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Autoionization of Water
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Transcript of Autoionization of Water
Autoionization of Water
Although water is a molecular substance, very low concentrations of hydronium ions and hydroxides ions are formed by autoionization.
Autoinization
H O
H
+ H O
H
H O+ H
H
+H O -
Autoionization of Water
Autoionization: the process in which water spontaneously
forms low concentrations of H+ and OH - ions by proton transfer from one water molecule to another
At any given time only a very small number of water molecules are ionized.• If every letter in our text represented a water
molecule, you would have to look through about 50 texts to find one H3O+!
2 H2O (l) H3O+ (aq) + OH- (aq)
Autoionization of Water
An equilibrium constant expression can be written for the autoionization of water:
2 H2O (l) H3O+ (aq) + OH - (aq)
Kw = [H3O+] [OH -] = [H+ ] [OH- ]
where Kw = ionization constant for water= ion-product constant= 1.00 x 10-14 at 25oC
Autoionization of Water
The value of Kw (and all other equilibrium constants) varies with temperature:Kw = 1.14 x 10-15
at 0oCKw = 1.00 x 10-14
at 25oCKw = 9.61 x 10-14 at 60oC
You should assume 25oC unless otherwise stated in a problem.
Autoionization of Water
By definition:
Neutral solution: [H+ ] = [OH- ]
Acidic solution: [H+ ] > [OH- ]
Basic solution: [H+ ] < [OH- ]
Autoionization of Water
Example: Calculate the [H+ ] and [OH- ] in a neutral solution at 25oC.
Autoionization of Water
Example: What is the [H+ ] at 25oC for a solution in which [OH- ] = 0.010 M.
Autoionization of Water
Example: What is the [OH- ] at 25oC in a solution in which [H+ ] = 2.5 x 10-6 M.
pH
Since the [H+] is usually very small in aqueous solutions, we normally express the [H+] in terms of pH.
pH = - log10 [H+]
Sig Figs and logs: only the digits after the decimal point are significant.
pH
Example: Calculate the pH of a solution with [H+] = 2.52 x 10-5.
pH
Example: Calculate the pH of a solution with [OH-] = 6.5 x 10-5.
pH
The negative log is also used to express the magnitude of other small quantities:
pOH = - log [OH- ]
pH and pOH are related by the following equation that is derived by taking the negative log of the expression for Kw
pH + pOH = 14.00 at 25oC
pH
Example: Calculate the pOH of a solution with [OH - ] = 2.5 x 10-3 M.
pH
Example: Calculate the pH of a solution with [OH - ] = 2.5 x 10-3 M.
Approach 1:
pH
Approach # 2:
pH
Given the pH of a solution, you can also find the [H+] and the [OH-].
Since pH = - log [H+],
[H+] = 10 –pH
Since pOH = - log [OH-],
[OH-] = 10 -pOH
pH
Example: What are the [H+] and [OH-] for a solution with a pH of 2.50 at 25oC?
Strong Acids
Strong acid:Strong electrolyteIonizes completely in aqueous solution
HNO3 (aq) H + (aq) + NO3 – (aq)
The only significant source of H+ ion in an aqueous solution of a strong acid is usually the strong acid.
HNO3 (aq) H + (aq) + NO3 – (aq)
In a 0.05 M HNO3 (aq) solution,
[H+] = 0.05 mol HNO3 x 1 mol H+ = 0.05 ML 1 mol HNO3
Strong Acids
Consequently, the [H+] in a solution of a strong monoprotic acid can be determined easily using the concentration of the strong acid itself.
Strong Acids
Example: What is the pH of a 0.25 M HCl (aq) solution?
Strong Bases
Strong Basestrong electrolyte ionizes completely in aqueous solution
NaOH (aq) Na+ (aq) + OH- (aq)
Common strong basesalkali metal hydroxideshydroxides of Ca, Sr, and Ba
Strong Bases
The pH of an aqueous solution of a strong base can be determined using the concentration of the strong base
NaOH (aq) Na+ (aq) + OH- (aq)
A 0.25 M solution of NaOH has an [OH-] of 0.25 M:
0.25 mol NaOH x 1 mol OH- = 0.25 ML 1 mol NaOH
Strong Bases
The pH of the base solution can then be found in two ways:
Calculate pOH use pH + pOH = 14.00 to determine pH
Calculate [H+]use [H+] [OH-] = 1.00 x 10-14
Then calculate pH
Step 1: Determine [OH-]
Strong Bases
Example: Calculate the pH of a 0.25 M Ca(OH)2 (aq) solution.
