4.3 Ionic Product of Water and Calculation of pH and pOH

Water is often thought of as a neutral substance, but it actually undergoes a very slight self-ionization (autoionization). This process is fundamental to acid-base chemistry because it establishes a constant relationship between the concentrations of hydrogen ions (H^+) and hydroxide ions (OH^-) in any aqueous solution. This relationship is quantified by the ionic product of water (K_w). Understanding K_w is the key to calculating pH and pOH.

Definition

• Ionic Product of Water (K_w): The product of the equilibrium concentrations of H^+ (or H_3O^+) and OH^- ions in pure water at a given temperature.
• K_w Expression: K_w = [H_3O^+][OH^-] (or [H^+][OH^-]).
• Value at 25°C (298 K): K_w = 1.0 × 10^-14.
• Unit of K_w: mol² dm^-6.

Core Concept

The core concept is that in pure water, the concentration of H^+ ions equals the concentration of OH^- ions. Since water is a weak electrolyte and only slightly ionized, the equilibrium lies far to the left. At 25°C, the concentration of each ion is 1.0 × 10^-7 M. When multiplied together, they give K_w = 1.0 × 10^-14. In any aqueous solution, no matter what is dissolved in it, the product [H^+][OH^-] must always equal K_w at a given temperature. This allows us to calculate the concentration of one ion if we know the concentration of the other.

Detailed Explanation

The Autoionization of Water

Water undergoes a very small degree of ionization: H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} This can be simplified to: H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}

Recall that the product of the concentration of H^+ and OH^- ions in pure water at room temperature (298 K) is: K_w = [H_3O^+][OH^-] Where K_w is the ionic product or dissociation constant of water.

In pure water, the concentration of H^+ and OH^- ions are equal, i.e., 1.0 × 10^-7 M. So, the value of K_w at room temperature is 1.0 × 10^-14: K_w = (1.0 × 10^-7)(1.0 × 10^-7) = 1.0 × 10^-14

The pKw

Just as pH and pOH are negative logarithms of concentrations, pK_w is the negative logarithm of K_w: pK_w = -log K_w As the value of K_w is 1.0 × 10^-14, so: pK_w = -log(1.0 × 10^-14) = 14

Relationship Between pH, pOH, and pKw

The ionic product of water is related to the dissociation constants pK_a and pK_b of an acid and its conjugate base, respectively: pK_a + pK_b = pK_w = 14 (at 298 K) Or: pH + pOH = pK_w = 14 (at 298 K)

Thus, if the pK_a value of an acid is known, the pK_b value of its conjugate base can be found by subtracting it from 14.

Solved Example from the Textbook

Example 4.3: If the concentration of NaOH in a solution is 2.5 × 10^-4 M, what is the concentration of H_3O^+ ion at 25°C?

Solution:

Data: [OH^-] = 2.5 × 10^-4 M

We can assume room temperature, so: 1.0 × 10^-14 = [H_3O^+][OH^-]

To find the concentration of H_3O^+, solve for [H_3O^+]: [H_3O^+] = 1.0 × 10^-142.5 × 10^-4 [H_3O^+] = 4.0 × 10^-11 M

Final Answer: The concentration of H_3O^+ is 4.0 × 10^-11 M.

Intermediate Level

The ionic product of water, K_w, is the equilibrium constant for the autoionization of water. At 25°C, K_w = [H^+][OH^-] = 1.0 × 10^-14. This relationship holds for all aqueous solutions, regardless of whether they are acidic, basic, or neutral. This means that if you know the concentration of H^+, you can calculate the concentration of OH^-, and vice versa. The pOH is defined as -log[OH^-], and at 25°C, pH + pOH = 14. This allows you to calculate pH from pOH or from [OH^-] directly. The value of K_w is temperature-dependent, so the sum pH + pOH will be different at temperatures other than 25°C.

Important Points

• - K_w = [H^+][OH^-] = 1.0 × 10^-14 at 25°C.
• - In pure water: [H^+] = [OH^-] = 1.0 × 10^-7 M.
• - pH + pOH = 14 at 25°C.
• - K_w is temperature-dependent.
• - pK_w = -log K_w = 14 at 25°C.
• - pK_a + pK_b = pK_w.
• - In any aqueous solution, the product [H^+][OH^-] is always constant at a given temperature.

Remember

• - K_w = 1.0 × 10^-14 at 25°C.
• - [H^+] = K_w / [OH^-] and [OH^-] = K_w / [H^+].
• - pH + pOH = 14 (at 25°C).
• - pK_a + pK_b = 14 (at 25°C).
• - K_w increases with increasing temperature (endothermic).
• - Pure water is neutral at all temperatures, even though its pH changes with temperature.

Common Mistakes

• - Assuming that K_w is always 10^-14. This is only true at 25°C. At other temperatures, the value is different, and the sum of pH and pOH is not 14.
• - Forgetting that K_w applies to all aqueous solutions, not just pure water. Even in an acidic or basic solution, the product [H^+][OH^-] equals K_w.
• - Confusing pH and pOH formulas. pH is for [H^+], pOH is for [OH^-]. Always double-check which one you are calculating.
• - Thinking that a solution with pH > 7 at a temperature other than 25°C is basic. The neutrality of a solution depends on the relative concentrations of H^+ and OH^-, not on the absolute pH value. At a higher temperature, pure water can have a pH of 6.13, but it is still neutral.
• - Forgetting that K_w is temperature-dependent. When performing calculations, be sure to use the K_w value for the given temperature.

Quick Revision

The ionic product of water (K_w) is the product of the equilibrium concentrations of H^+ and OH^- ions in water. At 25°C, K_w = [H^+][OH^-] = 1.0 × 10^-14. In pure water, the concentrations of H^+ and OH^- are equal (1.0 × 10^-7 M). The relationship pK_w = pH + pOH = 14 holds at 25°C. K_w is temperature-dependent and increases with temperature. This relationship allows you to calculate the concentration of H^+ or OH^- if one is known, and to convert between pH and pOH.

60-Second Revision

The ionic product of water, K_w = [H^+][OH^-], is 1.0 × 10^-14 at 25°C. In pure water, [H^+] = [OH^-] = 10^-7 M, so pH = 7. The relationship pH + pOH = 14 holds at 25°C. K_w is temperature-dependent. You can use K_w to calculate [OH^-] from [H^+] (or vice versa) using [H^+] = K_w / [OH^-].

Frequently Asked Questions (FAQs)

Q1. What is the ionic product of water (K_w)? Answer: The ionic product of water (K_w) is the product of the equilibrium concentrations of hydrogen ions (H^+) and hydroxide ions (OH^-) in pure water at a given temperature. It is expressed as K_w = [H^+][OH^-].

Q2. What is the value of K_w at 25°C? Answer: At 25°C (298 K), the value of K_w is 1.0 × 10^-14. This means that in pure water, [H^+] = [OH^-] = 1.0 × 10^-7 M.

• Q3. What is the relationship between pH and pOH at 25°C?
• Answer: At 25°C, the relationship is:
• pH + pOH = 14
• This means that if you know the pH of a solution, you can calculate the pOH, and vice versa.

• Q4. If the concentration of NaOH in a solution is 2.5 × 10^-4 M, what is the concentration of H_3O^+ ion at 25°C?
• Answer: NaOH is a strong base, so [OH^-] = 2.5 × 10^-4 M.
• Using K_w:
• [H_3O^+] = K_w[OH^-] = 1.0 × 10^-142.5 × 10^-4 = 4.0 × 10^-11 M
• So, the concentration of H_3O^+ is 4.0 × 10^-11 M.

• Q5. What is the pKw value at 25°C?
• Answer: The pKw is defined as -log K_w. At 25°C, K_w = 1.0 × 10^-14, so:
• pK_w = -log(1.0 × 10^-14) = 14

Q6. Why does the value of K_w change with temperature? Answer: The autoionization of water is an endothermic process (Δ H > 0). According to Le Chatelier's Principle, increasing the temperature shifts the equilibrium to the right, producing more H^+ and OH^- ions. This increases the value of K_w. Therefore, K_w is temperature-dependent.

• Q7. What is the relationship between pK_a and pK_b?
• Answer: The relationship is:
• pK_a + pK_b = pK_w = 14 (at 25^°C)
• This means that if you know the pK_a of an acid, you can find the pK_b of its conjugate base by subtracting it from 14.

Q8. What is the concentration of H^+ ions in pure water at 25°C? Answer: In pure water at 25°C, [H^+] = 1.0 × 10^-7 M. This is because K_w = [H^+][OH^-] = 1.0 × 10^-14, and in pure water, [H^+] = [OH^-].

• Q9. What is the pOH of a solution with a pH of 5 at 25°C?
• Answer: At 25°C, pH + pOH = 14. So, if pH = 5:
• pOH = 14 - 5 = 9

• Q10. How do you calculate [OH^-] if you know [H^+]?
• Answer: You use the ionic product of water:
• [OH^-] = K_w[H^+]
• At 25°C, if [H^+] = 1.0 × 10^-3 M, then:
• [OH^-] = 1.0 × 10^-141.0 × 10^-3 = 1.0 × 10^-11 M

Multiple Choice Questions (MCQs)

• Q1. What is the ionic product of water (K_w) at 25°C?
• A. 1.0 × 10^-7
• B. 1.0 × 10^-14
• C. 1.0 × 10^-10
• D. 1.0 × 10^-4
• Correct Answer: B
• Explanation: The ionic product of water, K_w = [H^+][OH^-], is 1.0 × 10^-14 at 25°C.

• Q2. In pure water at 25°C, what is the concentration of H^+ ions?
• A. 1.0 × 10^-14 M
• B. 1.0 × 10^-7 M
• C. 1.0 × 10^-10 M
• D. 1.0 × 10^-4 M
• Correct Answer: B
• Explanation: In pure water, [H^+] = [OH^-] = 1.0 × 10^-7 M, since K_w = 1.0 × 10^-14.

• Q3. What is the pKw value at 25°C?
• A. 7
• B. 14
• C. 10
• D. 1
• Correct Answer: B
• Explanation: pK_w = -log K_w = -log(1.0 × 10^-14) = 14.

• Q4. If the concentration of NaOH is 2.5 × 10^-4 M, what is the concentration of H_3O^+?
• A. 2.5 × 10^-4 M
• B. 4.0 × 10^-11 M
• C. 1.0 × 10^-7 M
• D. 1.0 × 10^-14 M
• Correct Answer: B
• Explanation: [H_3O^+] = K_w / [OH^-] = 1.0 × 10^-14 / 2.5 × 10^-4 = 4.0 × 10^-11 M.

• Q5. What is the relationship between pH and pOH at 25°C?
• A. pH + pOH = 7
• B. pH + pOH = 14
• C. pH - pOH = 14
• D. pH × pOH = 14
• Correct Answer: B
• Explanation: At 25°C, pH + pOH = pK_w = 14.

• Q6. If the pH of a solution is 5, what is its pOH at 25°C?
• A. 5
• B. 9
• C. 14
• D. 7
• Correct Answer: B
• Explanation: pOH = 14 - pH = 14 - 5 = 9.

• Q7. What is the relationship between pK_a and pK_b?
• A. pK_a + pK_b = 7
• B. pK_a + pK_b = 14
• C. pK_a - pK_b = 14
• D. pK_a × pK_b = 14
• Correct Answer: B
• Explanation: For a conjugate acid-base pair at 25°C, pK_a + pK_b = pK_w = 14.

• Q8. How does the value of K_w change with increasing temperature?
• A. It decreases.
• B. It increases.
• C. It remains constant.
• D. It becomes zero.
• Correct Answer: B
• Explanation: The autoionization of water is endothermic, so increasing the temperature shifts the equilibrium to the right, increasing K_w.

Q9. What is the concentration of OH^- ions in a solution with a pH of 3 at 25°C? A. 1.0 × 10^-3 M B. 1.0 × 10^-11 M C. 1.0 × 10^-7 M D. 1.0 × 10^-14 M Correct Answer: B Explanation: If pH = 3, then [H^+] = 10^-3 M. Using K_w, [OH^-] = 1.0 × 10^-14 / 1.0 × 10^-3 = 1.0 × 10^-11 M.

• Q10. What is the pOH of a solution with a hydroxide ion concentration of 1.0 × 10^-4 M?
• A. 4
• B. 10
• C. 7
• D. 14
• Correct Answer: A
• Explanation: pOH = -log[OH^-] = -log(1.0 × 10^-4) = 4.

Multiple Choice Questions (MCQs)

• Q1. What is the value of the ionic product of water (K_w) at 25°C?
• A. 1.0 × 10^-7
• B. 1.0 × 10^-14
• C. 1.0 × 10^-10
• D. 1.0 × 10^-4
• Correct Answer: B
• Explanation: K_w = [H^+][OH^-] = 1.0 × 10^-14 at 25°C.

• Q2. What is the concentration of H^+ in pure water at 25°C?
• A. 1.0 × 10^-14 M
• B. 1.0 × 10^-7 M
• C. 1.0 × 10^-10 M
• D. 1.0 × 10^-4 M
• Correct Answer: B
• Explanation: In pure water, [H^+] = [OH^-] = 1.0 × 10^-7 M.

• Q3. What is the pKw value at 25°C?
• A. 7
• B. 14
• C. 10
• D. 1
• Correct Answer: B
• Explanation: pK_w = -log(1.0 × 10^-14) = 14.

• Q4. If [OH^-] = 2.5 × 10^-4 M, what is [H_3O^+] at 25°C?
• A. 2.5 × 10^-4 M
• B. 4.0 × 10^-11 M
• C. 1.0 × 10^-7 M
• D. 1.0 × 10^-14 M
• Correct Answer: B
• Explanation: [H_3O^+] = K_w / [OH^-] = 1.0 × 10^-14 / 2.5 × 10^-4 = 4.0 × 10^-11 M.

• Q5. What is the relationship between pH and pOH at 25°C?
• A. pH + pOH = 7
• B. pH + pOH = 14
• C. pH - pOH = 14
• D. pH × pOH = 14
• Correct Answer: B
• Explanation: At 25°C, pH + pOH = 14.

• Q6. If the pH of a solution is 5, what is its pOH at 25°C?
• A. 5
• B. 9
• C. 14
• D. 7
• Correct Answer: B
• Explanation: pOH = 14 - 5 = 9.

• Q7. What is the relationship between pK_a and pK_b?
• A. pK_a + pK_b = 7
• B. pK_a + pK_b = 14
• C. pK_a - pK_b = 14
• D. pK_a × pK_b = 14
• Correct Answer: B
• Explanation: At 25°C, for a conjugate acid-base pair, pK_a + pK_b = 14.

• Q8. How does the value of K_w change with increasing temperature?
• A. It decreases.
• B. It increases.
• C. It remains constant.
• D. It becomes zero.
• Correct Answer: B
• Explanation: The autoionization of water is endothermic, so K_w increases with temperature.

Q9. What is the concentration of OH^- in a solution with a pH of 3 at 25°C? A. 1.0 × 10^-3 M B. 1.0 × 10^-11 M C. 1.0 × 10^-7 M D. 1.0 × 10^-14 M Correct Answer: B Explanation: If pH = 3, [H^+] = 10^-3 M. Using K_w, [OH^-] = 1.0 × 10^-14 / 1.0 × 10^-3 = 1.0 × 10^-11 M.

• Q10. What is the pOH of a solution with [OH^-] = 1.0 × 10^-4 M?
• A. 4
• B. 10
• C. 7
• D. 14
• Correct Answer: A
• Explanation: pOH = -log(1.0 × 10^-4) = 4.