Today is 6 March 2009 -- . This note is about the salt dissolving experiment

The experiment was to see if a change in dissolving time occurred as more and more salt entered the solution.  This is one aspect of solution chemistry.  Three different water temperatures were studied; cold, room temperature and hot.  Students added about 7 grams of table salt, sodium chloride or NaCl,  to 100 ml of water, stirred until it dissolved and timed how long it took.  They weighed the container of water before and after.  The difference in weight was how much salt they had added.  Then they repeated this with another approximately 7 grams, stirred, timed and weighed.  They repeated it a third time.  The results were to be graphed as the total weight of salt on one axis and the dissolving time for each salt addition on the the other axis.  For example, let us say the first 7 grams took 50 seconds to dissolve, the second 7 grams took 62 seconds and the third took 72 seconds.  The student would make marks at the intersection of 7 g and 50 sec,  14 g (7g +7g) and 62 seconds and 21 g (7g + 7g + 7g) and 72 seconds.  This example shows that as more salt is added to the solution, it takes longer to dissolve.  What would it mean if less time was taken?  

Some students used three different temperatures of water and attempted to plot that data.  That was incorrect and the data was worthless.  They were to use the same water temperature; either cold, hot or room temperature, and add salt to it.  By comparing data between different groups of students, testing at different temperatures, the class was supposed to determine if hot water dissolved the salt the fastest and cold water the slowest, or some other way.  

Today is 11 March 2010 -- We are studying solution chemistry

You can study this information on your own:  It is Chapter 13 in your text (HOLT, Chemistry).  Read and take notes in your lab notebook on Ch. 13, sections 1 and 2.  To prepare for this section, define the key terms on page 454 of your text.  Also, answer the following questions:

1.  We need a 30% solution of copper sulfate.  We have available 40 ml of 25% copper sulfate and an unlimited amount of 60% CuSO₄.  How many milliliters of the 60% solution should we mix to get the required 30% concentration?

.3(40ml + x) = .25(40ml) + .6x

12ml + .3x = 10ml + .6x

12ml - 10ml = .6x - .3x

2ml = .3x

2ml/.3 = x = 6.67 ml

How do I know this?

A "30%" solution means 30 grams of a solute in 100 ml of a solvent.  Similarly, "25%" is 25 grams in 100 ml.  Since we have 40 ml, then the amount of solute we have is:

 (25 grams) x (40 ml/100ml) = 10 grams

The 60% solution is 60 grams in 100 ml, but we don't know how much to add so we write:

(60 grams) x (x ml/100 ml) = 0.6x g

The result, 30%, means 30 grams in 100 ml. But we only know the volume as (40 ml + x ml) so we write: (30 grams /100 ml) x (40ml + x ml)

Put it together:

10 g + (.6x g) = (.3g/ml)x(40ml + x ml)

10 g + .6x g = 12 g + .3x g

Rearrange:

.6x g - .3x g = 12 g - 10 g

.3x g = 2 g

x = 2 g/.3g = 6.67

Prove this!

The atomic mass of CuSO₄ = 63.57 amu + 32.06 amu + (4)(15.99 amu) = 159.59 amu.

The number of moles of each solution is:

Conc	Moles/ 100 ml	Moles/ml	ml need  (or have)	moles
30%	30/159.59	0.00188	40 + x	0.0752+ 0.00188x
25%	25/159.59	0.00157	40	0.0628
60%	60/159.59	0.00376	x	0.00376x

0.0752 + 0.00188x = 0.0628 + 0.00376x

Rearrange

0.0752 - 0.0628 = 0.00376x - 0.00188x

0.0124 = 0.00188x

x= 0.0124/0.00188 = 6.60 ml of 60% solution

2.  It turns out that we need 140 ml of the 30% copper sulfate solution.  Did we make enough in the first problem?  If not, how many milliliters of each solution (25% and 60%) do we need to make 140 ml of 30% CuSO₄ solution?

Let x be the ml needed of one solution.  Then 140 - x = ml of other solution.

