calculating the ph of acids and bases
DESCRIPTION
Calculating the pH of Acids and Bases. Strong vs. Weak. Strong Acids & Bases. Dissociate completely in water Also known as strong electrolytes Electrolytes conduct electricity in aqueous solutions The more ions dissociated…the more electricity conducted. Strong Acids & Bases. HCl HNO 3 - PowerPoint PPT PresentationTRANSCRIPT
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Calculating the pH of Acids and Bases
Strong vs. Weak
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Strong Acids & Bases
Dissociate completely in water Also known as strong electrolytes
Electrolytes conduct electricity in aqueous solutions
The more ions dissociated…the more electricity conducted
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Strong Acids & Bases
HCl HNO3
HClO4
H2SO4
All alkali metal hydroxides
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Weak Acids & Bases
Do not completely dissociate in water The less dissociated they are…the weaker
electrolytes they are Any acid or base not on the aforementioned
list is considered weak Weak bases are often difficult to
recognize…look for the presence of –NH2, the amine group
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pH of Strong Acids
Write the dissociation of HCl
HCl(aq) H+(aq) + Cl-(aq)
Note the one-way arrow!
Let’s figure out the pH of a 12M solution of concentrated HCl.
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pH of Strong Acids
Make a chart for the dissociation
[HCl]
[H+] [Cl-]
i
f
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pH of Strong Acids
[HCl]
[H+] [Cl-]
i 12 0 0
f
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pH of Strong Acids
[HCl]
[H+] [Cl-]
i 12 0 0
-12 +12 +12
f
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pH of Strong Acids
[HCl]
[H+] [Cl-]
i 12 0 0
-12 +12 +12
f 0 12 12
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pH of Strong Acids
If the [H+] is 12M, then we can determine the pH of the solution
pH = -log [H+]
pH = -log [12]
pH = -1.08
Since it dissociates completely and an equilibrium is never reached, you really don’t need to make a chart.
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pH of Strong Acids
What is the pH of concentrated sulfuric acid, 18M?
Only one H+ dissociates at a time.
This is a diprotic acid…that is, it has two dissociable hydrogen ions
H2SO4(aq) H+(aq) + HSO41-(aq)
HSO41-(aq) H+(aq) + SO4
2-(aq)
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pH of Strong Acids
The first dissociation is complete, while the second is not. The second reaches equilibrium
pH = -log [18]
Thus, the first dissociation really determines the pH of a strong, multiprotic acid.
pH = -1.255
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pH of Strong Bases
Write the dissociation of NaOH
NaOH(aq) Na+(aq) + OH-(aq)
Note the one-way arrow!
Let’s figure out the pH of a 6M solution of NaOH.
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pH of Strong Bases
Make a chart for the dissociation
[NaOH] [Na+] [OH-]
i
f
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pH of Strong Bases
[NaOH] [Na+] [OH-]
i 6 0 0
f
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pH of Strong Bases
[NaOH] [Na+] [OH-]
i 6 0 0
-6 +6 +6
f
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pH of Strong Bases
[NaOH] [Na+] [OH-]
i 6 0 0
-6 +6 +6
f 0 6 6
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pH of Strong Bases
If the [OH-] is 6M, then we can determine the pH of the solution
pOH = -log [OH-]
pOH = -log [6]
pOH = -0.778
pH + pOH = 14
pH = 14 – (-0.778) pH = 14.8
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pH of Weak Acids
Write the dissociation of the weak acid HC2H3O2
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
Note the two-way arrow!
Let’s figure out the pH of a 17.4M solution of concentrated HC2H3O2.
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pH of Weak Acids
Since it’s weak, it will reach equilibrium.
Since it will reach equilibrium, it has an equilibrium constant.
The equilibrium constant of a weak acid is called a Ka.
Write the Ka expression for the dissociation of acetic acid.
