the equilibrium constant
Concentration of pure solid/ liquid is set to equal 1 in equilibrium expression CaCO 3 (s) CaO(s) + CO 2 (g) Products are solid and gas so expression is
Changes in Temperature Reversible reactions are exothermic in one direction and endothermic in the other Effect of changing temp depends on which direction is exothermic or endothermic
According to Le Chatelier s principle, addition of heat shifts equilibrium so heat is absorbed This favors endothermic rxn Removal of heat favors exothermic rxn
Rise in temp increase rate of any rxn In equilibrium system, rates of 2 rxns not equal So, value of equilibrium constant is affected by temp
Haber process is exothermic N 2 (g) + 3H 2 (g) 2NH 3 (g) + 92 kj High temp not desirable b/c it favors decomposition of NH 3 (endothermic) Low temp forward rxn too slow to be useful (doesn t make enough NH 3 )
Temp used in Haber process is balance between kinetic and eq requirements High enough that eq is created quickly Low enough that eq concentration of NH 3 is large Average temp (500 ) and high pressure (700-1000 atm) makes satisfactory amount of NH 3
Production of colorless dinitrogen tetroxide gas, N 2 O 4, from dark brown NO 2 gas is exothermic All 3 flasks contain same total mass of gas (mixture of NO 2 and N 2 O 4 )
2NO 2 (g) N 2 O 4 (g) Temp of system lowered to 0 This causes eq to shift right (exothermic rxn, makes more heat) to balance for lower temp Allows more colorless N 2 O 4 to form Makes light brown color
2NO 2 (g) N 2 O 4 (g) System is at equilibrium at 25 At eq, system has equal amount of NO 2 and N2O 4 Produces medium brown color
2NO 2 (g) N 2 O 4 (g) Raise temp to 100, equilibrium shifts to left Forward rxn = exothermic Reverse rxn = endothermic (absorbs excess heat) Causes more dark brown NO 2 gas to form
Endothermic Rxn For endothermic, heat shows up on reactant side 556 kj + CaCO 3 (s) CaO(s) + CO 2 (g) Increase in temp causes value of K to increase Eq shifts right
External Stress Predicted Equilibrium Shift Change in pressure Affects gases only. Concentration of reactant(s) increased Concentration of product(s) increased Temperature increased Temperature decreased Equilibrium shifts toward products (to the right). Equilibrium shifts toward reactants (to the left). Equilibrium shifts toward endothermic rxn Equilibrium shifts toward exothermic rxn
Rxns That Go to Completion Some rxns involving compounds formed by interaction of ions in solutions appear to go to completion (ions almost completely removed from solution)
Degree to which reacting ions removed depends on solubility of compound formed and if compound formed is soluble, the degree of ionization
A product that escapes as gas, precipitates as solid, or is only slightly ionized removes most of reacting ions from solution 1. Formation of a gas 2. Formation of precipitate 3. Formation of slightly ionized product
1. Formation of a Gas Unstable substances formed as products of ionic reactions decompose spontaneously Ex. Carbonic acid, H 2 CO 3, acid in carbonated water (Pepsi, Coke, etc.) H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Reaction practically goes to completion b/c CO 2 escapes as gas if container is open to air
2. Formation of Precipitate When solutions of NaCl and AgNO 3 are mixed, white precipitate of AgCl forms Na + (aq) + Cl (aq) + Ag + (aq) + NO 3 (aq)! Na + (aq) + NO 3 (aq) + AgCl(s) If equal amounts of reactants used, only Na+ and NO 3 - ions stay in solution Rxn goes to completion b/c an insoluble product is formed
3. Formation of Slightly Ionized Product Neutralization rxns between H 3 O + ions from aqueous acids and OH - ions from aqueous bases result in formation of H 2 O molecules H 2 O only slightly ionized H 3 O + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)! Na + (aq) + Cl - (aq) + 2H 2 O(l) Get rid of spectator ions H 3 O + (aq) + OH - (aq)! 2H 2 O(l)
