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<h1><!--mstheme--><font color="#000099"><img src="../../media/sol/soltitle.gif"
alt="solutionstitle.gif" width="220" height="42"><!--mstheme--></font></h1>
</center>
<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#type">Types of Solutions</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#ar">Molarity</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#al">Molality</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#mole">Mole Fraction</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a>
<!--msthemelist--></table><!--mstheme-->
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<a name="type"></a>
<center>
<img src="../../media/sol/typosol.jpg" alt="Types of Solutions" width="364" height="61">
</center>
<br>
<p>You may be asking yourself why solutions are necessary. Well, they
serve a lot of practical purposes. For instance, most chemical reactions
are run in solution. Solutions also usually have a lower melting and
freezing point than just a substance by itself.</p>
<br>
<p>Let's say that we have a handful of salt crystals, NaCl. When this salt
dissolves in water, there is a uniform dispersion of ions in this water.
What we have just formed is called a solution. A solution is defined as a
homogeneous mixture of two or more substances, consisting of ions or
molecules. A solution can be a solid, liquid, or gas. A solution
results from the combination of a solute and a solvent. A solute is, in
the case of a gas or solid dissolved in liquid, the gas or solid, but in
other cases it is the part that has the smaller amount. A solvent is, in
the case of a gas or solid dissolved in liquid, the liquid, but in other
cases it is the part that has the larger amount. In the example
concerning the salt, the solvent would be water and the salt would be the
solute.</p>
<br>
<p>So that we can understand the concept of solubility, let's consider the
process of dissolving sodium chloride in water. We know that NaCl is an
ionic compound, and that it dissolves in water as Na<sup>+</sup> and
Cl<sup>-</sup> ions. When we mix 50 grams of salt in 100 mL of water we
find something cool going on. The sodium and chlorine ions leave the
surface of the crystal and enter into the solution. These ions then move
about at random in the solution and occasionally bump into another
crystal and stick, returning to their crystalline state. As time passes,
more sodium chloride continues to dissolve into the solution, which means
that the ion concentration increases. This means that eventually more ions
are going to bump into each other and return to their crystalline state.
Over time, a dynamic equilibrium will be reached in which the rate at
which ions leave the crystals is equivalent to the rate at which ions
collide to form a crystal.</p>
<br>
<p>At this equilibrium, no more salt appears to dissolve. We have reached
what we call a saturated solution. A saturated solution is a solution
that is in equilibrium with respect to a given dissolved substance. An
unsaturated solution is a solution that is not in equilibrium with respect
to given dissolved substance, leaving room for more substance to
dissolve. The solubility of a substance is the amount that dissolves in a
given quantity of water at a given temperature to give a saturated
solution. Sometimes, through special circumstances we can obtain a
supersaturated solution. A supersaturated solution is a solution that
contains more dissolved solute than is normally possible at that
temperature.</p>
<br>
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<center>
<a href="#type">Types of Solutions</a> |
<a href="#ar">Molarity</a> |
<a href="#al">Molality</a> |
<a href="#mole">Mole Fraction</a> |
<a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a> |
<a href="#top">Top of Page</a>
</center>
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<a name="ar"></a>
<center>
<img src="../../media/sol/molarity.jpg" alt="Molarity" width="183" height="58">
</center>
<p>Molar Concentration or Molarity is defined as the moles of solute
dissolved in one liter of solution.
<!--mstheme--></font><table boerder="0">
<tr>
<td rowspan=2><!--mstheme-->Molarity = <!--mstheme--></font></td>
<td align=center><!--mstheme--><u>moles of solute</u><!--mstheme--></font></td>
</tr>
<tr>
<td align=center><!--mstheme-->liter of solution<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>Example:
<p>.0678 g of NaCl is placed in a 25.0 ml flask full of water. When the
NaCl dissolves, what is the molarity of the solution?
<p>.0464 M NaCl
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<center>
<a href="#type">Types of Solutions</a> |
<a href="#ar">Molarity</a> |
<a href="#al">Molality</a> |
<a href="#mole">Mole Fraction</a> |
<a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a> |
<a href="#top">Top of Page</a>
</center>
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<a name="al"></a>
<center>
<img src="../../media/sol/molality.jpg" alt="molality" width="172" height="58">
</center>
<p>The molality of a solution is the moles of solute per kilogram of
solvent.
<!--mstheme--></font><table boerder="0">
<tr>
<td rowspan=2><!--mstheme-->molality = <!--mstheme--></font></td>
<td align=center><!--mstheme--><u>moles of solute</u><!--mstheme--></font></td>
</tr>
<tr>
<td align=center><!--mstheme-->kilogram of solvent<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>Example:
<p>.20 mol of ethylene glycol / 2.0 Kg of solvent = .10 m ethylene glycol
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<center>
<a href="#type">Types of Solutions</a> |
<a href="#ar">Molarity</a> |
<a href="#al">Molality</a> |
<a href="#mole">Mole Fraction</a> |
<a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a> |
<a href="#top">Top of Page</a>
</center>
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<a name="mole"></a>
<center>
<img src="../../media/sol/molefraction.jpg" alt="Mole Fraction" width="295" height="56">
</center>
<p>Mole fraction is the number of moles of a substance divided by the
total
number of moles in the solution.
<!--mstheme--></font><table boerder="0">
<tr>
<td rowspan=2><!--mstheme-->Mole Fraction = <!--mstheme--></font></td>
<td align=center><!--mstheme--><u>moles of substance A</u><!--mstheme--></font></td>
</tr>
<tr>
<td align=center><!--mstheme-->Total Moles of solution<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>Example:
<p>If there is 4.0 moles of HCl in 12 moles of solution then the mole
fraction of HCl would be 1/3. When all mole fractions are added together
the total is always 1.