Step 1: Determine [OH-]
Strong Bases
Step 2: Calculate pOH
Step 3: Calculate pH
Strong Bases
Example: What is the pH of a solution prepared by mixing 10.0 mL of 0.015 M Ba(OH)2 and 30.0 mL of 7.5 x 10-3 M NaOH?
Strong Bases
Step 1: Find the total [OH-]
Strong Bases
Step 2: Find pOH
Step 3: Find pH
Weak Acids
Most acidic substances are weak acids:partially ionize in solutionthe solution contains an equilibrium
mixture of acid molecules and its component ions
CH3CO2H H+ (aq) + CH3CO2
- (aq)
Weak Acids
The extent to which a weak acid ionizes can be expressed using an equilibrium constant known as the acid-dissociation constant (Ka).
For a general reaction:HX (aq) H+ (aq) + X- (aq)
Ka = [H+][X-]
[HX]
Note: The rules for writing an expression for Ka are the same as those for Kc, Kp and Ksp.
Weak Acids
The magnitude of Ka indicates the tendency of the hydrogen ion in an acid to ionize.
The larger the value of Ka, the stronger the acid is.
The pH of a weak acid solution can be calculated using the initial concentration of the weak acid and its Ka.
Weak Acids
To calculate the pH of a weak acid solution:Write the ionization equilibrium for
the acid.Write the equilibrium constant
expression and its numerical value.Set up a table showing initial
concentration, change, equilibrium concentration.
Substitute equilibrium concentrations into the equilibrium constant expression.
Weak Acids
To calculate the pH of a weak acid solution (cont):Solve for the change in concentration.
Assume that the change in concentration is small (i.e. < 5%) compared to the initial concentration of the weak acid.
Check the validity of previous assumptions.If x/initial concentration x 100% >5.0%, you
must use the quadratic equation to solve for x.
Calculate the final concentrations and pH.
Weak Acids
Example: Calculate the pH of a 0.20 M solution of HCN. Ka = 4.9 x 10-10
Step 1: Write the equation for the ionization.
Step 2: Write the expression for Ka.
Weak Acids
Step 3: Set up a table.
Weak Acids
Step 4: Substitute equilibrium concentrations into the Ka expression.
Step 5: Assume that x << 0.20 M and solve for x.
Weak Acids
Step 6: Check the validity of our assumption.
Weak Acids
Step 7: Substitute value for x into the table to find the [H+].
Weak Acids
Step 8: Calculate the pH using the [H+]
Weak Bases
Many substances behave as weak bases in water.
Weak base + H2O conjugate acid + OH-
The extent to which a weak base reacts with water to form its conjugate acid and OH- ion can be expressed using an equilibrium constant known as the base-dissociation constant (Kb).
Weak Bases
Kb always refers to the equilibrium in which a base reacts with water to form its conjugate acid and OH- ion.
For the reaction:
NH3 (aq) + H2O (l) NH4+ (aq) + OH-
(aq)
Kb = [NH4+] [OH-]
[NH3]
Note: The rules for writing an expression for Kb are the same as those for Kc, Kp and Ksp.
Weak Base
To calculate the pH of a weak base solution:Write the ionization equilibrium for
the base.Write the equilibrium constant
expression and its numerical value.Set up a table showing initial
concentration, change, equilibrium concentration.
Substitute equilibrium concentrations into the equilibrium constant expression.
Weak Bases
To calculate the pH of a weak base solution (cont):Solve for the change in concentration.
Assume that the change in concentration is small (i.e. < 5%) compared to the initial concentration of the weak base.
Check the validity of previous assumption.
Calculate the [OH-] concentration and pOH
Use pOH to calculate pH.
Weak Bases
Example: Calculate the pH of a 0.20 M solution of methylamine, CH3NH2. Kb = 3.6 x 10-4.
Step 1: Write the equation for the ionization.
Step 2: Write the expression for Kb.
Weak Bases
Step 3: Set up a table.
Weak Bases
Step 4: Substitute equilibrium concentrations into the Kb expression.
Step 5: Assume that x << 0.20 M and solve for x.
Weak Bases
Step 6: Check the validity of our assumption.
Weak Bases
Step 7: Substitute value for x into the table to find the [OH-].
Weak Acids
Step 8: Calculate the pOH
Step 9: Calculate the pH