Solubility and Complex Ion Equilibria

Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution.

When the solid is first added to water, no ions are initially present.

As dissolution proceeds, the concentration of ions increases until equilibrium is established. This occurs when the solution is saturated.

The equilibrium constant, the K sp, is no more than the product of the ions in solution. (Remember, solids do not appear in equilibrium expressions.)

For a saturated solution of AgCl, the equation would be: AgCl (s)  Ag + (aq) + Cl - (aq) AgCl (s)  Ag + (aq) + Cl - (aq)

The solubility product expression would be: K sp = [Ag + ] [Cl - ] K sp = [Ag + ] [Cl - ]

The AgCl (s) is left out since solids are left out of equilibrium expressions (constant concentrations).

You can find loads of K sp ’s on tables. Find the K sp values & write the K sp expression for the following:

CaF 2(s)  Ca F - K sp = Ag 2 SO 4(s)  2 Ag + + SO 4 -2 K sp = Bi 2 S 3(s)  2 Bi S -2 K sp =

Determining K sp From Experimental Measurements In practice, K sp ’s are determined by careful laboratory measurements using various spectroscopic methods. Remember STOICHIOMETRY!!

Example Lead (II) chloride dissolves to a slight extent in water according to the equation: PbCl 2  Pb Cl - PbCl 2  Pb Cl -

Calculate the K sp if the lead ion concentration has been found to be 1.62 x M.

Solution If lead’s concentration is “x”, then chloride’s concentration is “2x”. So.... K sp = (1.62 x )(3.24 x ) 2 K sp = (1.62 x )(3.24 x ) 2 = 1.70 x = 1.70 x 10 -5

Exercise 12 Calculating K sp from Solubility I Copper(I) bromide has a measured solubility of 2.0 X mol/L at 25°C. Calculate its K sp value.

Solution K sp = 4.0 X 10 -8

Exercise 13Calculating Ksp from Solubility II Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 X mol/L at 25°C.

Solution K sp = 1.1 X

ESTIMATING SALT SOLUBILITY FROM K sp SOLUBILITY FROM K sp

Example The K sp for CaCO 3 is 3.8 x 10 25°C. Calculate the solubility of calcium carbonate in pure water in a) moles per liter a) moles per liter b) grams per liter b) grams per liter

The relative solubilities can be deduced by comparing values of K sp. BUT, BE CAREFUL! 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 while some become more soluble! Aragonite. Aragonite.

Exercise 14Calculating Solubility from K sp The K sp value for copper(II) iodate, Cu(IO 3 ) 2, is 1.4 X at 25°C. Calculate its solubility at 25°C.

Solution = 3.3 X mol/L

Exercise 15Solubility and Common Ions Calculate the solubility of solid CaF 2 (K sp = 4.0 X ) (K sp = 4.0 X ) in a M NaF solution.

Solution = 6.4 X mol/L

K sp and the Reaction Quotient, Q With some knowledge of the reaction quotient, we can decide 1) whether a ppt will form, AND 2) what concentrations of ions are required to begin the ppt. of an insoluble salt.

1. Q < K sp, the system is not at equil. (unsaturated) 2. Q = K sp, the system is at equil. (saturated) 3. Q > K sp, the system is not at equil. (supersaturated)

Precipitates form when the solution is supersaturated!!!

Precipitation of Insoluble Salts Metal-bearing ores often contain the metal in the form of an insoluble salt, and, to complicate matters, the ores often contain several such metal salts.

Dissolve the metal salts to obtain the metal ion, concentrate in some manner, and ppt. selectively only one type of metal ion as an insoluble salt.

Exercise 16Determining Precipitation Conditions A solution is prepared by adding mL of 4.00 X M Ce(NO 3 ) 3 to mL of 2.00 X M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 X ) precipitate from this solution?