H 3 O + (aq) + OH (aq) 2H 2 O(l) b/c water only slightly ionized, it exists almost entirely as covalently bonded molecules So, as long as they are originally in equal amounts, hydronium and hydroxide ions are almost entirely removed from solution
Common-Ion Effect Eq rxn can be pushed in desired direction by applying Le Chatelier s principle Ex. HCl gas bubbled into saturated solution of NaCl HCl is extremely soluble in water, and is almost completely ionized HCl(g) + H 2 O(l)! H 3 O + (aq) + Cl (aq)
HCl(g) + H 2 O(l) H 3 O + (aq) + Cl (aq) Eq for a saturated solution of NaCl is NaCl(s) Na + (aq) + Cl (aq) As HCl dissolves in high enough amount, it increases [Cl - ] This is stress on eq system
System balances, according to Le Chatelier s principle, by combining some Cl- ions with equal number of Na+ ions This causes some solid NaCl to precipitate out Relieves stress of added chlorine
New eq has higher [Cl-] but lower [Na+] BUT product of [Na+] and [Cl-] still have same value as before Common-ion effect! phenomenon, in which the addition of an ion common to two solutes brings about precipitation or reduced ionization
Common-ion effect also seen when one ion species of weak electrolyte is added in excess to solution A 0.1M CH 3 COOH (acetic acid) solution is only about 1.4% ionized to make hydronium ions (H 3 O + ) and acetate ions (CH 3 COO - ) CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq)
CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) Small additions of sodium acetate, NaCH 3 COO, to solution of acetic acid greatly increases acetate-ion concentration Eq shifts to use up some acetate ions More molecules of acetic acid are formed and concentration of H 3 O + reduced
Equilibria of Acids, Bases, and Salts
Ionization Constant of a Weak Acid About 1.4% solute molecules in 0.1M acetic acid solution are ionized at room temp Remaining 98.6% of acetic acid molecules stay unionized So, solution contains 3 species of particles in eq: CH 3 COOH molecules, H 3 O + ions, and acetate ions,ch 3 COO
Eq constant for this system expresses eq ratio of ions to molecules CH 3 COOH + H 2 O H 3 O + + CH 3 COO
At 0.1M concentration, water molecules are much more than the number of acetic acid molecules One can assume the molar concentration of H 2 O molecules stays constant So, b/c both K and H 2 O are constant, the product K[H 2 O] is constant
Left side can be simplified by setting K[H 2 O] = K a Acid-ionization constant! K a K a, like K, is constant for specified temp and has new value for each new temp
K a for weak acid is a small value To determine actual value for K a for acetic acid at specific temp, eq concentrations of H 3 O + ions, CH 3 COO ions, and CH 3 COOH molecules must be known
The ionization of a molecule of CH 3 COOH in water yields one H 3 O + ion and one CH 3 COO ion These concentrations can be found experimentally by measuring the ph of the solution
Ionization for some dilute acetic acid solutions at 25 Molarity % Ionized [H 3 O+] [CH 3 COOH] K a 0.100 1.33 0.00133 0.0987 1.79 x 10-5 0.0500 1.89 0.000945 0.0491 1.82 10 5 0.0100 4.17 0.000417 0.00958 1.81 10 5 0.00500 5.86 0.000293 0.00471 1.82 10 5 0.00100 12.6 0.000126 0.000874 1.82 10 5 Notice Ka is almost identical for every solution molarity shown
At constant temp, increase in [CH 3 COO - ] through addition of NaCH 3 COO disturbs eq Disturbance causes decreases in [H 3 O + ] and increase in [CH 3 COOH] Eventually, eq is reestablished with same value of Ka But higher [CH 3 COOH] and lower [H 3 O + ] than before extra CH 3 COO - was added
Changes in [H 3 O + ] affect ph Decrease in [H 3 O + ] means increase in ph
Buffers Previous solution contains weak acid (CH 3 COOH) and salt of weak acid (NaCH 3 COO) Solution can react with acid or base When small amounts of weak acid or bases added, ph remains nearly constant
Weak acid and common ion, CH 3 COO -, act as buffer against large changes in ph Buffered solution! solution that can resist changes in ph
