CHM 112-Dr. Rob Craig, Ph.D

YOUR FIRST LAB WENT VERY WELL(

• silver chloride (AgCl) with • Ksp = 1.8×10−10,

mercury(I) chloride (Hg2Cl2) with

• Ksp = 1.3×10−18

• LEAD chloride PbCl2

• (Ksp) is 1.7×10−5

THIS ALSO MEANS

• 1.9 mg ( moles) of AgCl will dissolve in each liter of water.

• PbCl2 in water is low (4.5 g/L at 20 °C)

• *MUCH MORE LEAD DISSOLVES IN WATER THAN SILVER –ACCORDING TO THE KSP

Ag+ and Pb2+.

• Of the cations encountered in our qualitative analysis scheme, only two form insoluble chlorides, Ag+ and Pb2+.

• These can be separated from a mixture of the other cations by adding a slight excess of hydrochloric acid

most famous reactions

• In one of the most famous reactions in chemistry, addition of colorless aqueous silver nitrate to an equally colorless solution of sodium chloride produces an opaque white precipitate of AgCl:

most famous reactions

• Ag+(aq) + Cl−(aq) → AgCl(s)**IT DECOMPOSES WHEN EXPOSED TO UV-lightOld film process

1.77 × 10−10 at room temperature

• The solubility product, Ksp, for AgCl in water is 1.77 × 10−10 at room temperature, which indicates that only 1.9 mg ( moles) of AgCl will dissolve in each liter of water.

• Additional HCl• AgCl(s) + Cl−(aq) → AgCl2

−(aq)

The Solubility Product Constant Ksp

• Salts differ in their solubilities. In general, compounds of alkali metals are soluble in water.

• Many ionic compounds however, are insoluble. (Most “ insoluble” salts will dissolve to some extent in water. These salts are said to be slightly soluble in water.

• For example, when the “insoluble” salt AgCl is mixed with water, a very small amount of it dissolves.

equilibrium expression

• AgCl(s) Ag+(aq) + Cl(aq)⇔• Because AgCl is a solid it is left out of the

equilibrium expression,

• Ksp = [Ag+][Cl-]

product of the concentration

• Ksp is equal to the product of the concentration terms each raised to the power of the coefficients of the substance in the dissociation equation.

• Ksp for AgCl at 25 degrees is 1.77 × 10−10

concentration of silver chloride

• What is the concentration of silver chloride ions in a saturated silver chloride solution at 25

• degrees? Ksp = 1.8 x 10-10• AgCls Ag+ aq + Claq⇔• Ksp = [Ag+][Cl-] = 1.8 x 10-10• Since [Ag+]= [Cl-]• [Ag+]2 = 1.8 x 10-10• [Ag+] = 1.3x10-5

how soluble a substance is.

• The equilibrium constant Ksp called the solubility constant indicates how soluble a substance is. If the Ksp is low, the solid is not very soluble.

• If the Ksp is high, the solid is soluble. As long as the product of the concentration of the ions does not exceed the Ksp value, no

precipitate forms.

Predicting Precipitate Formation

• When the product of the concentration of the ions exceeds the value of Ksp they cannot exist in equilibrium anymore.

• They will form a precipitate in order to reduce the concentrations

Problem:

• Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3 is added to a beaker containing 1 L of tap water with a Cl- ion concentration of 1.0 x 10-5 mol/L?

• [Cl-] = 1.0 x 10-5 mol/L• [Ag+] calculation. Start with the given info, 1

mL of 0.1 mol/L

Problem:

• Set up a ratio that states, if there are 0.1 moles of Ag+ in 1000 mL how many moles will be in 1 mL?

• 0.1 mol = x• 1000mL 1 mL• x = 0.0001 moles of Ag+ are present. (This is

how many moles there are, this is not The concentration)

Problem: trial ion product = “Q”

• The [Ag+] will be this many moles of Ag+ dissolved in the volume of solution in this question.

