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
 HNO3
 HClO4
 H2SO4
 All alkali metal hydroxides

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

pH of Strong Acids

Write the dissociation of HCl
HCl(aq) 
+
H (aq)
+
Cl (aq)
Note the oneway arrow!

Let’s figure out the pH of a 12M solution of
concentrated HCl.
pH of Strong Acids

Make a chart for the dissociation
[HCl [H+] [Cl-]
]
i

f
pH of Strong Acids
i

f
[HCl [H+] [Cl-]
]
12
0
0
pH of Strong Acids
i

f
[HCl [H+] [Cl-]
]
12
0
0
-12
+12 +12
pH of Strong Acids
i
[HCl [H+] [Cl-]
]
12
0
0

-12
f
0
+12 +12
12
12
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.
pH of Strong Acids

What is the pH of concentrated sulfuric
acid, 18M?

This is a diprotic acid…that is, it has two
dissociable hydrogen ions

Only one H+ dissociates at a time.

H2SO4(aq)  H+(aq) + HSO41-(aq)

HSO41-(aq)  H+(aq) + SO42-(aq)
pH of Strong Acids

The first dissociation is complete, while the
second is not. The second reaches
equilibrium

Thus, the first dissociation really determines
the pH of a strong, multiprotic acid.

pH = -log [18]

pH = -1.255
pH of Strong Bases

Write the dissociation of NaOH
NaOH(aq) 
+
Na (aq)
+
OH (aq)
Note the oneway arrow!

Let’s figure out the pH of a 6M solution of
NaOH.
pH of Strong Bases

Make a chart for the dissociation
[NaOH] [Na+]
i

f
[OH-]
pH of Strong Bases
[NaOH] [Na+] [OH-]
i

f
6
0
0
pH of Strong Bases
[NaOH] [Na+] [OH-]
i
6
0
0

-6
+6
+6
f
pH of Strong Bases
[NaOH] [Na+] [OH-]
i
6
0
0

-6
+6
+6
f
0
6
6
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
pH of Weak Acids

Write the dissociation of the weak acid
HC2H3O2
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
Note the twoway arrow!

Let’s figure out the pH of a 17.4M solution
of concentrated HC2H3O2.
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.
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
pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i

eq
17.4
0
0
pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i
17.4
0
0

-x
+x
+x
eq
pH of Weak Acids
[HC2H3O2] [H+] [C2H3O2-]
i
17.4
0
0

-x
+x
+x
eq
17.4 - x
x
x
pH of Weak Acids
 Plug
into the Ka expression
 1.76
x 10-5 =
 Test
the 5% rule
x
= 0.0175
[x][x]
[17.4 - x]
pH of Weak Acids
x
= 0.0175 = [H+]
 pH
= -log [0.0175]
 pH
= 1.76
pH of Weak Bases

Write the dissociation of the weak base
NH3
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)
Note the twoway arrow!

When dissociating a weak base, react it with
water to justify the acceptance of the H+
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.
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.
pH of Weak Bases
[NH3] [NH4+] [OH-]
i

eq
15.3
0
0
pH of Weak Bases
[NH3] [NH4+] [OH-]
i
15.3
0
0

-x
+x
+x
eq
pH of Weak Bases
[NH3] [NH4+] [OH-]
i
15.3
0
0

-x
+x
+x
eq
15.3 – x
x
x
pH of Weak Bases
 Plug
into the Kb expression
 1.75
x 10-5 =
 Test
the 5% rule
x
= 0.0164
[x][x]
[15.3 - x]
pH of Weak Bases
x
= 0.0164 = [OH-]
 pOH
= -log [0.0164]
 pOH
= 1.79
 pH
= 14 - 1.79
 pH
= 12.2
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) + C4H4O62- (aq)
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.
pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i

eq
0.333
0
0
pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i
0.333
0
0

-x
+x
+x
eq
pH of Multiprotic Weak Acids
[H2C4H4O6] [H+] [HC4H4O6-]
i
0.333
0
0

-x
+x
+x
eq
0.333 -x
x
x
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.
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.
pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O62-]
i

eq
pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O62-]
i

eq
0.0170
0.017
0
pH of Multiprotic Weak Acids
[HC4H4O61-]
[H+] [C4H4O62-]
i
0.0170
0.017
0

-x
+x
+x
eq
pH of Multiprotic Weak Acids
[HC4H4O61-] [H+] [C4H4O62-]
i
0.0170
0.017
0

-x
+x
+x
eq
0.0170 – x 0.017
+x
x
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.
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
pH of Multiprotic Weak Acids

What are the equilibrium concentrations of
all of the species?
[H2C4H4O6]eq = 0.333 – 0.017 = 0.316 M
[HC4H4O61-]eq = 0.017 – 4.31 x 10-5 = 0.01696 M
[C4H4O62-]eq = 4.31 x 10-5 M