Let s say a small amount of acid is added to acetic acid-sodium acetate solution Acetate ions react with most of added hydronium ions to form nonionized acetic acid molecules CH 3 COO (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l) From added acid [H 3 O + ] and ph remain amost unchaged
Now let s say a small amount of base is added to original solution The OH - ions of base react with and remove H 3 O + ions to form nonionized water molecules Acetic acid molecules then ionize and restore eq concentration of hydronium ions CH COOH(aq) + H O(l) H O + (aq) + 3 2 3 CH COO (aq) From reaction of OH- 3 of base with H3O+
A solution of a weak base containing a salt of the base behaves in a similar way [OH - ] and ph of solution stay constant with small additions of acids or bases
Suppose a base is added to aqueous NH 3 solution that also contains NH Cl 4 Ammonium ions donate proton to added OH - ions to form nonionized water molecules From added base NH 4+ (aq) + OH (aq) NH 3 (aq) + H 2 O(l)
If small amount of acid added to solution instead, OH - from solution accept protons from added H 3 O + from acid to form water NH 3 in solution then ionize and restore eq concentration of H 3 O + and ph of solution NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq)
Ionization Constant of Water The self-ionization of water is an eq reaction H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Eq is created with very low [H 3 O + ] and [OH - ] K w = [H 3 O + ][OH ] = 1.0 10 14
Hydrolysis of Salts Salts are formed during neutralization reaction between Br nsted acid and Br nsted base When a salt dissolves in water, it makes cations of the base and anions of the acid
Solution might be expected to be neutral Some salts (NaCl and KNO 3 ) are neutral When Na 2 CO 3 dissolves in water, solution turns red litmus paper blue Ammonium chloride in water turns blue litmus red
Variation in ph can be explained by looking at the ions formed when each of these salts dissociates If ions formed are from weak acids or bases, they react with water solvent, and ph will have value higher than 7 Hydrolysis! reaction between water molecules and ions of a dissolved salt
If anions react with water, process is anion hydrolysis and solution is more basic If cations react with water, process is cation hydrolysis and solution is more acidic
Anion Hydrolysis The anion of the salt is the conjugate base of the acid from which it was formed It is also a proton acceptor
If acid is weak, the conjugate base (the anion) will be strong enough to remove protons from some water molecules (proton donors) to form OH - ions Eq is created where the net effect of anion hydrolysis is increase in [OH - ] of solution (more basic)
Eq equation for typical weak acid in water, HA, forming hydronium and an anion, A- is as follows HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) From this equation, K a can be written
Hydrolysis reaction between water and anion, A - that is made by dissociation of weak acid, HA, is represented as A (aq) + H 2 O(l) HA(aq) + OH (aq) In forward reaction, anion gets proton from water to form HA (weak acid) and OH - left over from water molecule
Extent of OH - ion formation and position of eq depends on strength of anion Lower K a value of HA, stronger attraction for protons that anion (A - ) will have compared with OH -
Greater production of OH - As strength of A - increase, eq position is more to the right
Aq solutions of sodium carbonate are strongly basic Sodium ions do not undergo hydrolysis in aq solution, but carbonate ions react as a Br nsted base
Carbonate anion receives proton from H 2 O to form slightly ionized hydrogen carbonate ion, HCO -, 3 and the OH - ion CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq)
CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) +OH (aq) [OH - ] increases until eq is established As a result the [H 3 O + ] decreases so that the product [H 3 O + ][OH - ] stays equal to K w ph is higher than 7, and solution is basic