Therefore[Ag+] = 0.0001 mole = 0.0001 mole = 1.0 x 10-4 mol/L1000 mL + 1 mL 1.001 LT.I.P. = [Ag+][Cl-]= (1.0 x 10-4)(1.0 x 10-5)= 1.0 x 10-9The Ksp for AgCl is 1.8 x 10-10 therefore T.I.P > Ksp so a ppt. will form.

STEPS TO FOLLOW1. Determine which ions will form what compound2. Calculate the number of moles of each ion (n = C x V)3. Determine the total volume of the solutions after mixing4. Calculate the concentration of each ion (C = n / V)5. Substitute the concentrations of the ions into the Ksp expression to solve for Q

Comparing Q and K does this

6. Compare Q and KGiven the initial concentration, we can predict which way a reaction will goComparing Q and K does thisQ > K :A precipitate forms and a saturated solution exists Reaction goes to leftQ = K :A saturated solution exists Equilibrium is establishedQ < K : No precipitate forms and an unsaturated solution exists. Reaction goes to right

MUST MEMORIZE FOR MCAT

• Q > K :A precipitate forms and a saturated solution exists Reaction goes to left

• Q = K :A saturated solution exists Equilibrium is established

• Q < K : No precipitate forms and an unsaturated solution exists. Reaction goes to right

LEAD AND MERCURY

• While Lead(II) chloride is abundant in many natural water reserves, it is unsafe for human consumption and must be filtered out.

LEAD AND MERCURY

• The solubility of PbCl2 in water is low (4.5 g/L at 20 °C) and for practical purposes it is considered insoluble.

• Its solubility product constant (Ksp) is 1.7×10−5.

LEAD AND MERCURY

• It is one of only four commonly insoluble chlorides, the other three being silver chloride (AgCl) with Ksp = 1.8×10−10, copper(I) chloride (CuCl) with Ksp = 1.72×10−7 and

• mercury(I) chloride (Hg2Cl2) with Ksp = 1.3×10−18

With heat

• PbCl2(s) → heat→ Pb+2 (aq) + 2Cl-

• Then you added acetic acid and CrO4-2

(aq

• Pb+2 (aq) + CrO4

-2 (aq) → Pb

• Pb+2 (aq) + CrO4

-2 (aq) → Pb CrO4(s)

AgCl(s) can be removed from Hg2Cl2(s)

• AgCl(s) Ag+(aq) + Cl(aq)⇔

• AgCl(s) + Cl−(aq) → AgCl2−(aq)

• **YOU ADDED NH4OH

• AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl−(aq)

Hg22+(aq) + 2Cl-(aq) <==> Hg2Cl2(s)

Hg22+(aq) + 2Cl-(aq) <==> Hg2Cl2(s)

you added HCL Then add NH4OHHg2Cl2(s) + 2NH3(aq) --> Hg(l) + HgNH2Cl(s) + NH4

+

(aq) + Cl-(aq)

Also. In the presence of chloride ions, it dissolves to give the tetrahedral coordination complex [HgCl4]2−.

A disproportionation reaction-

Hg2Cl2(s) + 2NH3(aq) -->

Hg(l) + HgNH2Cl(s) + NH4+(aq) + Cl-(aq)

•

• Aqueous ammonia reacts with Hg2Cl2 to produce metallic mercury (black) and mercury(II) amidochloride (white), a disproportionation reaction:

• Hg2Cl2(s) + 2NH3(aq) --> Hg(l) + HgNH2Cl(s) + NH4

+(aq) + Cl-(aq)

Lead chloro complexes

• PbCl2(s) + Cl- → [PbCl3]-(aq)

• PbCl2(s) + 2 Cl- → [PbCl4]2-(aq)

• Addition of chloride ions to a suspension of PbCl2 gives rise to soluble complex ions. In these reactions the additional chloride (or other ligands) break up the chloride bridges that comprise the polymeric framework of solid PbCl2(s).

• Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)

• Pb(CH3COO)2(aq) + HCl(aq) → PbCl2(s) + 2 CH3COOH(aq)

• basic PbCO3 + 2 HCl(aq) → PbCl2(s) + CO2(g) + H2O[7]

• Pb(NO3)2(aq) + 2 HCl(aq) → PbCl2(s) + 2 HNO3(aq)