Chemistry : Chapter 16 : Hydrolysis of Acids

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Hydrolysis of Acids

>> Parts of this equation/concept include:

A -	Strong Acids
B -	Weak Acids
C -	Polyprotic Acids >> Sulfuric Acid >> Weak Polyprotic Acids

A. Strong Acids

There are seven strong acids: hydrochloric (HCl), hydrobromic (HBr), hydroiodic (HI), nitric (HNO₃), perchloric (HClO₄), chloric (HClO₃) and sulfuric (H₂SO₄). Since each of these acids completely ionize in water, for all except sulfuric acid, the concentration of the acid is the concentration of H₃O⁺. Sulfuric acid is a special case of the polyprotic acids detailed in that section.

>> Example 1

What are the pH and pOH of the following solutions?

0.034 M HCl

9.26 M HNO₃

Solution:

HCl + H₂O H₃O⁺ + Cl^–

0.034 mol HCl/L

1 mol H₃O⁺

1 mol HCl

= 0.034 M H₃O⁺

pH = 1.47

pOH = 14.00 – 1.47 = 12.53

HNO₃ + H₂O H₃O⁺ + NO₃^–

[H₃O⁺] = 9.26 M

pH = –0.967

Despite persistent rumors, pH can be less than zero, and will be if the concentration of H₃O⁺ is greater than 1. Since the concentration has three significant figures, the pH has three decimal places.

pOH = 14.00 – (–0.967) = 14.97 (Remember to count decimal places when determining significant figures from subtraction.)

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B. Weak Acids

The extent of reaction is determined by the equilibrium constant. The equilibrium constant, K_a, refers to the reaction of a weak acid with water, making H₃O⁺ as one of the products.

Most weak acids ionize to a very small extent. Under those circumstances, the ionization of the weak acid will not change the concentration of the acid. If the change in concentration is called x, you can assume that any x added to or subtracted from a number is zero. You cannot make this assumption about multiplied xs. This assumption is usually safe to make if K_a is less than 10^–5.

This assumption makes the math significantly easier. Even if the assumption is incorrect, the answer from this assumption is an approximation of the correct answer obtained from the quadratic formula.

The general formula for the equilibrium reaction is

HA + H₂O H₃O⁺ + A^–

>> Example 2

What are the pH and degree of ionization for 0.22 M HNO₂?

Solution:

HNO₂ + H₂O H₃O⁺ + NO₂^–

K_a =

[H₃O⁺][NO₂^–]

[HNO₂]

= 4.5 x 10^–4

The K_a value is from the appendix at the back of the textbook.

The ICE table (Chapter 15) can be used to determine the amount of each substance at equilibrium.

HNO₂ H₃O⁺ NO₂^–

Initial 0.22 0 0

Change –x +x +x

Equilibrium 0.22 – x x x

Using the equilibrium values in the equilibrium constant expression

4.5 x 10^–4 =

(x)(x)

(0.22 – x)

Assuming x is small, the equation becomes

4.5 x 10^–4 =

x²

0.22

Solving for x,

9.9 x 10^–3 = x

To check the assumption that –x does not change the concentration of HNO₂

0.22 – 0.0099 = 0.2101

Rounding to two significant figures, the concentration does change. This is not surprising, since K > 10^–5. So the quadratic equation must be used to solve the equation.

4.5 x 10^–4 =

x²

(0.22 – x)

9.9 x 10^–5 – 4.5 x 10^–4x = x²

0 = x² + 4.5 x 10^–4 – 9.9 x 10^–5

Quadratic:

x =

–4.5 x 10^–4± [(4.5 x 10^–4)² – 4(1)(–9.9 x 10^–5)]^1/2

2

x =

–4.5 x 10^–4± [2.025 x 10^–7 + 3.96 x 10^–4]^1/2

2

x =

–4.5 x 10^–4± [3.96202 x 10^–4]^1/2

2

x =

–4.5 x 10^–4± [0.0199]

2

Since a negative value of concentration is impossible, only the positive root is used.

x = 9.7 x 10^–3

This is pretty close to the approximate value of 9.9 x 10^–3!

