Mr. Bartelt Presents Solutions and their properties Solutions part 2
Expressions of concentration u Molarity—we already know this one. Sadly, there are many other ways to express concentration of solutions. 1. Mass Percent 2. Mole Fraction 3. Molality
Mass percent u As the name suggests this is mass of solute (what’s being dissolved) over mass solution (whole thing) u Commonly used in the place of molarity for stuff you can buy at Wal. Mart – Vinegar (acetic acid and water) – Hydrogen peroxide (H 2 O 2 and water)
Examples u 1. 2. Find the molarity of a 5% solution of acetic acid (CH 3 COOH) in water. This is the vinegar you buy at the store. (assume D= 1. 00 g/ml) Assume a solution mass of 100. g Now you know that you’ve got 5. 0 g of acetic acid and 100. ml of water.
Example u M=mol/Volume – Volume = 100. ml or 0. 100 L – mol= g/MM – We have 5 grams of acetic acid – Acetic acid’s MM is 60. 05 g/mol – 5. 0 g / 60. 05 g/mol= 0. 0833 mol
Next example u u u You can buy 3. 00 % Hydrogen peroxide at the store. I have 1. 00 M hydrogen peroxide in the lab. Which is more concentrated?
Mole fraction u This is similar to what we looked at in our unit on the gas laws. u u The above equation assumes a solution of only two components. If there were more then the denominator would have additional terms
Example What is the mole fraction if you mixed equal masses of water and ethylene glycol [C 2 H 4(OH)2]? u Assume 100. grams of each u
Mole fraction example u u u Concentrated sulfuric acid is 18. 0 M H 2 SO 4. What is the mole fraction water in 18. 0 M sulfuric acid. Assume density is the same as sulfuric acid 1. 84 g/ml. Assume 1. 00 liter Now we have 18. 0 mol H 2 SO 4
You try u Based on the calculations above, determine the mass percent of concentrated sulfuric acid.
Molality At this point it may seem like I’m just making these names up. u This is actually a very important expression of concentration that will prove extremely useful when we determine freezing point depression and boiling point elevation in lab on Wednesday. u Molality is defined as moles solute over kilograms of solvent. u
Example u What is the molality of battery acid? – u Assume 100. g of solution Hence – u Battery acid is sulfuric acid and water with a mass % of 33. 5% acid. 33. 5 g H 2 SO 4 and 66. 5 g water I now need – – moles H 2 SO 4 (easy) Kilograms water (easy)
Example u I now need – – moles H 2 SO 4 (easy) Kilograms water (easy)
Note u These are terms that need to be memorized u Molarity (M) = moles/liter u Mass % = masssample/masstotal u Mole fraction = nsample/ntotal u Molality (m) = nsample/kgsovent – You need to work many problems to get this down
Like dissolves like u u We have not discussed Lewis dot structures yet but we need to discuss the fundamental principal of solubility. That priciple is “like dissolves like” – – u u This means that polar solvents are good at dissolving polar molecules and ionic compounds Additionally, non-polar solvents are good at dissolving non-polar molecules. We will investigate this in a quick activity But first the structures
Entropy u Let’s look at water and hexane (C 6 H 14) Water has O-H bonds which are highly polar Hexane has C-H bonds which are non-polar
Determining “likes” u You don’t really need to know how to draw the Lewis structure to determine if a substance is polar or non-polar. u A good rule of thumb is – if your molecule is exclusively carbon and hydrogen it’s non-polar – if it contains carbon, hydrogen, and nitrogen or oxygen as well, it’s polar.
Determining “likes” u There are countless examples contrary to this rule, but you can’t be expected to know them so you should be fine. u We will look at something you should be able to figure out now however. u If the molecule is extremely large and only contains one oxygen or nitrogen the region of polarity will not be large enough to make it polar.
Polar “tails” u Fatty acids are long chains of hydrocarbons with polar “tails” on the ends. The gray and black spheres are hydrogen and carbon respectively. The red spheres are oxygen. NOTE: The size of the non-polar “tail” dominates the polarity of the molecule and hence the molecule is considered non-polar. Hence oil and water don’t mix.
Another notable exception u CO 2 has 2 carbon oxygen bonds which is highly polar, but they happen to be exactly opposite each other in the molecule. u Hence they “cancel” each other’s polarity making the molecule nonpolar.
Polar or non-polar u CH 3 OH (methanol)
Polar or non-polar u CH 4 (methane)
Polar or non-polar u Carbon tetrachloride
Polar or non-polar u C 12 H 22 O 11 (sugar)
Polar or non-polar u Any ionic compound
Polar or non-polar u Graphite (carbon)
Now we’re ready Today’s lab will focus on the principle of “like dissolves like”. You will note this in your lab and answer questions that relate to the topic of like dissolves like.