Cation Hydrolysis Cation of salt is conjugate acid of base from which it was formed Also a proton donor
If base is weak, cation strong enough to donate proton to water molecule to form H 3 O + Eq created when net effect of cation hydrolysis is increase in [H 3 O + ] of solution (more acidic)
A typical weak base, B, used to get general expression for K b (base dissociation constant) B(aq) + H 2 O(l) BH + (aq) + OH (aq)
Hydrolysis rxn between water and cation, BH +, made by dissociation of weak base, B, is represented as BH + (aq) + H 2 O(l) H 3 O + (aq) + B(aq) In forward reaction, BH + donates proton to water to form H3O + Extent of H 3 O + formation and position of eq depend on strength of BH +
Lower Kb value of B, the stronger the donation of protons that BH + will have and more H O + ions made 3 As strength of BH + increases, eq position lies to the right
Ammonium chloride, NH 4 Cl, dissociates in water to make NH 4 + and Cl - and an acidic solution Cl - are conjugate base of a strong acid, HCl, so they don t have tendency to hydrolyze in aqueous solution
NH + are conjugate acid of weak base, 4 NH 3 NH + donate protons to water molecules 4 Eq created with increase in [H O + ] so 3 ph is lower than 7
Hydrolysis in Acid-Base Reactions Hydrolysis helps explain why end point of neutralization can happen at a ph other than 7 Hydrolysis properties of salts determined by relative strengths of acids and bases from which the salts were formed Salts can be placed in 4 general categories, depending on their hydrolysis properties
1.Strong acid-strong base 2.Strong acid-weak base 3.Weak acid-strong base 4.Weak acid-weak base
Salts of strong acids and strong bases produce neutral solutions Neither the cation of strong base nor anion of strong acid hydrolyze much in aqueous solutions
HCl is strong acid Cl - doesn t hydrolyze in water NaOH is strong base Na + doesn t hydrolyze in water So, aq solutions of NaCl are neutral
Similarly, KNO 3 is salt of strong acid HNO 3 and strong base KOH ph of aqueous KNO 3 is always very close to 7
Strong acid-weak base Aq solutions of salts formed from rxns between weak acids and strong bases are basic
Anions of dissolved salt hydrolyzed in water ph raised, indicating [OH-] increase Anions from weak base (strong proton acceptor)
1 Equivalence point 2 4 3 At point 1 on the titration curve, only acetic acid is present. The ph depends on the weak acid alone. At 2 there is a mixture of CH 3 COOH and CH 3 COO. Adding NaOH changes the ph slowly. At point 3 all acid has been converted to CH 3 COO.This hydrolyzes to produce a slightly basic solution. At 4 the ph is due to the excess OH that has been added.
Strong acid-weak base Salts of strong acids and weak bases produce acidic aqueous solutions Cations of dissolved salt hydrolyzed in water
ph is lowered, indicated [H 3 O + ] has increased Cations are positive ions from a weak base (strong proton donor)
At point 1 on the titration curve, only aqueous ammonia is present. The ph is determined by the base alone. At 2 there is a mixture of NH 3 and NH 4+. Adding HCl changes the ph slowly. At point 3 all aqueous ammonia has been converted to NH4 +. At 4 the ph is determined by the excess H 3 O + that is being added. 1 2 Equivalence point 3 4
Weak acid-weak base Salts of weak acids and weak bases can make either acidic, neutral, or basic solutions, depending on salt dissolved b/c both ions of dissolved salt are hydrolyzed greatly If both ions hydrolyzed equally, solution stays neutral
In cases where both acid and base are very weak, the salt can undergo complete decomposition to the products of hydrolysis Ex. When Al 2 S 3 is placed in water, both precipitate and gas are formed Al 2 S 3 (s) + 6H 2 O(l) 2Al(OH) 3 (s) + 3H 2 S(g) Both products hardly soluble in water and removed from solution
Solubility Equilibrium