[H₃O⁺] = 9.7 x 10^–3 M

pH = 2.01

The degree of ionization (percent ionization) is the amount of acid that is ionized (the change box in the acid column without the sign) divided by the initial amount of acid times 100.

percent ionized =

x

[acid]

100

percent ionized =

9.7 x 10^–3

0.22

100

percent ionized = 4.4%

>> Example 3

What is the pH and percent ionized of 0.56 M HOI?

Solution:

HOI + H₂O H₃O⁺ + OI^–

K_a =

[H₃O⁺][OI^–]

[HOI]

= 2.3 x 10^–11

HOI H₃O⁺ OI^–

Initial 0.56 0 0

Change –x +x +x

Equilibrium 0.56 – x x x

K_a = 2.3 x 10^–11 =

(x)(x)

(0.56 – x)

Assuming x was small,

2.3 x 10^–11 =

x²

0.56

3.6 x 10^–6 = x

Checking the assumption,

0.56 – 3.6 x 10^–6 = 0.56

So

[H₃O⁺] = x = 3.6 x 10^–6 M

pH = 5.44

>> Example 4

The pH of a 0.11 M solution of a weak acid is 4.51. What is its K_a?

Solution:

The general reaction of a weak acid in water is

HA + H₂O H₃O⁺ + A^–

So the K_a expression is

K_a =

[H₃O⁺][A^–]

[HA]

Using the ICE table,

HA H₃O⁺ A^–

Initial 0.11 0 0

Change –x +x +x

Equilibrium 0.11 – x x x

At equilibrium, the pH is 4.51. From pH, the [H₃O⁺] can be calculated:

[H₃O⁺] = 10^–4.51 = 3.1 x 10^–5

According to the ICE table, [H₃O⁺] = x. Using that value for x in the K_a expression,

K_a =

[H₃O⁺][A^–]

[HA]

K_a =

(x)(x)

(0.11 – x)

K_a =

(3.1 x 10^–5)²

(0.11 – 3.1 x 10^–5)

K_a = 8.7 x 10^–9

>> Example 5

What is the K_a of a weak acid that is 1.49% ionized in water at a concentration of 0.19 M?

Solution:

The general reaction of a weak acid in water is

HA + H₂O H₃O⁺ + A^–

So the K_a expression is

K_a =

[H₃O⁺][A^–]

[HA]

Using the ICE table,

HA H₃O⁺ A^–

Initial 0.19 0 0

Change –x +x +x

Equilibrium 0.19 – x x x

The equation for percent ionized is

percent ionized =

x

(initial concentration HA)

100

Solving that equation for x,

x

(0.19)

100 = 1.49%

x = 2.8 x 10^–3

Using that value for x in the K_a expression,

K_a =

[H₃O⁺][A^–]

[HA]

K_a =

(x)(x)

(0.19 – x)

K_a =

(2.8 x 10^–3)²

(0.19 – 0.0028)

K_a = 4.2 x 10^–5

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C. Polyprotic Acids

Polyprotic acids can donate more than one proton. These acids must donate one proton (H⁺) at a time. The equilibrium constant for the first acidic H⁺ is K_a1. The equilibrium constant for the second acidic H⁺ is always much smaller than K_a1 and is called K_a2. If there is a third acidic proton; its equilibrium constant is K_a3.

Because these reactions happen in sequence, they can be solved in the same sequence.

>> Sulfuric Acid

Sulfuric acid is a special case of a polyprotic acid, because the first acidic proton ionizes in a complete reaction

H₂SO₄ + H₂O HSO₄^– + H₃O⁺

The second acidic proton only partially reacts with water but is a rather high K_a

HSO₄^– + H₂O SO₄^2– + H₃O⁺ K_a = 0.012

>> Example 6

What is the pH of 0.034 M H₂SO₄?

Solution:

From the first reaction,

H₂SO₄ + H₂O HSO₄^– + H₃O⁺ So [H₃O⁺] = [HSO₄^–] = 0.034 M

From the second reaction,

HSO₄^– + H₂O SO₄^2– + H₃O⁺

K_a =

[H₃O⁺][SO₄^2–]

[HSO₄^–]

= 0.012

Using the ICE table, the answers from the first reaction become the initial values for the second equilibrium.