Henry’s law We will not do any calculations with Henry’s law in this class. u Henry’s law states that the concentration of dissolved gas in a solution is directly proportional to the partial pressure of the gas above the solution. u This explains how carbonation works. u When brewing beer yeast breaks sugar down into alcohol and carbon dioxide. If the bottle is capped and the CO 2 is trapped and builds up pressure. u This pressure drives CO 2 into the solution and makes the beer carbonated. u
Temperature effects on solutions u Typically an increase in temperature will result in greater solubility of solids, but not always. u The opposite is true of gasses. The higher the temperature, the less gas can be maintained in solution. u Click here if you must know more.
Why do things dissolve? u Disorder!!! If something is dissolved in something else u The dissolution process makes a more disordered system u Now we need to look at Le Chatelier’s Principle
Le Chatelier’s Principle u Le Chatelier’s Principle applies to systems that are at equilibrium (ΔG=0) u We’ll look at boiler scale in great detail to learn this new principle.
Equilibrium systems u Ca. CO 3 is “insoluble” – This doesn’t mean that NO Ca. CO 3 will be in solution, it just means that very little will be in solution. – But still, a little is dissolved Ca 2+ + CO 32 - Ca. CO 3 u If water contains Ca 2+ ions (and nearly all tap water does) it will bond with any CO 32 - present and form limestone in a pipe. Not good.
More systems If water contains carbonate ions this reaction will also take place u CO 32 -(aq) + CO 2(aq) + H 2 O(l) 2 HCO 3 u What happens to the solubility of CO 2 at higher temperatures? u If CO 2 is removed from our system, the equation will shift to the left to compensate for the loss of CO 2. u Now we have extra CO 32 - and this reaction proceeds Ca 2+ + CO 32 - Ca. CO 3 u
Le Chatelier's Principle u If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change. CO 32 -(aq) + CO 2(aq) + H 2 O(l) 2 HCO 3 Ca 2+ + CO 32 - Ca. CO 3 If T goes up CO 2 goes down CO 32 and H 2 O increase to compensate. The addition of CO 32 - then drives the production of more Ca. CO 3 and your pipes get clogged up.
Colligative properties are properties of solutions that depend on the number of particles in a given volume of solvent and not on the mass of the particles. u We can use colligative properties to determine the molar mass of substances. u This is very useful. u We’re going to look specifically at: u – Boiling point elevation – Freezing point depression
You can change the freezing point? u Of course you can. Ever added salt to snow to make it melt? u Probably not because you only see snow twice a decade, but back in my homeland of Chicago it’s done every winter. u The salt is added to increase the freezing point of the water and hence melts the ice.
Freezing point depression u ΔT = Kfmsolution m=molality u ΔT stands for the change in freezing point. u Kf is a substance specific constant u We’re only doing this for water so just memorize that – Kf = 1. 86 (ºC • kg/mol) for freezing point – Kb = 0. 51 (ºC • kg/mol) for boiling point
Examples u What freezing point would you expect from a solution of 120. 0 ml of water and 25. 0 g of sugar (C 12 H 22 O 11) u ΔT = Kfmsolution u First find molality of solution – Need moles sugar – Need kg water
Examples u ΔT = Kfmsolution – 0. 0731 mol sugar – 0. 120 m water u ΔT = Kfkgsolution Δ– Need moles sugar T = (1. 86 ºC • kg/mol)(0. 609 mol/kg) – = 1. 13 ºC u ΔTNeed kg water 120. ml 0. 120 kg water u The freezing point will drop by 1. 13 ºC u
Boiling point These problems are the exact same except the Kf is now a Kb. The b stands for boiling. u ΔT = Kbmsolution Kb = 0. 51 (ºC • kg/mol) u What boiling point elevation would you expect from the addition of 20. 0 grams of Na. Cl to 200. 0 ml of water? u Answer you’ll get: ΔT = 2. 62 ºC But this answer is wrong
Why’s it wrong When you dissolve sugar, the molecule doesn’t break up. The moles of sugar are the moles dissolved. u When you dissolve salt, you produce both Na+ and Cl- ions. u The BPE and FPD are affected by the # of particles, and thus when one mole of Na. Cl is dissolved an mole of Na+ and Clions are produced for a total of 2 moles of particles in solution. u
New equation ΔT = Kb • i • msolution u i in this case would be 2 – For Na 2 SO 4 it would be 3 – For Al(NO 3)3 it would be 4 u What boiling point elevation would you expect from the addition of 20. 0 grams of Na. Cl to 200. 0 ml of water?
We’ll confirm this in lab u We’re going to dissolve 20. 0 g of salt in 200. 0 ml of water in lab tomorrow and determine the freezing point. u Now you know what the freezing point should be. u But this is only the “calculated” freezing point, we’ll see what the experimental freezing point really is.
But why does the FP depress? u Think back to Shiva u What has more disorder, salt water or regular water?
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