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pH of Weak Acids
Ka = [H+][C2H3O2-]
[HC2H3O2]
The Ka value for acetic acid is
1.76 x 10-5 M
Make a chart for the dissociation
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pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i 17.4 0 0
eq
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pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i 17.4 0 0
-x +x +x
eq
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pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i 17.4 0 0
-x +x +x
eq 17.4 - x x x
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pH of Weak Acids
1.76 x 10-5 = [x][x] [17.4 - x]
Plug into the Ka expression
Test the 5% rule
x = 0.0175
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pH of Weak Acids
x = 0.0175 = [H+]
pH = -log [0.0175]
pH = 1.76
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pH of Weak Bases
Write the dissociation of the weak base NH3
NH3(g) + H2O(l) NH4+(aq) + OH-(aq)
Note the two-way arrow!
When dissociating a weak base, react it with water to justify the acceptance of the H+
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pH of Weak Bases
Since it’s weak, it will reach equilibrium.
Since it will reach equilibrium, it has an equilibrium constant.
The equilibrium constant of a weak base is called a Kb.
Write the Kb expression for the dissociation of ammonia.
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pH of Weak Bases
Kb = [NH4+][OH-]
[NH3]
The Kb value for ammonia is
1.75 x 10-5 M
Make a chart for the dissociation of a 15.3M solution of NH3 in water.
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pH of Weak Bases
[NH3] [NH4+] [OH-]
i 15.3 0 0
eq
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pH of Weak Bases
[NH3] [NH4+] [OH-]
i 15.3 0 0
-x +x +x
eq
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pH of Weak Bases
[NH3] [NH4+] [OH-]
i 15.3 0 0
-x +x +x
eq 15.3 – x x x
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pH of Weak Bases
1.75 x 10-5 = [x][x] [15.3 - x]
Plug into the Kb expression
Test the 5% rule
x = 0.0164
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pH of Weak Bases
x = 0.0164 = [OH-]
pOH = -log [0.0164]
pOH = 1.79
pH = 14 - 1.79
pH = 12.2
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pH of Multiprotic Weak Acids
Write the dissociations of tartaric acid, H2C4H4O6 (found in cream of tartar)
H2C4H4O6(aq) H+(aq) + HC4H4O61- (aq)
HC4H4O61-(aq) H+(aq) + C4H4O6
2- (aq)
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pH of Multiprotic Weak Acids
Let’s figure out the pH of a 100-mL solution that contains 5.00g of H2C4H4O6. Ka1 is 9.20 x 10-4 and Ka2 is 4.31 x 10-5.
First determine the initial molarity of the tartaric acid.
5g x 1mol x 1 = 0.333M
150.1g 0.1L
Make a chart for the 1st dissociation.
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pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i 0.333 0 0
eq
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pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i 0.333 0 0
-x +x +x
eq
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pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i 0.333 0 0
-x +x +x
eq 0.333 -x x x
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pH of Multiprotic Weak Acids
Write the Ka1 expression
Ka1 = [H+][HC4H4O61-]
[H2C4H4O6]
Plug your values into the expression.
9.20 x 10-4 = [x][x]
[0.333 - x]
Test the 5% rule.
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pH of Multiprotic Weak Acids
Use the quadratic to solve for x.
x = 0.0170 = [H+]
More hydrogen ion will dissociate in the next dissociation…this is just the amount from the first dissociation.
Make a chart for the 2nd dissociation.
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pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O6
2-]
i
eq
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pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O6
2-]
i 0.0170 0.017 0
eq
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pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O6
2-]
i 0.0170 0.017 0
-x +x +x
eq
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pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O6
2-]
i 0.0170 0.017 0
-x +x +x
eq 0.0170 – x 0.017 + x
x
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pH of Multiprotic Weak Acids
Write the Ka2 expression
Ka2 = [H+][C4H4O62-]
[HC4H4O61-]
Plug your values into the expression.
4.31 x 10-5 = [0.0170 + x][x]
[0.0170 - x]
Test the 5% rule.
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pH of Multiprotic Weak Acids
5% rule works!
x = 4.31 x 10-5 = [C4H4O62-]
So the total amount of hydrogen ion is represented by 0.0170 + 4.31 x 10-5…
[H+] = 0.0170431
pH = 1.77
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pH of Multiprotic Weak Acids
What are the equilibrium concentrations of all of the species?
[HC4H4O61-]eq = 0.017 – 4.31 x 10-5 =
[C4H4O62-]eq = 4.31 x 10-5 M
[H2C4H4O6]eq = 0.333 – 0.017 = 0.316 M
0.01696 M