(140ml)(30%) = (25%)(x) + (60%)(140ml - x)

4200ml% = 25x% + 8400ml% - 60x%

4200ml% - 8400ml% = 25x% - 60x%

-4200ml% = -35x%

x = 4200ml%/35% = 120 ml (of 25% solution)

140 - x = 40 ml of 60% solution

Today is 30 July 2009 -- Today we learned about solutions

In particular, we learned about making a supersaturated sodium acetate solution.  (This post segues into the discussion of "enthalpy.")  The chemical equation for the reaction between acetic acid (which is 5% in the vinegar that we used) and sodium bicarbonate produced sodium acetate, a salt, in addition to the carbon dioxide that we studied last week, and water. 

The reason we are using sodium acetate is because it stays fluid, even though you would expect it to crystalize when its solubility limit is reached.  Crystallization will occur if a "seed crystal" is dropped into the solution, or if the system is disturbed in some other way.  Then, heat is released as the crystals grow.  The heat release is enough to allow sodium acetate to be used in hand warmers.  (They can be "recharged" by placing in boiling water for a few moments.)

The hand warmers are made by sealing the supersaturated sodium acetate in pouches made from "lay-flat tubing."  I brought an Impulse sealer for this purpose, and tubing.  To activate the change from fluid to solid, a small strip of stainless steel sheet is sealed in the pouch with the solution.  When the strip is flexed, it creates "nucleation sites" for crystallization to occur. 

The following information was provided to the students as a note:

"A reusable hand warmer can be made of a supersaturated solution of sodium acetate, which releases heat on crystallization.  Sodium acetate trihydrate crystals melt at 58 °C, dissolving in their water of crystallization. When they are heated to around 100 °C, and subsequently allowed to cool, the aqueous solution becomes supersaturated. This solution is capable of supercooling to room temperature without forming crystals. By clicking on a metal disc in the heating pad, a nucleation center is formed which causes the solution to crystallize into solid sodium acetate trihydrate again. The bond-forming process of crystallization is exothermic, hence heat is emitted. The latent heat of fusion is about 264–289 kJ/kg. Unlike some other types of heat packs that depend on irreversible chemical reactions, sodium acetate heat packs can be easily recharged by boiling until all crystals are dissolved. Therefore they can be recycled indefinitely.

"Sodium acetate is inexpensive, and is usually purchased from chemical suppliers, instead of being synthesized in the laboratory. It is sometimes produced in a laboratory experiment by the reaction of acetic acid with sodium carbonate, sodium bicarbonate, or sodium hydroxide. These reactions produce aqueous sodium acetate, and water. Carbon dioxide is produced in the reaction with sodium carbonate and bicarbonate, and it leaves the reaction vessel as a gas (unless the reaction vessel is pressurized):

CH3–COOH + Na+[HCO3]– _ CH3–COO– Na+ + H2O + CO2

"This is the basis for the well-known "volcano" reaction between baking soda and vinegar. 84 grams of sodium bicarbonate (baking soda) react with 750 ml of 8% vinegar to make 82 g sodium acetate in water. By boiling off most of the water, one can refine either a concentrated solution of sodium acetate or crystals."

I also gave my students the following question:

Given that the solubility of sodium acetate (Molar mass=82g/mol) is 76 grams per 100 grams of water.?

What is "super saturated?"

a) 8.5 moles of sodium acetate dissolved in 1L of water

b)1.8 moles of sodium acetate dissolved in 300 ml of water

c)5.5 moles of sodium acetate dissolved in 500 ml of water

d) 1.2 moles of sodium acetate dissolved in 200 ml of water

e) all of them

The reason for this question is to help students think logically to a conclusion.  To answer the question, you need to calculate how many moles of sodium acetate will saturate 1000 grams of water. (Since 1 gram of water occupies a volume of 1 ml, we are determining how many moles of sodium acetate it takes to saturate a liter of water.)  Then, you need to see which answer results in MORE sodium acetate in the water (in other words, "supersaturated").

