Lab 3Lab 3

SolubilitySolubility

Concentration and ConductivityConcentration and Conductivity

DemonstrationDemonstration

Sodium chloride was added to 300 mL of water Sodium chloride was added to 300 mL of water in small incrementsin small increments 11stst scoop dissolves scoop dissolves

and solution barely conducts electricity (dim light)and solution barely conducts electricity (dim light) 22ndnd scoop dissolves and light gets brighter scoop dissolves and light gets brighter 33rdrd scoop dissolves and lights brighter scoop dissolves and lights brighter 44thth scoop dissolves but light gets only a bit brighter scoop dissolves but light gets only a bit brighter 55thth scoop does not appear to dissolve scoop does not appear to dissolve

Increasing ConductivityIncreasing Conductivity

Suggests increasing number of ions in Suggests increasing number of ions in same volume of water to allow solution to same volume of water to allow solution to conduct a higher voltageconduct a higher voltage

+ +

+

+

++

+

-

-

-

--

--

-

--

-

-

+

+

+

++

Least concentrated

Most concentrated

Compare the number of ions in the same volume

Graph to compare conductivity and Graph to compare conductivity and concentrationconcentration

conductivity

concentration

Increasing concentration —

Increasing conductivity

Conductivity no longer increases

Eventually Constant ConductivityEventually Constant Conductivity

Conductivity no longer changesConductivity no longer changes

This suggests that the concentration is no This suggests that the concentration is no longer changinglonger changing

Solution must have reached a maximum Solution must have reached a maximum concentrationconcentration

Undissolved solid remains at the bottomUndissolved solid remains at the bottom

The solution is said to be SATURATED.The solution is said to be SATURATED.

Lab ProceduresLab Procedures

A saturated solutions of several ionic A saturated solutions of several ionic compounds were preparedcompounds were prepared

Salt was added until solid remained at the bottomSalt was added until solid remained at the bottom

The solution was decantedThe solution was decanted The solution was poured down a stirring rod leaving the solid in The solution was poured down a stirring rod leaving the solid in

behind in the beakerbehind in the beaker The goal was to isolate the saturated solutionThe goal was to isolate the saturated solution

10.0 mL of the saturated solution heated until all 10.0 mL of the saturated solution heated until all the water was removedthe water was removed

This allowed us to measure the amount of salt that was This allowed us to measure the amount of salt that was dissolved in the 10.0 mL of solutiondissolved in the 10.0 mL of solution

Calculating the concentration of the Calculating the concentration of the saturated solutionsaturated solution

Use the data of # grams in 10.0 mL of solutionUse the data of # grams in 10.0 mL of solution

Convert grams to moles Convert grams to moles

Convert mL to LConvert mL to L

MOLARITYmoles of solute

liter of solution

Sample DataSample Data

EOS

The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution

Equilibrium in a Saturated SolutionEquilibrium in a Saturated Solution

When the solution is saturated, there is no When the solution is saturated, there is no net change- conditions remain constantnet change- conditions remain constantThe solution and solid are in equilibriumThe solution and solid are in equilibriumBoth reactions Both reactions (dissolving and crystallizing)(dissolving and crystallizing) are occurring at equal ratesare occurring at equal ratesExampleExample

NaCl(s) NaCl(s) Na Na++(aq) + Cl(aq) + Cl--(aq)(aq)The ions are dissociating as fast as the ions The ions are dissociating as fast as the ions crystallizecrystallize

Dynamic Nature of EquilibriumDynamic Nature of EquilibriumWhen a system reaches When a system reaches equilibriumequilibrium, the forward and reverse , the forward and reverse reactions continue to occur … but at equal rates.reactions continue to occur … but at equal rates.

We are usually concerned with the situation after equilibrium is reached.

After equilibrium the concentrations of reactants and products remain constant.

Dynamic Equilibrium Dynamic Equilibrium IllustratedIllustrated

NaCl dissolves and

recrystallizes continuously.

After a time, the solution contains

radioactive Na+ …

NaCl containing radioactive Na+ is added to a saturated NaCl solution.

