INTRO CHEMISTRY II

CHEM 1305

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A solution is a homogeneous mixture of two or more substances.

Some types of solutions are familiar.  For example, salt solution is salt dissolved in water.  A gaseous solution we all use is air which is a solution of various gases like oxygen, nitrogen, carbon dioxide, water and other gases.  It is possible to have solid solutions.  Metal alloys are solid solution.  An example is brass which is a solution of copper and zinc.  Two liquids may dissolve to give a liquid solution.  Examples include gasoline, alcoholic beverages, vinegar, bleach etc.

A solution is generally composed of two substances.

The solute is the substance being dissolved and is in smaller quantity.

The solvent is the substance that dissolves the solute and is in larger quantity.

Properties of Solution

Factors That Affect Solubility

What determines if a solute will dissolve in a given solvent?

Henry's Law

Henry's Law -   the solubility of a sas in a liquid is directly proportional to the applied pressure.

This means that higher the pressure, the more gas dissolves in a liquid.

Carbonated soft drinks are examples of a gas (carbon dioxide) dissolved in a liquid (water) at high pressure.  When a soft drink can is opened, the dissolved carbon dioxide escapes ( bubbles). If this can is left open for some time, the drink goes "flat".  All of the carbon dioxide has escaped.

What can be done to increase the rate at which a solute dissolves in a solvent?

Factors That Affect the Rate of Dissolution

Relative Terms for Expressing Solute Concentration

Solubility is the amount of solute that can be dissolved in a given amount of solvent.

Non-quantitaive terms: this means it does not contain any numbers but is based on relative strengths.

Note: All of these are temperature dependent.  A solution that is saturated at 25oC may be unsaturated at 45oC.

Quantitative Terms for Expressing Solute Concentration

Solutions are made up of solute and solvent.  Concentration terms are basically a relationship between the amount of solute present and the amount of solution.  Terms like percent, parts per million (ppm), molarity and normality each describe the concentration of a solute in a solution.  The particular concentration term to use is dependent on the eventual use of the solution.

Molarity (moles of solute per liter of solution) and Normality (equivalents of solute per liter of solution) are generally used for calculations involving chemical reactions (stoichiometry) since moles and equivalents are essentially counting units.  Remember that chemical reactions are the result of ions, atoms or particles "colliding" with each other.

Normality and equivalents are frequently used in volumetric analyses because they simplify the calculations.  They are regularly used in industrial laboratories. However, many chemistry journals do not permit the use of normality and equivalents in their articles and require the use of molarity and moles.

Percent, parts per thousand (ppt), parts per million (ppm) and parts per billion (ppb) are useful for describing concentrations on a mass or volume basis.  Concentrated acids and bases are given in percent concentration.

To convert these types of concentrations to molarity, you must know the density or specific gravity of the solution.  For extremely dilute solutions, the density is taken to be that of water (1.0 g/ml).

Percent Concentration of Solute

The concentration of a solution can be expressed as the percentage of the solute based on the entire solution.  The solution is the total of the solute and the solvent.

The concentration can be expressed as the percent by mass (click here to practice percent by mass).

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Example:  20 grams of NaCl is is mixed with 180 ml of water, what would be concentration as %NaCl by mass in this solution?

First, write the formula for % mass which in this case would be the % NaCl since NaCl would be the solute and solutions are named after the solute.

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Next, make a table of all the data needed for the formula.

% NaCl (w/w) =  ?

mass water  =  180 g   (remember that the density of water = 1.0 g/ml)

mass NaCl  =  20 g

Next, plug the numbers into the formula.

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When dealing with two liquids, the concentration can also be expressed as the percent by volume (click here to practice percent solution).

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Example:  200 ml of antifreeze is is mixed with 0.8 liters of water, what would be % concentration of antifreeze  by volume in this solution?

First, write the formula for % volume which in this case would be the % anitfreeze since the antifreeze would be the solute and solutions are named after the solute.

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Next, make a table of all the data needed for the formula.