How do you know the amount of sodium acetate to make?  Based on experiments shown in class, students should have about 25 ml of supersaturated solution in a pouch about 4 inches long.  Using the answer to the problem, above, my students are guided to determine how much sodium acetate is needed for 25 ml.  Set up a "proportion" equation as shown in class.  Then, use the concept of stoichiometry (i.e. -- mole ratio) to determine how much bicarbonate of soda is needed to produce it, and how much acetic acid is needed to react with the bicarb.  Remember, there is only 5% acetic acid in vinegar, so the amount of vinegar to be used must be calculated.   

Once the acetic acid and bicarb have reacted, a clear solution results.  If it is cloudy, then not all of the bicarb has reacted and the acid was the "limiting reagent."  Remember though that bicarb is somewhat soluble in water so you need to add enough acetic acid to react with that amount, as well as with the undissolved bicarb.  You can't just look for a clear solution. 

Today is 26 March 2010 -- Solution Chemistry Exam

This exam covers solution chemistry.  Partial answers are provided.  You will need to follow the logic and complete the problems for credit.   Remember that "(aq)" means aqueous (e.g. -- "in water"). 

i.  "Define Molarity"

YOU Check the text definition.

ii. Define "Molality"

YOU Check the text definition.

iii.  Define "parts per million."

YOU Check the text definition.

1. A 0.750 L aqueous solution contains 90.0 g of ethanol, C2H5OH. Calculate the molar concentration of the solution in mol/Liter

Want molarity (M) = moles/liter.  Moles = grams/atomic wt.  So M = moles/0.75 liter

2. What mass of NaCl is dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M?

Want mass NaCl.  M = moles/liter = 0.364 molar = moles/0.152 Liter.  Solve for moles.

Then, since moles = mass/At. Wt. Rearrange for (At. Wt.)(moles) = mass

3. What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution?

Want mass  C6H12O6.  M = moles/liter  0.248 M = moles/0.325 liter

mass = moles x at. wt.

4. A mass of 98 g of sulfuric acid, H2SO4, is dissolved in water to prepare a 0.500 M solution. What is the volume of the solution?

Want Volume.  Find moles; moles = mass H2SO4/At. Wt.  Then Volume = moles/0.5 M

5. A solution of sodium carbonate, Na2CO3, contains 53.0 g of solute in 215 mL of solution. What is its molarity?

Want molarity.  Find moles; moles = 53 g Na2CO3/At. Wt Na2CO3. 

Then M = moles/0.215 Liter

6. What is the molarity of a solution of HNO3 that contains 12.6 g of solute in 5.00 L of solution?

Want molarity.  Find moles; moles = 12.6 g HNO3/At. Wt. HNO3.  Then M = moles/5.0 liter.

7. What mass of copper(II) nitrate, Cu(NO3)2, is present in 50.00 mL of a 4.55 x 10-3 M aqueous solution?

Want mass.  M = moles/liter; 4.55 x 10-3 M = moles/0.050 liter.  Then mass = moles x at. wt. of Cu(NO3)2

8.  What volume of 0.778 M Na2CO3 (aq) solution should be diluted to 150.0 ml with water to reuduce its concentration to 0.0234 M Na2CO3 (aq)?

Want volume.  First find moles: M = moles/Liter.  Rearrange for (0.778 M Na2CO3) x 1 liter = moles.  Then divide by 1000 ml/liter for moles per milliliter.  Then calculate how many moles needed for dilute solution; M = moles/liter = 0.0234 M Na2CO3 = moles/0.150 liter.  Since M means "moles per liter," the liter unit cancels out when you do the math.  You have moles needed which you divide by the answer to the first part.  That is in moles per milliliter.  Moles unit cancels and you have answer in milliliters.

9.  A 25 ml sample of HCl (aq) is diluted to a volume of 500 ml.  If the concentration of the diluted solution is 0.085 M, what was the concentration of the original solution?

Want Molarity.  Dil HCl = 0.085 M = moles HCl/liter.  Rearrange to find moles HCl.  Then find how many moles in 0.5 liter:  0.5 liter x moles/liter = moles available.

Original concentration = M = moles available/orig. vol. = concentration, M