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 one-way 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

[HCl]

[H+] [Cl-]

i 12 0 0

f

pH of Strong Acids

[HCl]

[H+] [Cl-]

i 12 0 0

-12 +12 +12

f

pH of Strong Acids

[HCl]

[H+] [Cl-]

i 12 0 0

-12 +12 +12

f 0 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?

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)

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

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.

pH of Strong Bases

Make a chart for the dissociation

[NaOH] [Na+] [OH-]

i

f

pH of Strong Bases

[NaOH] [Na+] [OH-]

i 6 0 0

f

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 two-way 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 17.4 0 0

eq

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

1.76 x 10-5 = [x][x] [17.4 - x]

Plug into the Ka expression

Test the 5% rule

x = 0.0175

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 two-way 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 15.3 0 0

eq

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

1.75 x 10-5 = [x][x] [15.3 - x]

Plug into the Kb expression

Test the 5% rule

x = 0.0164

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) + C4H4O6

2- (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 0.333 0 0

eq

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+] [C4H4O6

2-]

i

eq

pH of Multiprotic Weak Acids

[HC4H4O61-] [H+] [C4H4O6

2-]

i 0.0170 0.017 0

eq

pH of Multiprotic Weak Acids

[HC4H4O61-] [H+] [C4H4O6

2-]

i 0.0170 0.017 0

-x +x +x

eq

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

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?

[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