•In the section entitled “Chemical Reactions” we discussed precipitation and acid/base reactions within a water-based medium (i.e., aqueous solutions). Water is by far the most important solvent for chemical reactions. A high percentage of naturally occurring chemical reactions occur with water as the solvent. Water has several important properties that make it an interesting solvent: In particular we consider its polarity:
•We have already seen how the polarity of water affects the solution of ionic compounds. What happens with aqueous solutions of molecular compounds? Two situations arise in this case: (1) if the molecular compound is a solid (e.g. table sugar (sucrose), vitamin C (ascorbic acid), wax etc.) or (2) the molecular compound is is a liquid (e.g., methanol, benzene, carbon tetrachloride etc.)
•Case I: Solid Molecular Compound
•If the molecules in the solid are polar then it is possible that the solid can dissolve in water. On the other hand if the molecules of the solid are nonpolar then there is no chance that the solid will dissolve in water. This is easy to understand because if the solid has polar molecules then there is an electrostatic attraction between the water and solid molecules. This attraction can lead to solution of the solid (similar to the solution of ionic compounds). On the other hand, if the solid is nonpolar, then water molecules are attract only to one another and basically ignore the solid. consequently the solid does not dissolve. These situations are shown in the figures next slide.
•Water is represented by the red/black ovals and the polar molecular solid by the gray/white ovals. The dashed lines represent electrostatic attractions. The electrostatic attractions between water and the polar solid make the solution of the solid possible.
•Same as above except the solid is nonpolar. Note that there are no attractions between water and solid. Thus the solid will not dissolve.
•Case 2: Liquid Molecular Compound
•If the molecular compound is a liquid two things can happen when it comes into contact with water: it can either mix with the water or it can separate from the water. When two liquids mix they are said to me “miscible”. When two liquids not not mix (separate) they are said to be “immiscible”. When will molecular liquids be miscible and when will they be immiscible? The answer is similar to the one for solids above: If the liquid molecules are polar (e.g. methanol (rubbing alcohol)), then it will be miscible with water. If the liquid molecules are nonpolar (e.g., oil), then the liquid will be immiscible with water. These situations are shown in the figures next slide.
•When both liquid molecules are polar then they can attract one another – which leads to mixing (miscibility)
•When the molecular liquid is nonpolar, then the water molecules attract only one another while ignoring the nonpolar liquid. the result is that the two liquids are immiscible.
•For liquids we come to the conclusion that polar-polar liquids are miscible, polar-nonpolar liquids are immiscible and,as an obvious extension, nonpolar-nonpolar liquids are miscible. This observation leads to the often-said statement “like dissolves like”.
•Solubility of Ionic Compounds in Aqueous Solutions
•When a salt is placed in an aqueous medium (i.e. water) a “tug-of-war” immediately begins to take place between the polar water molecules and the ions in the salt. The forces in the tug-of-war are electrostatic and between the ion-ion attractions and the water-ion attractions. If the water-ion attraction win the war then the salt is soluble. If the ion-ion attractions win then the salt is insoluble. Consider table salt (NaCl), it is highly soluble and its solubility can be written as
•At the atomic level the ions in the salt are being “solvated” by the water molecules as the animation below suggests (place mouse over the image). Note the orientation of the water molecules to the positive and negative ions. The cage of water molecules around the solvated ion is known as a “hydration” cage.
•Aqueous Solutions: Concentration
Most chemical reactions reaction occur in a liquid solvent/solute (i.e., solution) environment. Typically the solute(s) in a solution will be the reactants. Since stoichiometric calculations require amounts of reactants, we need a way to express amounts of reactants when they are in solution. The way chemists do this is through the concept of concentration. Concentration is a way to express the amount of solute per unit amount of solution/solvent. There are several ways that concentration can be expressed.
Ppm (parts per million) Volume/Volume Percent
Ppb (parts per billion) Percent By Mass
Weight/Weight Percent Molarity
•Concentration: ppm and ppb
•Parts per million (ppm) and parts per billion (ppb) are examples of expressing concentrations by mass. These units turn out to be convenient when the solute concentrations are very small (almost trace amounts). For example, if a solution has 1 ppm solute this would mean that 1 g of solution would have one “millionth” gram of solute. Equivalently, 1 kg of this solution will have 1 mg of solute etc.. By definition we have:
•For example, suppose a 155.3 g sample of pond water is found to have 1.7×10-4 g of phosphates. What is the concentration of phosphates in ppm?