HSO₄^– H₃O⁺ SO₄^2–

Initial 0.034 0.034 0

Change –x +x +x

Equilibrium 0.034 – x 0.034 + x x

Because the initial concentration of sulfate was zero, it must increase in concentration. (you can't lose what you don't have.) If the equilibrium shifts to produce sulfate, it must also produce H₃O⁺. Consequently, its change is also positive.

Using the equilibrium line in the K_a expression,

0.012 =

(0.034 + x)(x)

(0.034 – x)

Assuming x is small (it cancels as both +x and –x),

0.012 =

(0.034)(x)

(0.034)

0.012 = x

Obviously, this is not small (0.034 – 0.012 = 0.022). Using the quadratic instead,

4.08 x 10^–4 – 0.012x = 0.034x + x²

0 = x² + 0.046x – 4.08 x 10^–4

x =

–0.046± [(0.046)² – 4(1)(–4.08 x 10^–4)]^1/2

2

x =

–0.046± [(2.116 x 10^–3) + 1.632 x 10^–3]^1/2

2

x =

–0.046± [3.748 x 10^–3]^1/2

2

x =

–0.046± [0.06122]

2

The negative value is impossible, so

x = 0.0076

Recall from the ICE table that

[H₃O⁺] = 0.034 + 0.0076 = 0.0416 = 0.042 M

pH = 1.38

>> Weak Polyprotic Acids

Each equilibrium is solved sequentially, with the equilibrium concentrations of the first reaction becoming the initial concentrations of the next reaction.

If the question asks for pH, only the first equilibrium needs to be solved. The subsequent equilibria will not significantly change the concentration of H₃O⁺.

>> Example 7

What is the equilibrium concentration of each species in an aqueous solution of 0.57 M H₂C₂O₄?

Solution:

The first equilibrium is

H₂C₂O₄ + H₂O HC₂O₄^– + H₃O⁺

K_a =

[HC₂O₄^–][H₃O⁺]

[H₂C₂O₄]

= 5.9 x 10^–2

Using the ICE table,

H₂C₂O₄ H₃O⁺ HC₂O₄^–

Initial 0.57 0 0

Change –x +x +x

Equilibrium 0.57 – x x x

5.9 x 10^–2 =

(x)(x)

(0.57 – x)

Assuming x is small,

5.9 x 10^–2 =

x²

(0.57)

0.18 = x

Obviously, x is not small (0.57 – 0.18 = 0.39).

Using the quadratic,

0.059 =

x²

(0.57 – x)

0.03363 – 0.059x = x²

0 = x² + 0.059x – 0.03363

x =

–0.059± [(0.059)² – 4(1)(–0.03363)]^1/2

2

x =

–0.059± [0.3715]^1/2

2

Discarding the negative value,

x = 0.16

So

[H₂C₂O₄] = 0.57 – 0.16 = 0.41 M

[H₃O⁺] = [HC₂O₄^–] = 0.16 M

For the second equilibrium,

HC₂O₄^– + H₂O H₃O⁺ + C₂O₄^2–

K_a2 =

[H₃O⁺][C₂O₄^2–]

[HC₂O₄^–]

= 6.4 x 10^–5

Using the ICE table,

HC₂O₄^– H₃O⁺ C₂O₄^2–

Initial 0.16 0.16 0

Change –y +y +y

Equilibrium 0.16 – y 0.16 + y y

6.4 x 10^–5 =

(0.16 + y )(y)

(0.16 – y)

Assuming y is small,

6.4 x 10^–5 =

(0.16)(y)

(0.16)

6.4 x 10^–5 = y

Checking the assumption, 0.16 – 6.4 x 10^–5 = 0.16, so the assumption is good.

Therefore

[H₃O⁺] = [HC₂O₄^–] = 0.16 M and [C₂O₄^2–] = 6.4 x 10^–5 M

and

pH = 0.80

Note that the pH is the same regardless of whether or not the second ionization is considered.

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