… and the added salt now contains some

stable Na+.

Calculating an Equilibrium ConstantCalculating an Equilibrium Constant

Since conditions remain constant at equilibrium, Since conditions remain constant at equilibrium, an equilibrium constant (Keq) can be calculatedan equilibrium constant (Keq) can be calculated

Keq is calculated Keq is calculated as a ratio of concentrations of products compared to as a ratio of concentrations of products compared to

reactants at equilibriumreactants at equilibrium the coefficients in the equation become the exponents the coefficients in the equation become the exponents

in the Keqin the Keq

Since the concentrations of solids and liquids Since the concentrations of solids and liquids are constant, they are both left out of the Keqare constant, they are both left out of the Keq

Ksp- Ksp- the equilibrium constant for the equilibrium constant for saturated solutions of ionic compoundssaturated solutions of ionic compounds

NaCl(s) NaCl(s) Na Na+ + (aq) + Cl(aq) + Cl- - (aq)(aq)

Ksp = [NaKsp = [Na++][Cl][Cl--]] If the concentration of dissolved NaCl in a If the concentration of dissolved NaCl in a

saturated solution is 5 Msaturated solution is 5 M Then the concentration of sodium ions must Then the concentration of sodium ions must

be 5 M and the concentration of chloride ions be 5 M and the concentration of chloride ions must be 5 Mmust be 5 M

Ksp = [NaKsp = [Na++][Cl][Cl--] = (5 M)(5M) = 25 ] = (5 M)(5M) = 25

Calculating Concentration given the Calculating Concentration given the KspKsp

AgCl(s) AgCl(s) Ag Ag+ + (aq) + Cl(aq) + Cl- - (aq)(aq)

Ksp = [AgKsp = [Ag++][Cl][Cl--]] The concentration is unknownThe concentration is unknown

Let x = [AgCl], then x also x = [AgLet x = [AgCl], then x also x = [Ag++] and x = [Cl] and x = [Cl--]] Ksp = [AgKsp = [Ag++][Cl][Cl--] = (x)(x) = x] = (x)(x) = x2 2 = 1.6 x 10= 1.6 x 10-10-10

So x = 1.3 x 10So x = 1.3 x 10-5 -5 MM

Much less silver chloride is present in saturated Much less silver chloride is present in saturated solution than sodium chloridesolution than sodium chloride The lower Ksp shows the compound is less solubleThe lower Ksp shows the compound is less soluble

Given the Concentration Given the Concentration calculate the Kspcalculate the Ksp

AgAg22S(s) S(s) 2Ag 2Ag+ + (aq) + S(aq) + S2- 2- (aq)(aq)

Ksp = [AgKsp = [Ag++]]22[S[S2-2-] ] The concentration of silver sulfide in a saturated solution The concentration of silver sulfide in a saturated solution is 3.1 x 10is 3.1 x 10-17-17M = [AgM = [Ag22S]S]The concentration of sulfide ions must beThe concentration of sulfide ions must be

3.1 x 103.1 x 10-17-17M = [SM = [S2-2-] ] The concentration of silver ions must be The concentration of silver ions must be

2 x 3.1 x 102 x 3.1 x 10-17-17M = 6.2 x10M = 6.2 x10-17-17M = [AgM = [Ag++]]Ksp = [AgKsp = [Ag++]]22[S[S2-2-] ]

= (6.2 x10= (6.2 x10-17-17))22(3.1 x 10(3.1 x 10-17-17))= 3.2 x 10= 3.2 x 10-49-49

A very small Ksp indicates that this compound is barely solubleA very small Ksp indicates that this compound is barely soluble

Ksp and solubilityKsp and solubility

Ksp is calculated using the concentrations Ksp is calculated using the concentrations of dissolved ions in the saturated solution, of dissolved ions in the saturated solution, thus Ksp indicates the relative solubility of thus Ksp indicates the relative solubility of compoundscompounds

A Ksp greater than 0.1 indicates a A Ksp greater than 0.1 indicates a compound is solublecompound is soluble