•A similar procedure would be followed to calculate ppb. In the above example the pond water would be 1,100 ppb.
•Now suppose we have 400 g sample of pond water and it has a concentration of 3.5 ppm dissolved nitrates. What is the mass of dissolved nitrates in this sample?
•Concentration: Weight/Weight %
•This concentration unit is similar to ppm or ppb except it focuses on the solute as a percent (by mass) of the total solution. It is appropriate for relatively large solute concentrations. by definition we have
•As an example consider 5 g sugar dissolved in 20 g of water. What is the w/w% concentration of sugar in this solution?
•Now suppose we have 450 g of an NaCl solution that is 35 NaCl w/w %. What is the mass of NaCl ?
•Concentration: Weight/Volume %
•Another, somewhat less often used, concentration unit is weight/volume percent (w/v %). It is defined as
•For example, suppose we dissolve 1.2 g of NaCl in enough water to make 160 mL of (saline) solution, what is the w/v % of NaCl?
•We interpret this number as 0.75 g of NaCl in 100 mL of solution. 500 mL of 0.75 w/v % NaCl solution would contain 0.75 x 5 g = 3.75 g of NaCl.
•Concentration: Volume/Volume %
•When the solute is a liquid sometimes it is convenient to express its concentration in volume/volume percent (v/v %). The definition of v/v % is
•Wine has about 12 mL of alcohol (ethanol) per 100 mL of solution. Wine would have the following v/v % alcohol content
•Alcoholic beverages often have alcohol content cited as “proof”. The proof value is twice the v/v % value. Therefore the above wine would be 24 % proof.
•Molarity is the most useful concentration for chemical reaction in solution because it directly relates moles of solute to volume of solution. The definition of molarity is
•As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution?
•Now, suppose we have a beaker with 175 mL of a 0.55 M HCl solution. How many moles of HCl is in this beaker?
•As a final example suppose we have a solution of 0.135 M NaCl and we need 1.2 moles of NaCl. What volume of the NaCl solution is required?
•Often it is necessary to take a concentrated solution and dilute it. However, we want to dilute it in a controlled way so that we know the concentration after dilution. The way this is done can be extracted from the following figure of dilution:
•The solute (denoted by red disks) is concentrated in the beaker on the left. Adding water dilutes the solution as shown with the beaker on the right. However, note that although the concentration changes upon dilution, the number of solute molecules does not. In other words the number of moles of solute is the same before and after dilution.
•Since Moles = Molarity x Volume (i.e., moles= M x V) we end up with the following equation relating molarity and volume before and after dilution:
Mi x Vi = Mf x Vf
•Suppose we need 150 mL of 0.25 M NaCl. On the shelf we find a bottle of 2M NaCl. What do we do? The concentrated molarity is Mi and the volume needed is Vi. We need to determine Vi and can do so by rearranging the above equation and doing the resulting calculation:
•Thus, we need 18.8 mL of the 2M NaCl solution, put it in a beaker and add enough water to make 150 mL of solution. the resulting solution will have a molarity of 0.25 M NaCl.
•Example 2: We take 25 mL of a 0.45 M AgNO3 solution and dilute it to 300 mL. What is the molarity of the resulting solution?
•Molarity allows us to do mole/mole stoichiometric calculations when the reaction occurs in solution. Consider the chemical reaction:
•Suppose we want to know what mass of CaCO3 is required to react with 25 mL of 0.75 M HCl. We can solve this problem by using the same mole/mol stoichiometric concepts already discussed.
In a titration a known volume of a reactant (titrant), with known concentration, is slowly added to a vessel with another reactant until the reaction is complete. The point at which the reaction is complete is known as the “end-point”. This situation is depicted in the figure on the left. Typically the end-point is determined visually through either a color change or formation of a precipitate.
Consider the acid/base reaction:
NaOH(aq) + HCl(aq) —-> NaCl(aq) + H2O(l)
Suppose we have flask of HCl but do not know what its concentration is. We are told that we need to know how many moles of HCl there is in this solution. We do a titration experiment: In our laboratory we find a bottle labeled 0.035 M NaOH. We fill a buret with this NaOH solution and put 1 drop of phenolphthalein* indicator in the flask with the HCl. We slowly add the NaOH to the HCl until we notice that the solution in the flask turn a slight pink color. The amount of NaOH added was 27.5 mL. From this information and the balanced reaction above we can determine the moles of HCl: