Edvantage Science AP Chemistry 2 WorkbookPLUS Chapter 5.3

5.3

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5.3 The Ionization of Water Warm Up 1. Describe the difference between a weak acid and a strong acid. ________________________________________________________________________________________ ________________________________________________________________________________________ 2. Compare the relative electrical conductivities of 1.0 M HCl, 1.0 M CH3COOH, and distilled water. Explain your reasoning. ________________________________________________________________________________________ ________________________________________________________________________________________ 3. On your Ka table, find the equation for water acting as an acid and reproduce it here. Write the Ka expression for water.

The Ion-Product Constant of Water

In the previous Section, you saw that water can act as either a Brønsted-Lowry weak acid or weak base. Water is amphiprotic and so can form both hydronium and hydroxide ions. In a Brønsted-Lowry equilibrium, we can see how one water molecule donates a proton to another water molecule (Figure 5.3.1). This is called autoionization.

+ H

O

…

H

H

H

O H

O

– H

+

O

H

H

 H3O+(aq) + OH−(aq)

H2O(l) + H2O(l) Figure 5.3.1 Autoionization of water

As with any equilibrium equation, we can write a corresponding equilibrium constant expression: Kw = [H3O+][OH−] where Kw is the water ionization constant or ion product constant.

Remember that the value of an equilibrium constant depends on the temperature.

At 25°C, Kw = 1.00 × 10–14 = [H3O+][OH−]

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Chapter 5 Acid-Base Equilibrium 289

From this information, we can calculate the [H3O+] and [OH−] in water at 25°C. It is obvious from the equilibrium that the [H3O+] = [OH−] at any temperature. Because of this, water is neutral.

H2O(l) + H2O(l)



H3O+(aq) + OH−(aq) x x

Kw = 1.00 × 10–14 = [H3O+][OH−]

= (x)(x) = x2 x = 1.00 × 10−7 M [H3O+] = [OH−] = 1.0 × 10−7 M

Quick Check 1. The autoionization of water is endothermic. If the temperature of water is increased, what happens to the concentration of hydronium ion, hydroxide ion, and Kw?

_______________________________________________________________________________________

_______________________________________________________________________________________ 2. At higher temperatures, how will the concentration of hydronium ion and hydroxide ion compare to each other? _______________________________________________________________________________________ _______________________________________________________________________________________ 3. At 10°C, Kw = 2.9 × 10−15. Calculate [H3O+] and [OH−] in water at this temperature.

Adding Acid or Base to Water

When we add an acid or base to water, we cause the equilibrium present in water to shift in response.

In 1.0 M HCl, HCl, which is a strong acid, thus ionizes 100% to produce 1.0 M H3O+ and 1.0 M Cl− ions: HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)

1.0 M 1.0 M The added hydronium ions cause the water equilibrium to shift left and the concentration of hydroxide to decrease.

Likewise, when a strong base dissolves in water, the added hydroxide ions cause a shift left, and a corresponding hydronium ion concentration to decrease.

An increase in [H3O+] causes a decrease in [OH−], and vise versa. The [H3O+] is inversely proportional to [OH−].

Both H3O+ and OH− ions exist in all aqueous solutions: • If [H3O+] > [OH−], the solution is acidic.

• If [H3O+] = [OH−], the solution is neutral.

• If [H3O+] < [OH−], the solution is basic.

290 Chapter 5 Acid-Base Equilibrium

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Sample Problem 5.3.1 — Calculating [H3O+] and [OH−] in Solutions of a Strong Acid or Strong Base What is the [H3O+] and [OH−] in 0.50 M HCl? Justify that the solution is acidic.

What to Think About

How to Do It

1. HCl is a strong acid and so will ionize 100%. Thus [HCl] = [H3O+].

[H3O+] = 0.50 M

2. The temperature is not specified. Assume 25°C.

Kw = 1.00 × 10–14 = [H3O+][OH−]

3. Substitute into Kw and solve for [OH−].

1.00 × 10–14 = (0.50 M)[OH−] [OH−] = 2.0 × 10–14 M Because the [H3O+] > [OH−], the solution is acidic.

Practice Problems 5.3.1 — Calculating [H3O+] and [OH−] in Solutions of a Strong Acid or Strong Base (Assume the temperature in each case is 25°C.) 1. Calculate the [H3O+] and [OH−] in 0.15 M HClO4. Justify that this solution is acidic.

2. Calculate the [H3O+] and [OH−] in a saturated solution of magnesium hydroxide. (Hint: this salt has low solubility, but is a strong base.)

3. A student dissolved 1.42 g of NaOH in 250. mL of solution. Calculate the resulting [H3O+] and [OH−]. Justify that this solution is basic.

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Chapter 5 Acid-Base Equilibrium 291

Mixing Solutions of Strong Acids and Bases

When we react strong acids with strong bases, we need to consider two factors: solutions will dilute each other when mixed, and then acids will neutralize bases. The resulting solution will be acidic, basic, or neutral depending on whether more acid or base is present after neutralization.

Sample Problem 5.3.2 — What Happens When a Strong Acid Is Added to a Strong Base?

What is the final [H3O+] in a solution formed when 25 mL of 0.30 M HCl is added to 35 mL of 0.50 M NaOH?

What to Think About 1. When two solutions are combined, both are diluted. Calculate the new concentrations of HCl and NaOH in the mixed solution.

2. The hydronium ions and hydroxide ions will neutralize each other. Since there is more hydroxide, there will be hydroxide left over.

Calculate how much will be left over.

3. Use Kw to calculate the hydronium ion concentration from the hydroxide ion concentration.

How to Do It

HCl is a strong acid, so [HCl] = [H3O+] NaOH is a strong base, so [NaOH] = [OH−] [HCl] = 0.30 mol × 0.025 L = 0.125 M L 0.060 L 0.50 mol [NaOH] = × 0.035 L = 0.292 M L 0.060 L + [H3O ]initial = 0.125 M [OH−]initial = 0.292 M [OH−]excess = [OH−]initial – [H3O+]initial

= 0.292 M – 0.125 M

= 0.167 M = 0.17 M Kw = [H3O+][OH−] 1.00 × 10−14 = [H3O+](0.17 M) [H3O+] = 6.0 × 10−14 M

Practice Problems 5.3.2 — What Happens When a Strong Acid Is Added to a Strong Base?

1. Calculate the final [H3O+] and [OH−] in a solution formed when 150. mL of 1.5 M HNO3 is added to 250. mL of 0.80 M KOH.

2. Calculate the mass of solid NaOH that must be added to 500. mL of 0.20 M HI to result in a solution with [H3O+] = 0.12 M. Assume no volume change on the addition of solid NaOH.

3. Calculate the resulting [H3O+] and [OH−] when 18.4 mL of 0.105 M HBr is added to 22.3 mL of 0.256 M HCl.

292 Chapter 5 Acid-Base Equilibrium

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