A compound is designated as insoluble is A compound is designated as insoluble is the Ksp is less than 0.1 the Ksp is less than 0.1

Calculating the Concentration of Calculating the Concentration of Calcium fluoride in a saturated solutionCalcium fluoride in a saturated solution

Ksp = 5.3 x 10-11Ksp = 5.3 x 10-11 Reversible reaction at equilibrium Reversible reaction at equilibrium CaFCaF22(s) (s) Ca Ca2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Ksp = [CaKsp = [Ca2+2+][F][F--]]22

If x = [CaFIf x = [CaF22], then x = [Ca], then x = [Ca2+2+] and 2x = [F] and 2x = [F--]] Ksp = [CaKsp = [Ca2+2+][F][F--]]2 2 == 5.3 x 105.3 x 10-11-11= (x)(2x)= (x)(2x)22 (x)(4x(x)(4x22) = 4x) = 4x33 = 5.3 x 10 = 5.3 x 10-11-11

Divide by 4 and then cube rootDivide by 4 and then cube root

Preparing a solution of known molarityPreparing a solution of known molarity

Note the final volume of the solution is exactly 250mL.Note the final volume of the solution is exactly 250mL.

Calculate the molarity of the is solutionCalculate the molarity of the is solution

Again, a molar solution

The volume is exactly 1.0 L

Note that not all of the initial 1.0 L of water was used to prepare the solution

Dilution of a solutionDilution of a solution

Calculate the molarity of the new solution Calculate the molarity of the new solution using Musing M11VV11 = M = M22VV22

Other Units of ConcentrationsOther Units of Concentrations

MolalityMolalityMoles of solute per kg of solventMoles of solute per kg of solvent

Mole fractionMole fractionMoles of solute per total # molesMoles of solute per total # moles

Percent by MassPercent by Mass

mass of solute/mass of solution x 100mass of solute/mass of solution x 100

Mole Fraction and Mole Mole Fraction and Mole PercentPercent

The mole fraction (xi) of a solution component i is the fraction of all the molecules in the solution that are molecules of i

total()iinmolixtotalmoln==

EOS

The mole percent of a solution component is its mole fraction multiplied by 100

Sum of xi must equal 1

Calculating other concentrationsCalculating other concentrations

1.0L of the 0.100M CuSO1.0L of the 0.100M CuSO44 solution was solution was prepared by adding 25.0g or 0.100 moles of prepared by adding 25.0g or 0.100 moles of solid to 992mL of water.solid to 992mL of water.Calculate the molality of the solutionCalculate the molality of the solution Looking for moles of CuSOLooking for moles of CuSO44 per kg of water per kg of water

Calculate the mole fraction of the solution.Calculate the mole fraction of the solution. Looking for moles of CuSOLooking for moles of CuSO44 per total moles per total moles Total moles = moles of CuSOTotal moles = moles of CuSO44 + moles of water + moles of water

Calculate the percent by mass of the solution.Calculate the percent by mass of the solution. Looking for mass of CuSOLooking for mass of CuSO44 per total mass per total mass

DilutionDilution

The desired molarity solutions are often The desired molarity solutions are often prepared from concentrated stock solutions prepared from concentrated stock solutions (routinely used solutions prepared in (routinely used solutions prepared in concentrated form) by adding water. This concentrated form) by adding water. This process is called dilution.process is called dilution.

Moles of solute before dilution Moles of solute before dilution

= moles of solute after dilution= moles of solute after dilution

MM11VV11 = M = M22VV22

Figure 4.11 (a) A Measuring Pipet (b) A Volumetric (transfer) Pipet

Example: What volume of 16 M sulfuric acid

must be used to prepare 1.5 L of 0.10 M H2SO4

solution?

V1 = Volume before dilution = ?

M1 = Concentration before dilution = 16 M

V2 = Volume after dilution = 1.5 L

M2 = Concentration after dilution = 0.10 M

M1V1 = M2V2

V1 =M2V2 / M1 = (0.10 M x 1.5 L)/ 16 M = 0.0094 L = 9.4 mL