Solution yes

Exercise 17 Precipitation A solution is prepared by mixing mL of 1.00 X M Mg(NO 3 ) 2 and mL of 1.00 X M NaF. Calculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4 X ).

Solution [Mg 2+ ] = 2.1 X M [F - ] = 5.50 X M

SOLUBILITY AND THE COMMON ION EFFECT

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

LeChatelier’s Principle, that’s why! Be reasonable and use approximations when you can!!

Just remember what happened earlier with acetic acid and sodium acetate. The same idea here!

pH can also affect solubility. Evaluate the equation to see who would want to “react” with the addition of acid or base.

Would magnesium hydroxide be more soluble in an acid or a base? Why? Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH - (aq) Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH - (aq) (milk of magnesia) (milk of magnesia)

Why Would I Ever Care About K sp ??? Keep reading to find out ! Actually, very useful stuff!

Solubility, Ion Separations, and Qualitative Analysis …introduce you to some basic chemistry of various ions. …illustrate how the principles of chemical equilibria can be applied.

Objective: Separate the following metal ions: silver,lead, cadmium and nickel

From solubility rules, lead and silver chloride will ppt, so add dilute HCl. Nickel and cadmium will stay in solution.

Separate by filtration: Lead chloride will dissolve in HOT water… filter while HOT and those two will be separate.

Cadmium and nickel are more subtle. Use their K sp ’s with sulfide ion. Who ppt’s first???

Exercise 18 Selective Precipitation A solution contains 1.0 X M Cu + and 2.0 X M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp = 1.4 X ) or CuI (K sp = 5.3 X ) precipitate first?

Specify the concentration of I - necessary to begin precipitation of each salt.

Solution CuI will precipitate first. Concentration in excess of 5.3 X M required.

**If this gets you interested, lots more information on this topic in the chapter. Good bedtime reading for descriptive chemistry!

THE EXTENT OF LEWIS ACID-BASE REACTIONS: FORMATION CONSTANTS

When a metal ion (a Lewis acid) reacts with a Lewis base, a complex ion can form. The formation of complex ions represents a reversible equilibria situation.

A complex ion is a charged species consisting of a metal ion surrounded by ligands.

A ligand is typically an anion or neutral molecule that has an unshared electron pair that can be shared with an empty metal ion orbital to form a metal-ligand bond. Some common ligands are H 2 O, NH 3, Cl -, and CN -.

The number of ligands attached to the metal ion is the coordination number. The most common coordination numbers are: 6, 4, 2

Metal ions add ligands one at a time in steps characterized by equilibrium constants called formation constants. Ag + + 2NH 3  [Ag(NH 3 ) 2 ] +2 Ag + + 2NH 3  [Ag(NH 3 ) 2 ] +2 acid base acid base

Stepwise Reactions: Ag + (aq) + NH 3(aq)  Ag(NH 3 ) + (aq) K f1 = 2.1 x 10 3 K f1 = 2.1 x 10 3 [Ag(NH 3 ) + ] = 2.1 x 10 3 [Ag(NH 3 ) + ] = 2.1 x 10 3 [Ag + ][NH 3 ] [Ag + ][NH 3 ]

Ag(NH 3 ) + +NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f2 = 8.2 x 10 3 K f2 = 8.2 x 10 3 [Ag(NH 3 ) 2 + ] = 8.2 x 10 3 [Ag(NH 3 ) 2 + ] = 8.2 x 10 3 [Ag(NH 3 ) + ][NH 3 ] [Ag(NH 3 ) + ][NH 3 ]

In a solution containing Ag and NH 3, all of the species NH 3, Ag +, Ag(NH 3 ) +, and Ag(NH 3 ) 2 + exist at equilibrium. Actually, metal ions in aqueous solution are hydrated.

More accurate representations would be Ag(H 2 O) 2 + instead of Ag +, and Ag(H 2 O)(NH 3 ) + instead of Ag(NH 3 ) +.

