Part 1: Comparing Strength to Concentration
| | Will all acids that have the same concentration also have the same pH?
Click the test tube on the left and perform a short lab to answer this question.
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Why is it that acid solutions of the same concentration have different pH's?
Chemistry explain this fact by stating that acids (and bases) can have different
strengths.
Strength in chemistry is very different than
concentration. We define strength as follows
| | Strength: | Strength is defined as the degree of ionization or dissociation of the acid or base in aqueous solution. It can be measured using the equilibrium constant expression for the dissolving of the substance in water. The degree of ionization is the extent to which the acid, or ionic solids become ions. The equilibrium constant expression (Keq) for acid are termed Ka and for bases termed Kb. Tables of acid and base strength are listed in many textbooks. |
A. Comparing strong to weak acids
Strong acids are completely dissociated (100%). If the acid is less than 100% dissociated it is termed a weak acid. The kA of these strong acids is very high. (> 20). The following is a
table of acids indicating their relative strengths.
When chemists write the dissolving of
strong acids in water they only show a one way arrow.
| Example: | HCl (l) H + (aq) + Cl - (aq) |
| | - indicates the acid is completely ionized (100%)
- kA ~ 1.3 * 10 6
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Weak acids are partially ionized or dissociated and therefore form few ions. The kA values of these acids are small in comparison to strong acids. Chemists use a double arrow to indicate their dissolving in water.
| Example: | CH 3COOH (aq) CH3COO - (aq) + H + (aq) |
| | - indicates that acid is partially ionized ( in this case 1.3%)
- kA = 1.8 * 10 -5
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| | Concentration: | Concentration is a ratio of the amount of solute to the amount of solvent in a solution. Measures of concentration can include PPM, ppb, % , and molarity. |
B. Comparing concentrated to dilute solutions
Concentrated and dilute are relative terms but chemists tend to agree that:
Concentrated acids or bases have high molarities usually in excess or 6 mol/l.
Dilute acids and bases have smaller molarities usually less than 6 mol/l.
click eye to view answers
Part 2 : Calculating concentrations of acid and base solutions How do we calculate the acidity of an acid or base in an aqueous solution? In other words how do we calculate the [H+] and
[OH-] of an acid or base?
To answer this question we have to consider what happens to the water when an acid or base is added to it.
When acids and bases are dissolved into water to form solutions they disturb an equilibrium of ions that already exists in the water. The water equilibrium forms because water ionizes to a small degree to form H30+ and OH- ions, according to the following equilibrium.
2H2O (l) H30 + (aq) + OH - (aq) |
Only about 1 molecule in a billion ionizes in this way and therefore there are so few ions in water that it does not conduct a measurable current (unless you use very sensitive instruments).
Because the reactions are reversible and an equilibrium exists we can write an equilibrium constant expression (Kc ) for the above equation. It is:
| K = [H3O + (aq)] * [OH- (aq)]
[H2O]2 |
Since water is a liquid the [H20] does not change. Since the water can't change concentration this term is not included in the above expression and the equation becomes:
| Kw = [H3O + (aq)] * [OH- (aq)] |
Rearranging the above formula we can produce two others, one to calculate [H3O + ] and one to calculate [OH- (aq)].
| [H3O + ] | = | Kw | | [OH- (aq)] | = | Kw |
| | | [OH- ] | | | | [H3O + ] |
This equilibrium constant is labeled Kw and its value for pure water at 25 0C has been calculated. As with other equilibrium constants the units may be dropped. The value of Kw has been calculated as;
Calculating the [H3O +] in neutral water
Let's examine the ionization of water.
| 2H2O(l) | ------> | H3O+ (aq) | + | OH- (aq) |
The equilibrium constant expression is Kw = [H3O +] [OH-]
1) The mole ratio of H3O + to OH- is 1:1 , Therefore [H3O +] = [OH-]
2) Since the concentrations are the same let us give them a value of x .
3) The Kw expression becomes K w = (x)(x) or x2
4) Placing in the value of Kw we get
1.0 x 10 -14 = x2 , or x = 1.0 x 10 -7
5) Therefore [H3O +] and [OH-] = 1.0 x 10 -7 mol/l in neutral water
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Calculating [H3O +] in strong acids solutions
Chemists use the fact that an equilibrium exists in water to calculate the [H
3O +] and [OH-] in acid and base solutions
What happens when strong acids ( a source of H3O+ ions) or strong bases ( a source of OH- ions) are added to water?
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| | Answer:
The addition of H3O+ or OH- ions disturbs the water's equilibrium. We can predict the changes that will occur using Le Chatelier's principle. The following presentation illustrates how the water equilibrium reacts.
Click on the flash symbol to watch a presentation on how the equilibrium reacts.
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Using the above flash presentation as a guide predict what will happen when a strong base ( source of OH- ions ) is added to water
Answer: Water equilibrium and strong bases
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