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<center>
<a href="#type">Types of Solutions</a> |
<a href="#ar">Molarity</a> |
<a href="#al">Molality</a> |
<a href="#mole">Mole Fraction</a> |
<a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a> |
<a href="#top">Top of Page</a>
</center>
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<a name="kfkb"></a>
<center>
<img src="../../media/sol/kfkb.jpg" alt="K<sub>f</sub> and K<sub>b</sub>" width="172" height="63">
</center>
<p>K<sub>f</sub> and K<sub>b</sub> are the freezing point depression
constant and boiling point elevation constant. When a solute
is added to a solvent, the boiling point of the solution is always greater
than the boiling point o the pure solvent. Adding a solute also lowers
the freezing point.
<!--mstheme--></font><table border=1 bordercolordark="#000000" bordercolorlight="#000000">
<tr>
<th><!--mstheme-->Solvent<!--mstheme--></font></th>
<th><!--mstheme-->Formula<!--mstheme--></font></th>
<th><!--mstheme-->Melting Point (<sup>o</sup>C)<!--mstheme--></font></th>
<th><!--mstheme-->Boiling Point(<sup>o</sup>C)<!--mstheme--></font></th>
<th><!--mstheme-->K<sub>f</sub>(<sup>o</sup>C/m)<!--mstheme--></font></th>
<th><!--mstheme-->K<sub>b</sub>(<sup>o</sup>C/m)<!--mstheme--></font></th>
</tr>
<tr>
<td><!--mstheme-->Benzene<!--mstheme--></font></td>
<td><!--mstheme-->C<sub>6</sub>H<sub>6</sub><!--mstheme--></font></td>
<td><!--mstheme-->5.455<!--mstheme--></font></td>
<td><!--mstheme-->80.2<!--mstheme--></font></td>
<td><!--mstheme-->5.065<!--mstheme--></font></td>
<td><!--mstheme-->2.61<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->Ethanol<!--mstheme--></font></td>
<td><!--mstheme-->C<sub>2</sub>H<sub>5</sub>OH<!--mstheme--></font></td>
<td><!--mstheme-->--<!--mstheme--></font></td>
<td><!--mstheme-->78.3<!--mstheme--></font></td>
<td><!--mstheme-->--<!--mstheme--></font></td>
<td><!--mstheme-->1.07<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->Water<!--mstheme--></font></td>
<td><!--mstheme-->H<sub>2</sub>0<!--mstheme--></font></td>
<td><!--mstheme-->0.000<!--mstheme--></font></td>
<td><!--mstheme-->100.000<!--mstheme--></font></td>
<td><!--mstheme-->1.858<!--mstheme--></font></td>
<td><!--mstheme-->0.521<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>To determine the amount of change in boiling point you will need this
equation:
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>b</sub> = K<sub>b</sub> * m
<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>b</sub> change in boiling point
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->K<sub>b</sub> boiling point elevation constant
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->m molality of solution
<!--msthemelist--></table><!--mstheme-->
<p>Then <img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>b</sub> is added to the normal boiling point of the
pure solvent. Note that the identity of the solute is not important, just
its concentration (expressed in molality). Therefore, boiling point
elevation is a colligative property.
<p>To determine the freezing point of a solution, you need to calculate the
decrease in freezing point caused by the addition of a solute to the
solvent. Use the equation:
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>f</sub> = K<sub>f</sub> * m
<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>f</sub> change in freezing point
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->K<sub>f</sub> freezing point depression constant
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->m molality of solution
<!--msthemelist--></table><!--mstheme-->
<p>Then the <img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>f</sub> is subtracted from the normal freezing
point of the pure solvent. Freezing point depression is also a
colligative property.
<p>Example:
<p>Calculate the boiling point and freezing point of a solution of .30 g
of glycerol (C<sub>3</sub>H<sub>8</sub>O<sub>3</sub>) in 20.0 g of water.
<p>moles glycerol = (.30 g) (1 mole / 92 g) = .0033 moles
molality of solution = .0033 moles / .020 kg = .16 m
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>b</sub> = (.521 <sup>o</sup>C/m)(.16 m) = .083 <sup>o</sup>C
<p>Boiling point = 100.00 + .083 = 100.083 <sup>o</sup>C
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>f</sub> = (1.858 <sup>o</sup>C/m)(.16m) = .30 <sup>o</sup>C
<p>Freezing point = 0.00 <sup>o</sup>C - .30 <sup>o</sup>C = -.30
<sup>o</sup>C
<p>In an ionic solution, the total concentration of ions is important.
Therefore, another factor (i) is included in the equations.
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>b</sub> = K<sub>b</sub> * m * i
<p><img src="../../media/sol/lildelta.gif" alt="delta" width="8" height="10"> T<sub>f</sub> = K<sub>f</sub> * m * i
<p>"i" is the number of ions from each formula unit. In the previous
example, if NaCl had been the solute your change in boiling point and
freezing point would have needed to be multiplied by 2. (i=2 because NaCl
consists of a Na<sup>+</sup> ion and a Cl<sup>-</sup> ion
<img src="../../media/misc/blueline.gif" width=100% alt="blueline.gif">
<center>
<a href="#type">Types of Solutions</a> |
<a href="#ar">Molarity</a> |
<a href="#al">Molality</a> |
<a href="#mole">Mole Fraction</a> |
<a href="#kfkb">K<sub>f</sub> and K<sub>b</sub></a> |
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