Solubility Product Saturated solution contains maximum amount of solute possible at given temp with an undissolved excess of substance Not necessarily concentrated Depends on solubility of solute
General rule used to express solubilities Substance is said to be soluble if solubility is greater than 1 g per 100 g of water Insoluble! less than 0.1 g in 100 g water Slightly soluble! between 0.1-1.0 g per 100 g water
Silver chloride is so slightly soluble in water that it s sometimes called insoluble Solution reaches saturation at very low concentration of ions All Ag + and Cl - ions in excess of this concentration will precipitate as AgCl
Let s look at eq system in saturated solution of AgCl containing excess of solid salt (AgCl) System represented as: AgCl(s) Ag + (aq) + Cl (aq)
b/c concentration of pure substance as solid or liquid stays constant, adding more solid AgCl doesn t change concentration of undissolved AgCl So, [AgCl] does not appear in final expression
Rearrange equation so both constants on same side gives solubility-product constant, K sp K[AgCl] = [Ag + ][Cl - ] Solubility-product constant! of a substance the product of molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the chemical equation
K sp = [Ag + ][Cl - ] This equation is solubility-equilibrium expression for the reaction Expresses that K sp of AgCl is product of molar concentrations of its ions in saturated solution
Calcium fluoride CaF 2 is another slightly soluble salt CaF 2 (s) Ca 2+ (aq) + 2F (aq) K sp = [Ca 2+ ][F ] 2
K sp can be determined from solubility data Indicates a maximum of 8.9 x 10-5 g AgCl can dissolve in 100 g water at 10 1 mol AgCl has mass of 143.32 g Saturation concentration (solubility) of AgCl can be expressed in moles per liter of water
= 6.2 x 10-6 mol/l AgCl dissociates in solution, giving equal numbers of Ag+ and Cl- ions So, concentrations of ions are [Ag+] = 6.2 x 10-6 [Cl-] = 6.2 x 10-6
Ksp = [Ag + ][Cl - ] Ksp = (6.2 x 10-6 )(6.2 x 10-6 ) Ksp = 3.8 x 10-11 This is solubility-product constant of AgCl at 10
Practice Problem Calculate the solubility-product constant, K sp, for copper(i) chloride, CuCl, given that the solubility of this compound at 25 C is 1.08 x 10 2 g/100. g H2O.
1. Analyze Given: solubility of CuCl = 1.08 10 2 g CuCl/ 100. g H 2 O Unknown: K sp
2. Plan Start by converting the solubility of CuCl in g/100. g H O to mol/l 2 You will need the molar mass of CuCl to get moles CuCl from grams CuCl Then use the solubility of the [Cu + ] and [Cl ] ions in the K expression and solve for K sp sp
CuCl(s) Cu + (aq) + Cl (aq) Ksp = [Cu + ][Cl ] [Cu + ] = [Cl ] = solubility in mol/l
3. Compute = 1.09 x 10-3 mol/l CuCl [Cu + ] = [Cl - ] = 1.09 x 10-3 mol/l K sp = (1.09 x 10-3 ) (1.09 x 10-3 ) = 1.19 x 10-6
Practice Problem Calculate the solubility-product constant, K sp, of lead(ii) chloride, PbCl 2, which has a solubility of 1.0 g/100. g H 2 O at a temperature other than 25 C. 1.9 10 4
Five grams of Ag SO will 2 4 dissolve in 1 L of water. Calculate the solubility product constant for this salt. 2 10 5
Calculating Solubilities Once known, K sp can be used to determine solubility Suppose you want to know how much BaCO 3 can be dissolved in 1 L of water at 25 From table, K sp for BaCO 3 is 5.1 x 10-9
BaCO 3 (s) Ba 2+ (aq) + CO 3 2 (aq) Given value of K sp, can write solubilityequilibrium expression as K sp = [Ba 2+ ][CO 3 2 ] = 5.1 10 9 BaCO 3 dissolves until [Ba 2+ ] and [CO 3 2 ] = 5.1 10 9
Solubility-eq equation shows that Ba +2 and CO -2 3 enter solution in equal numbers So, they have same concentration Let [Ba 2+ ] = x, and [CO 3 2 ] = x
Let [Ba 2+ ] = x, and [CO 3 2 ] = x [Ba 2+ ][CO 2 ] = K = 5.1 10 9 3 sp (x)(x) = 5.1 10 9 x = 5.1 10 9 The solubility of BaCO 3 is 7.14 10 5 mol/l Solution concentration is 7.14 10 5 M for Ba 2+ ions and 7.14 10 5 M for CO 3 2 ions
Sample Problem Calculate the solubility of silver acetate,agch 3 COO, in mol/l, given the K sp value for this compound listed in Table 18-3.