The equations would be: Ag(H 2 O) 2 + (aq) + NH 3 (aq)  Ag(H 2 O)(NH 3 ) + (aq) + H 2 O (l) Ag(H 2 O)(NH 3 ) + (aq) + H 2 O (l) K f1 = 2.1 x 10 3 K f1 = 2.1 x 10 3 [Ag(H 2 O)(NH 3 ) + ] = 2.1 x 10 3 [Ag(H 2 O) 2 + ][NH 3 ] [Ag(H 2 O) 2 + ][NH 3 ]

Ag(H 2 O)(NH 3 ) + (aq) + NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2H 2 O (l) Ag(NH 3 ) 2 + (aq) + 2H 2 O (l) K f2 = 8.2 x 10 3 K f2 = 8.2 x 10 3 [Ag(NH 3 ) 2 + ] = 8.2 x 10 3 [Ag(H 2 O)(NH 3 ) + ][NH 3 ] [Ag(NH 3 ) 2 + ] = 8.2 x 10 3 [Ag(H 2 O)(NH 3 ) + ][NH 3 ]

The sum of the equations gives the overall equation, so the product of the individual formation constants gives the overall formation constant:

Ag + + 2NH 3  Ag(NH 3 ) 2 + Ag + + 2NH 3  Ag(NH 3 ) 2 + or or Ag(H 2 O) NH 3  Ag(NH 3 ) H 2 O K f1 x K f2 = K f K f1 x K f2 = K f (2.1 x 10 3 ) x (8.2 x 10 3 ) = 1.7 x 10 7 (2.1 x 10 3 ) x (8.2 x 10 3 ) = 1.7 x 10 7

Exercise 19 Calculate the equilibrium concentrations of Cu +2, NH 3, and [Cu(NH 3 ) 4 ] +2 when 500. mL of 3.00 M NH 3 are mixed with 500. mL of 2.00 x M Cu(NO 3 ) 2. K formation = 6.8 x

Solubility and Complex Ions

Complex ions are often insoluble in water. Their formation can be used to dissolve otherwise insoluble salts. Often as the complex ion forms, the equilibrium shifts to the right and causes the insoluble salt to become more soluble.

If sufficient aqueous ammonia is added to silver chloride, the latter can be dissolved in the form of [Ag(NH 3 ) 2 ] +.

AgCl (s)  Ag + (aq) + Cl - (aq) AgCl (s)  Ag + (aq) + Cl - (aq) K sp = 1.8 x K sp = 1.8 x Ag + (aq) + 2 NH 3(aq)  [Ag(NH 3 ) 2 ] + (aq) K formation = 1.6 x 10 7 K formation = 1.6 x 10 7

Sum: K = K sp x K formation = 2.0 x = {[Ag(NH 3 ) 2 + }[Cl - ] {[Ag(NH 3 ) 2 + }[Cl - ] [NH 3 ] 2 [NH 3 ] 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.

Exercise 20 Complex Ions Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) 2 3- in a solution prepared by mixing mL of 1.00 X M AgNO 3 with mL of 5.00 M Na 2 S 2 O 3.

The stepwise formation equilibria are: Ag + + S 2 O 3 2-  Ag(S 2 O 3 ) - K 1 = 7.4 X 10 8 Ag(S 2 O 3 ) - + S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- K 2 = 3.9 X 10 4

Solution [Ag + ] = 1.8 X M [Ag(S 2 O 3 ) - ] = 3.8 X M

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

The solubility of PbS in water: PbS (s)  Pb +2 + S -2 K sp = 8.4 x

The Hydrolysis of the S -2 ion in Water S -2 + H 2 O  HS - + OH - K b = 0.077

Overall Process: PbS + H 2 O  Pb +2 + HS - + OH - K total = K sp x K b = 6.5 x

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 K sp !

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 K sp.

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 CaCO 3, then hydroxide ions from the anion hydrolysis is 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 acid if the anion is the conjugate base of a strong acid!!