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Dec. 19 Hour Exam 1 (C-109 Start time 8AM)Coverage is Chapter 12 and 13.

Chapter 13 homework is game. Please don’t fall behind.

If you are having trouble please visit me ASAP

Solute-solvent, temperature and pressure are factors that affect the solubility of solutes in solvents.

Increasing temperature increases solubility for most ionic solutes in water.

Increasing temperature decreasessolubility for most gases in liquids.

Temperature ˚C

Temperature ˚C

Solu

bilit

y (g

/100

g H

2O)

1. Place a gas (say CO2) over the liquid and seal it with a piston.

2. We find that increasing the pressure increases the solubility of the gas in the liquid relative to atmospheric pressure.

3. Container is sealed and under pressure until opened.

4. At constant temperature, the concentration of gas solubilized in the solvent depends on the pressure above the liquid and the nature of the gas.

The solubility of a gas in a liquid depends on the gas, the liquid (solvent), the pressure and temperature.

Henry’s law:

Henry’s law: At constant temperature, the molarity of a given gas dissolved in a liquid is proportional to the partial pressure of that gas in equilibrium above the liquid.

Molarity of the dissolved gasin solution.

P is the partial vapor pressure of the gas above the solution.

Henry Law constant that depends on T, substance and the solvent. Must be given! (units: mol/L atm)

Cgas = kH Pgas

✴ Pressure has no or little impact on the solubility of solids in liquids or liquids in solids.

y = m x + b

Gas kH

(Mol/L atm)

N2 6.4 X 10-4

O2 1.3 X 10-3

CO2 3.3 X 10-2

Henry Law Constants at 25˚C vary with the gas and the nature of the solvent under question. They must be given or looked up in a handbook.

Let’s make a plot using Henry’s Law Equation: Plot solubility vs gas pressure we get a straight line. The slope is the the Henry Law constant for that gas.

Cgas = kH Pgas

y = m x + bSolubility

The partial vapor pressure of CO2 gas inside a bottle of liquid Coke is 4 atm at 25˚C. Assume constant temperature and calculate the solubility of CO2 at this pressure, and also when the cap on the coke is removed in ppm? The Henry Law Constant k for CO2 in water is 3.3 X 10-2 mol/L atm at 25˚C, and the partial pressure of CO2 in the atmosphere is 0.00033.

CCO2 = kH PCO2

ppm CO2 =0.1 mol CO2

L! 44.01 CO2

1 mol CO2! 103 mg

1 g= 4401 ppm = 4000 ppm

ppm CO2 =1.1! 10!5 mol CO2

L! 44.01 CO2

1 mol CO2! 103 mg

1 g= .48 ppm

CCO2 = 3.3! 10!2 mol

L atm! 0.00033 atm = 1.1! 10!5 mol

L

CCO2 = 3.3! 10!2 mol

L atm! 4 atm = 0.1

mol

L

PT =nART

VT+

nBRT

VT+

nCRT

VT

PT = (nA + nB + nC)RT

VT

PA

PT=

nART

VT

nTRT

VT

PA

PT=

nA

nT=

nA

n1 + n2 + n3 + ....=

We define and call the fraction;nA

nTthe mole fraction XA

PA =nA

nTPT

PB =nB

nTPT

XA + XB = 1

PT =nART

VT+

nBRT

VT+

nCRT

VT

PT = (nA + nB + nC)RT

VT

PA

PT=

nART

VT

nTRT

VT

PA

PT=

nA

nT=

nA

n1 + n2 + n3 + ....=

We define and call the fraction;nA

nTthe mole fraction XA

PA =nA

nTPT

PB =nB

nTPT

XA + XB = 1

Pi = Xi PT

this step is where most students get confused. The question is “how to express PA in terms of PT?”

Dalton’s Law of Partial Pressure

What is the concentration of O2 in a fresh water stream in equilibrium with air at 25˚C and 1 atm. Express the answer in ppm (mg/L). Note that the partial pressure of oxygen in the atmosphere is 0.21 atm and the Henry constant @ 25˚C is 1.3 x 10-3 mol/L atm.

Convert Molarity to ppm. Recall ppm means in this case mg/L.

ppm O2 = 2.73! 10!4 mol O2

L! 31.98 g O2

1 mol O2! 103 mg

1 g= 8.7 ppm

CO2 = 1.3! 10!3 mol O2

L atm! 0.21 atm = 2.7! 10!4 M

Cgas = kH Pgas

= moles soluteliters solution

Molarity (M)

Chemists use various definitions to quantify the amount of solutes in solutions or solvents.

Molality (m) = moles solutekg solvent

Why two units? Molality does not depend on volume and is not subject to temperature changes.

L solution is not the volume of solvent!

Two Key Riffs For Problem Solving: 1. Mass Solution = Mass Solvent + Mass Solute2. Use #1 and solution density to convert solution volume to solution mass

The world uses different units of concentration that you should know.

% by mass = x 100mass of solutemass of solution

x 100mass of solutemass of solute + mass of solvent=

x 100volume of solutevolume of solution=% by volume

x 100volume of solutevolume of solute + volume of solvent=

Mole Fraction (XA) = moles of A

total number of moles of solvent and solute

Analogies would be: 1 minute in 2 years or 1 cent in $10,000

Analytical and environmental chemists express minute concentrations as “parts per million” (ppm) and “parts per billion” (ppb).

1 ppm = 1 gram in mL water X 106 (mg/L)1 ppb = 1 gram in mL water X 109 (µg/L)

ppm =g solute

g solution! 106 ppb =

g solute

g solution! 109

If the solvent is water then many times the g solution is replaced by L solution the density of water 1 g/ml.

What is the molality of a 5.86 M ethanol (C2H5OH) solution? The density of the aqueous EtOH solution is measured to be 0.927 g/mL.

m = moles of solutekg solvent =

?1

?2= ?

What is the molality of a 5.86 M ethanol (C2H5OH) solution? The density of the aqueous EtOH solution is measured to be 0.927 g/mL.

Assume we have 1 L of 5.86 EtOH solution. We know from the given density that 1 Liter of EtOH solution has a mass of 927 g soln and also 5.86 moles of EtOH in it.

m = moles of solutekg solvent = 5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

Mass Solution = Mass Solvent + Mass SoluteMass Solvent = Mass Soln - Mass SoluteMass Solvent = 927 g Soln - 270 g EtOH = 657 g = 0.657 kg

1.00 L soln! 5.86 mol C2H5OH1 L soln

! 46.068 g EtOH

1 mol C2H5OH= 270 g EtOH

m = moles of solutekg solvent =

?1

?2= ?

The density of a 25.0% w/w solution of sulfuric acid (H2SO4) in water is 1.1783 g/mL at 25.0ºC. What is the molality and molarity of this solution?

Key: use the % mass to get something we can understand:

% by mass x 100mass of solutemass of solution=

25.0% w/w H2SO4 ==> 25.0 g H2SO4/100 g solution

25 g H2SO4

100 g soln! 1 mol H2SO4

98.1 g H2SO4

! 1.1783 g solution

1 mL! 1000 mL

1L= 3.00M

Molarity =

Molality =25 g H2SO4

100 g soln! 1 mol H2SO4

98.1 g H2SO4

! 1000 g soln

1 kg soln= 2.55 m H2SO4

(a) Find the concentration of calcium in parts per million in a 3.50-g pill that contains 40.5 mg of Ca.

(b) The label on a 0.750-L bottle of Italian chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain?

(c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water?

(c)

(a)3.5 g Ca

103 mgg Ca40.5 mg Ca x

106x = 1.16 x 104 ppm Ca

(b) 11.5 L alcohol100 L chianti0.750 L chianti x = 0.0862 L alcohol

2.39 mol C2H8O2

2.39 mol C2H8O2 + 3.22 mol H2O

= 0.423

χ C2H6O2 = 3.22 mol H2O

2.39 mol C2H8O2 + 3.22 mol H2O

= 0.577

χ H2O =

mol EG = 142 g EG 1 mole EG60.09 g EG

= 2.36 mol C2H6O2

moles H2O = 38.0 g H2O 1 mole H2O18.02 g H2O

= 3.22 mol H2O

X

X

An ethanol-water solution is prepared by dissolving 10.00 mL of C2H5OH (density etOH = 0.789 g/mL) in a sufficient volume of water to produce 100.0 mL of a solution. The density of the solution is found to be 0.982 g/mL.

What is the concentration of etOH expressed as a) %volume % b) mass % c) mass/volume % d) mole fraction e) mole percent f) molarity g) molality h) ppm?

1. Identify the masses of solute and the solvent ==> solution mass

2. Understand that “solution” is solute + solvent3. Understand and practice the definitions on the previous slide

4. Plug and chug

Sample ProblemA 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20˚C. What is the concentration as (a) mole fraction H2SO4, (b) % mass H2SO4, (c) molality H2SO4 (MW H2SO4 = 98.086 g/mol) ?

1. Assume 1.0 L solution: The mass in grams of 1 L of solution is 1046 g soln that 1 Liter contains 0.750 mol H2SO4.

Mass Solution = Mass Solvent + Mass soluteMass H2O = 1046 g soln - 73.56 g H2SO4 = 972 g H2OMol H2O = 972 g/18 g H2O = 54 mol H2O

Mass solute = g H2SO4 = 0.750 mol H2SO4 X 98.086 g/mol H2SO4

Mass solute = g H2SO4 = 73.56 g H2SO4

Mole fraction H2SO4 = 0.750 mol H2SO4/(0.750 mol H2SO4 + 54.0 mol H20)

= 0.0137

Sample ProblemA 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20ºC. What is the concentration in (a) mole fraction, (b) mass percent, (c) molality (MM = 98.086 g/mol) ?

(b) Mass % H2SO4 = 0.0736 kg H2SO4/1.046 kg total = 7.04%

(c) Since 0.972 kg water has 0.750 mol H2SO4 in it, 1 kg water would have 0.772 mol H2SO4 dissolved in it:

1.00 kg H2O × 0.750 mol H2SO4/0.972 kg H2O = 0.772 mol H2SO4

Thus, molality of sulfuric acid is 0.772 m