1. Analyze Given: K sp = 1.9 10 3 Unknown: solubility of AgCH 3 COO
2. Plan AgCH 3 COO Ag + (aq) + CH 3 COO (aq) K sp = [Ag + ][CH 3 COO ] [Ag + ] = [CH 3 COO ], so let [Ag + ] = x and [CH 3 COO ] = x
3. Compute K sp = [Ag + ][CH 3 COO ] K sp = x 2 x 2 = 1.9 10 3 x = 1.9 10 3 Solubility of AgCH 3 COO = 1.9 10 3 = 4.4 10 2 mol/l
Practice Problem Calculate the solubility of cadmium sulfide, CdS, in mol/l, given the K sp value listed in Table 18-3. 8.9 10 14 mol/l
Determine the concentration of strontium ions in a saturated solution of strontium sulfate, SrSO 4, if the K sp for SrSO 4 is 3.2 10 7. 5.7 10 4 mol/l
Precipitation Calculations In earlier example, BaCO 3 served as source for Ba 2+ and CO 3-2 b/c each mole of BaCO 3 yields 1 mol Ba +2 and 1 mol CO 3-2, concentrations were equal Equilibrium conditions doesn t require that the two ion concentrations be equal Eq still established to that [Ba 2+ ][CO 3 2 ] doesn t go over K sp
Suppose unequal amounts of BaCl 2 and Na 2 CO 3 are dissolved in water and solutions are mixed If ion product [Ba 2+ ][CO 3 2 ] goes over K sp for BaCO 3, a precipitate forms Precipitation continues until ion concentrations decrease to point where [Ba 2+ ][CO 3 2 ] = K sp
Sample Problem Will a precipitate form if 20.0 ml of 0.010 M BaCl 2 is mixed with 20.0 ml of 0.0050 M Na 2 SO 4?
1. Analyze Given: concentration of BaCl 2 = 0.010 M volume of BaCl 2 = 20.0 ml concentration of Na 2 SO 4 = 0.0050 M volume of Na 2 SO 4 = 20.0 ml Unknown: whether a precipitate forms
2. Plan The two possible new pairings of ions are NaCl and BaSO 4 Of these, BaSO 4 is a sparingly soluble salt It will precipitate if the ion product [Ba 2+ ][SO 4 2 ] in the combined solution exceeds K sp for BaSO 4 From the list of solubility products in Table 18-3, the K sp is found to be 1.1 10 10
BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) The solubility-equilibrium expression is written as K sp = [Ba 2+ ][SO 4 2 ] = 1.1 10 10 First [Ba 2+ ] and [SO 4 2 ] in the above solution must be found Then the ion product is calculated and compared with the K sp
3. Compute Calculate the mole quantities of Ba 2+ and SO 4 2 ions
Calculate the total volume of solution containing Ba 2+ and SO 2 4 ions 0.020 L + 0.020 L = 0.040 L Calculate the Ba 2+ and SO 2 4 ion concentrations in the combined solution
Trial value of the ion product: [Ba 2+ ][SO 4 2 ] = (5.0 10 3 )(2.5 10 3 ) = 1.2 10 5 The ion product is much greater than the value of K sp, so precipitation occurs.
Practice Problem Does a precipitate form when 100. ml of 0.0025 M AgNO 3 and 150. ml of 0.0020 M NaBr solutions are mixed? AgBr precipitates.
Does a precipitate form when 20. ml of 0.038 M Pb(NO ) and 30. ml of 0.018 3 2 M KCl solutions are mixed? PbCl 2 does not precipitate