% antifreeze (v/v)  =   ?

volume water  =  800 ml  (remember 0.8 liters = 800 ml)

volume antifreeze    =  200 ml

Next, plug the numbers into the formula.

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Sometimes, the concentration is expressed as a combination of the mass of the solute and the volume of the solution.

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Example:  150 ml of ethanol is mixed with 1.5 liters of water, what would be % concentration of ethanol by mass/volume in this solution?

First, write the formula for % mass/volume which in this case would be the % ethanol since the ethanol would be the solute and solutions are named after the solute.

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Next, make a table of all the data needed for the formula.

% NaOH (m/v) = ?

volume water = 1500 ml  (remember 1.5 liters = 1500 ml)

mass ethanol  = 118.5 g  (density = 0.79 g/ml and mass = density x vol)

Next, plug the numbers into the formula.

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The concentration can be expressed as parts by mass of solute per million parts of solution.

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Example:  A drinking water sample has 15 mg of Pb per 500 ml of solution.  Calculate the ppm of Pb in the water.

First, write the formula for ppm which in this case would be the ppm of Pb since lead would be the solute and solutions are named after the solute.

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Next, make a table of all the data needed for the formula. This is a dilute solution so assume the density of the solution is that of water.

ppm Pb  =   ?

mass Pb  =   15 mg

mass solution  =   500 ml  =  500 g  =  500,000 mg

Next, plug the numbers into the formula.

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Molarity (M)

Molarity is often used when more precise measurements of the concentration of solutions are needed and the solutions are being use in reactions and volumetric measurements (click here to practice molarity).

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In the laboratory, the molarity of a solution may have to be calculated from a known mass and volume in order to reproduce this concentration at a later date.

Example:  120 grams of NaOH is diluted to 750 ml.  What would be the molarity of this solution.

First, write the molarity formulas

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Next, make a table of all the data needed for the formula.

M = ?
moles =
liters = 
grams = 120 g
formula mass =

Next, convert ml to liters (750 ml / 1000 = liters)

0.75 liters (750 ml = 0.75 l)

Next, determine the formula mass of NaOH

Na = 23   O = 16   H = 1
NaOH = 40 amu

M = ?
moles =
liters =  0.75
grams = 120 g
formula mass = 40 amu

Next, calculate the moles of NaOH

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Finally, calculate the molarity of NaOH

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Alternately, in the laboratory, a certain volume of a molar solution may have to be made.

Example:  500 ml of a 2.5 M solution of Ca(OH)2 needs to be made.  What would be the mass of Ca(OH)2 needed to dilute to 800 ml.

First, write the molarity formulas

5-molarity.gif (1710 bytes)

5-moles.gif (1800 bytes)

Next, make a table of all the data needed for the formula.

M = 2.5 M
moles =
liters = 
grams = ?
formula mass =

Next, convert ml to liters (500 ml / 1000 = liters)

0.5 liters (500 ml = 0.5 l)

Next, determine the formula mass of Ca(OH)2

Ca = 40.1 x 1  =  40.1
O  = 16.0  x 2  =  32.0
H  =   1.0  x 2  =   2.0

Ca(OH)2 =  74.1 amu

Next, calculate the moles of Ca(OH)2

M = 2.5 M
moles = 1.25 moles
liters = 0.5 l
grams = ?
formula mass = 74.1 amu

Finally, calculate the grams of Ca(OH)2

To make the 2.5 M Ca(OH)2 solution, weigh out 92.6 grams of Ca(OH)2 and dilute with water to 800 ml.

Normality (N)

Normality is often used when more precise measurements of the concentration of solutions are needed and the solutions are being use in reactions of acids and bases.

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 Note that equivalents can be defined in several ways.  Here are just a few.

 

In the laboratory, the normality of a solution may have to be calculated from a known mass and volume in order to reproduce this concentration at a later date.

Example:  296.4 grams of Ca(OH)2 is diluted to 600 ml.  What would be the normality of this solution.

First, write the normality formulas

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Next, make a table of all the data needed for the formula.

