Solutions

Chapter 15

Mixtures

Heterogeneous mixture- unevenly mixed substance (separation can be seen)

Homogeneous mixture- evenly mixed substance (no separation can be seen)

Suspensions

~Small but visible particles suspended or floating in a gas or liquid (heterogeneous mixture)

Like a snow globe or dust or “shake before using”

the particles are too big to float forever without being stirred

If a suspension sits, the particles will settle Can be filtered out

Colloids or Colloidal Suspension

~mixture that appears uniform unless under a high powered microscope.

Particles are a little larger than the wavelength of light

Extremely light particles float almost indefinitely.

Milk, blood, smoke These can be separated in a centrifuge

Tyndall Effect

~Scattering of light by a colloid or suspension

Both a colloid and a suspension have particles larger than the wavelength of light, so when light shines through it should be deflected every which way.

This will make the beam of light visible.

Solutions

Particles are smaller than the wavelength of light. Therefore, it will not scatter light.

With solutions, no separation can be seen even under a high powered microscope.

Cannot be separated by any filter or by a centrifuge.

Can be separated by boiling/ melting points. salt water, metal alloys, air

Tyndall Effect

Colloid/suspension solution

Parts of a solution

Solvent- what the substance is dissolved in Solute- what is being dissolved Water is called the “universal solvent” because it dissolves a lot of substances and

is very common. Water solutions are called aqueous.

Mass and volume

In a solution, mass is conserved, however, volume is not.

That is to say, the mass of a solution = mass of the solute + solvent.

The volume of a solution may not equal the volume of the solute +solvent.

Example

It is easy to think of sand and water (not a solution, but it works for the general concept)

If you mix a liter of sand and a liter of water you get…

A mixture that is more than one liter but less than 2 liters.

Now this applies to solutions, if you mix 1 L of water with .5 liter of Na2 CO3 the resultant solution is more than 1 L but less than 1.5 L

Density of solutions

Increasing the mass of the solution and not increasing the volume comparatively will increase the density.

Dissolving solids into water almost always increases the density.

How much the density increases, depends on how much is dissolved.

Solution misconceptions Solutions don’t have to be a solid in a liquid. carbonated water is CO2 dissolved in water,

streams have dissolved O2 in them. The solvent doesn’t have to be water or even a

liquid. Alloys (two or more metals) are a solution as is

air. Several things dissolve in oils.

Gases

Gases dissolved in water tend to decrease the density of the solution.

Again the volume of the solution does NOT increase anywhere near the volume of the gas + water, but it does increase at a greater rate than the mass.

Liquids

Liquids may increase or decrease the density of the solution dependent on whether they are more or less dense than the solvent.

Rubbing alcohol will decrease the density of a water solution, where acetic acid will increase the density of a water solution.

Coke v. Diet Coke Coke cans sink in water, diet coke floats. That means a coke can is more dense than water,

diet coke is less dense. Aluminum is more dense than water, but there is

head space, a little air pocket, at the top of the can. Diet Coke (and all diet beverages) use artificial

sweeteners like Nutrasweet. Nutrasweet is 200x sweeter than sugar, so you need

to dissolve less in the solution, making it less dense

Concentration

~How much solute is present in a solution compared to the solvent.

Molarity (M)- moles of solute per liter of solution. M = mol/L

2.1 M AgNO3 means 2.1 mol of AgNO3 for every one liter of solution

Other measures of concentration

Name Abbrev. What it is

molality m mol solute/kg solvent

parts per million ppm g solute/g solvent x 106

parts per billion ppb g solute/g solvent x 109

mole fraction x mol solute/mol solution

percent by mass % g solute/g solution x 100

percent by volume % mL solute/mL solvent x100

Molarity Problems

Molarity = mol/L

Molarity = moles of solute / Liters of solution

Molarity problems

How many moles of HCl are in 125 mL of 2.5 M HCl?

2.5 mol HCl

1 L of soln.

.125 L of soln = .31 mol HCl

Here we go What concentration solution would be prepared if 39 g

of Ba(OH)2 were mixed in a 450 mL solution?

39 g Ba(OH)2 1 mol Ba(OH)2

171.316 g Ba(OH)2

=.2276 mol Ba(OH)2

.2276 mol Ba(OH)2 .45 L of solution

M = mol/L=.51 M Ba(OH)2

More

For a lab in this chapter, I need to make .60 L of 3.0 M NaOH, what mass of NaOH did I need?

.6 L x 3.0 M NaOH = 1.8 mol NaOH 1.8 mol NaOH x 39.998 g/mol = 72 g NaOH

Molarity Problems

A 0.24 M solution of Na2SO4 contains 0.36 moles of Na2SO4. How many liters were required to make this solution? 0.36 mol Na2SO4 1 L soln

0.24 mol= 1.5 L Na2SO4

Getting tougher AgNO3 + BaCl2 AgCl + Ba(NO3)2

Balance the equation. If 1.2 L of .50 M AgNO3 is reacted completely, what molarity solution of Ba(NO3)2 will be created if the volume increased to 1.5 L?

2 2

1.2 L x .5 M AgNO3 = .6 mol AgNO3 . 6 mol AgNO3 1 mol Ba(NO3)2

2 mol Ag NO3

= .3 mol Ba(NO3)21.5 L

= .20 M Ba(NO3)2

HNO3 + Zn H2 + Zn(NO3)2 If you have .65 L of 1.2 M HNO3 and you

react it completely what volume of H2 gas will you produce at STP?

2

1.2 M HNO3 x .65 L = .78 mol HNO3 . 78 mol HNO3 1 mol H2

2 mol HNO3 =.39 mol H2

.39 mol H2 22.4 L at STP

1 mol H2 = 8.7 L at STP

. 78mol HNO3 1 mol Zn(NO3)2

2 mol HNO3 = .39 mol Zn(NO3)2

.75 L

= .52 M Zn(NO3)2

HNO3 + Zn H2 + Zn(NO3)2 If you have .65 L of 1.2 M HNO3 and you

react it completely, what conc. of Zn(NO3)2 will be left if the volume increases to .75 L?

1.2 M HNO3 x .65 L = .78 mol HNO3

2

Fe + H2SO4 Fe2(SO4)3 + H2

If 350 mL of 2.3 M H2SO4 is completely reacted, what is the volume of hydrogen gas produced at 24o C and 114 kPa?

2 3 3

.35 L x 2.3 M = .805 mol H2SO4 . 805 mol H2SO4 1 mol H2

1 mol H2SO4

=.805 mol H2

PV = nRT114 kPa V = .805 mol (8.31) 297 K =17 L H2