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Assistant Lecture Aayad Amaar Concentration Most chemical ...

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College of Dentistry<br />

Inorganic Chemistry<br />

<strong>Assistant</strong> <strong>Lecture</strong><br />

<strong>Aayad</strong> <strong>Amaar</strong><br />

<strong>Concentration</strong><br />

<strong>Most</strong> <strong>chemical</strong> reactions occur in a liquid solvent/solute (i.e.,<br />

solution) environment. Typically the solute(s) in a solution will be<br />

the reactants. Since stoichiometric calculations require amounts of<br />

reactants, we need a way to express amounts of reactants when<br />

they are in solution. The way chemists do this is through the<br />

concept of concentration. <strong>Concentration</strong> is a way to express the<br />

amount of solute per unit amount of solution/solvent. There are<br />

several ways that concentration can be expressed. A few of these<br />

are:<br />

ppm (parts per million)<br />

ppb (parts per billion)<br />

Weight/weight percent (w/w %)<br />

Weight/volume percent (w/v %)<br />

Volume/volume percent (v/v %)<br />

molarity<br />

ppm and ppb<br />

Parts per million (ppm) and parts per billion (ppb) are examples<br />

of expressing concentrations by mass. These units turn out to be<br />

convenient when the solute concentrations are very small (almost<br />

trace amounts). For example, if a solution has 1 ppm solute this<br />

would mean that 1 g of solution would have one "millionth" gram<br />

of solute. Equivalently, 1 kg of this solution will have 1 mg of<br />

solute etc... By definition we have:<br />

For example, suppose a 155.3 g sample of pond water is found to<br />

have 1.7x10 -4 g of phosphates. What is the concentration of<br />

phosphates in ppm?


College of Dentistry<br />

Inorganic Chemistry<br />

A similar procedure would be followed to calculate ppb. In the<br />

above example the pond water would be 1100 ppb.<br />

Now suppose we have 400 g sample of pond water and it has a<br />

concentration of 3.5 ppm dissolved nitrates. What is the mass of<br />

dissolved nitrates in this sample?<br />

Weight/Weight %<br />

This concentration unit is similar to ppm or ppb except it<br />

focuses on the solute as a percent (by mass) of the total solution.<br />

It is appropriate for relatively large solute concentrations, by<br />

definition we have<br />

As an example considers 5 g sugar dissolved in 20 g of water.<br />

What is the w/w% concentration of sugar in this solution?<br />

Now suppose we have 450 g of NaCl solution that is 35 NaCl<br />

w/w %. What is the mass of NaCl?<br />

Weight/Volume Percent<br />

The concentration of a solution is defined as the amount of solute<br />

dissolved in a specified amount of solution,<br />

If we define the amount of solute as the mass of solute (in grams)<br />

and the amount of solution in volume units (milliliters),<br />

concentration is expressed as the ratio


College of Dentistry<br />

Inorganic Chemistry<br />

This concentration can then be expressed as a percentage by<br />

multiplying the ratio by the factor 100%. This result in<br />

The percent concentration expressed in this way is called<br />

weight/volume percent, or % (W/V). Thus<br />

Example<br />

Volume/Volume %<br />

When the solute is a liquid sometimes it is convenient to<br />

express its concentration in volume/volume percent (v/v %). The<br />

definition of v/v % is<br />

Wine has about 12 mL of alcohol (ethanol) per 100 mL of<br />

solution. Wine would have the following v/v % alcohol content:<br />

Molarity<br />

Molarity is the most useful concentration for <strong>chemical</strong> reaction<br />

in solution because it directly relates moles of solute to volume of<br />

solution. The definition of molarity is


College of Dentistry<br />

Inorganic Chemistry<br />

As an example, suppose we dissolve 23 g of ammonium chloride<br />

(NH 4 Cl) in enough water to make 145 mL of solution. What is the<br />

molarity of ammonium chloride in this solution?<br />

Now, suppose we have a beaker with 175 mL of a 0.55 M HCl<br />

solution. How many moles of HCl is in this beaker?<br />

As a final example suppose we have a solution of 0.135 M<br />

NaCl and we need 1.2 moles of NaCl. What volume of the NaCl<br />

solution is required?<br />

Dilution<br />

Often it is necessary to take a concentrated solution and dilute<br />

it. However, we want to dilute it in a controlled way so that we<br />

know the concentration after dilution. The way this is done can be<br />

extracted from the following figure of dilution:<br />

The solute is concentrated in the beaker on the left. Adding<br />

water dilutes the solution as shown with the beaker on the right.<br />

However, note that although the concentration changes upon<br />

dilution, the number of solute molecules does not. In other words


College of Dentistry<br />

Inorganic Chemistry<br />

the number of moles of solute is the same before and after<br />

dilution. Since Moles = Molarity x Volume (i.e., moles= M x V)<br />

we end up with the following equation relating molarity and<br />

volume before and after dilution:<br />

Mi x Vi = Mf x Vf<br />

Where i and f stand for initial and final. Suppose we need 150 mL<br />

of 0.25 M NaCl. On the shelf we find a bottle of 2M NaCl. What<br />

do we do? the concentrated molarity is Mi and the volume needed<br />

is Vi. We need to determine Vi and can do so by rearranging the<br />

above equation and doing the resulting calculation:<br />

Thus, we need 18.8 mL of the 2M NaCl solution, put it in a<br />

beaker and add enough water to make 150 mL of solution. The<br />

resulting solution will have a molarity of 0.25 M NaCl.


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