N = ?
eq =
liters = 
grams = 296.4 g
formula mass =
eq mass =
number replaceable OH =

Next, convert 600 ml to liters (600ml / 1000 = liters)

0.6 liters (600 ml = 0.6 l)

Next, determine the formula mass of Ca(OH)2

Ca = 40.1 x 1  =  40.1
O  = 16.0  x 2  =  32.0
H  =   1.0  x 2  =   2.0

Ca(OH)2 =  74.1 amu

Next, calculate the number of replaceable OH- in Ca(OH)2

Ca(OH)2 has 2 replaceable hydroxides

Next, calculate the equivalent mass of Ca(OH)2

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N = ?
eq =
liters = 0.6 liters
grams = 296.4 g
formula mass = 74.1 amu
eq mass = 37.05
number replaceable OH = 2

Next, calculate the equivalents of Ca(OH)2

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Finally, calculate the normality of Ca(OH)2

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Alternately, in the laboratory, a certain volume of a normal solution may have to be made.

 

Example:  600 ml of a 2.5 M solution of Ca(OH)2 needs to be made.  What would be the mass of Ca(OH)2 needed to dilute to 600 ml.

First, write the normality formulas

5-normal.gif (1878 bytes)

5-equiv.gif (1966 bytes)

5-eqmass.gif (2284 bytes)

Next, make a table of all the data needed for the formula.

N = 2.5 N
eq =
liters = 
grams = ?
formula mass =
eq mass =
number replaceable OH =

Next, convert 600 ml to liters (600 ml / 1000 = liters)

0.6 liters (600 ml = 0.6 l)

Next, determine the formula mass of Ca(OH)2

Ca = 40.1 x 1  =  40.1
O  = 16.0  x 2  =  32.0
H  =   1.0  x 2  =   2.0

Ca(OH)2 =  74.1 amu

Next, calculate the number of replaceable OH- in Ca(OH)2

Ca(OH)2 has 2 replaceable hydroxides

Next, calculate the equivalent mass of Ca(OH)2

12-norm-1e.gif (2669 bytes)

N = 2.5 N
eq =
liters = 0.6 liters
grams = ?
formula mass = 74.1 amu
eq mass = 37.05
number replaceable OH = 2

Next, calculate the equivalents of Ca(OH)2

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Finally, calculate the normality of Ca(OH)2

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To make the 2.5 M Ca(OH)2 solution, weigh out 55.6 grams of Ca(OH)2 and dilute with water to 600 ml.

Dilution Formula

The dilution formula is used to dilute a concentrated stock solution (these stock solutions are concentrated so as to save room during storage and costs in shipping) to a desired concentration.

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C1  =   the initial concentration
C2  =   the final concentration
V1  =  the initial volume
V2  =  the final volume

 

Example:  A stock solution of hydrochloric acid is has a concentration of 12M (C1).  100 ml (V2) of a 6M HCl (C2) solution needs to be made from the stock solution.  How many ml (V1) of the stock solution needs to be diluted to 100 ml (V2) to make the 6M (C2)  solution?

First using the dilution formula, solve for  V1 which is the volume of the 12M stock solution needed to make the 6M solution.

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Next,  make a table of the variables in the formula.

C1  =   12M
C2  =   6M
V1  =  ?
V2  =  100 ml

Next,  substitute the actual values for the variables in the formula.

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The 50 ml indicates that you would take 50 ml of the 12M HCl solution and dilute it to 100 ml with water to make 100 ml of a 6M  HCl solution.

p-Values

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The value X can be any chemical species or value.    The square brackets around a chemical species indicates molar concentration.    Some examples of p-values are pH,  pOH,  pCl, and  pKa.

The equation pH = -log[H+] means that the pH is equal to the negative logarithm of the molar concentration of the hydrogen ion.

Likewise, the equation pCl = -log[Cl-] means that the pCl is equal to the negative logarithm of the molar concentration of the chloride ion.  

For values such as Ka , the equation pKa = - log(Ka) means that the pKa is equal to the negative logarithm of the acid dissociation constant, Ka.

click here to practice solutions.

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