Ionic Product of Water (Kw)

The ionic product of water, also known as the water dissociation constant, is a fundamental concept in chemistry that describes the equilibrium constant for the self-ionization of water. It is crucial for understanding the pH and pOH of aqueous solutions.

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1. Definition

Ionic Product of Water (Kw): The ionic product of water ($K_w$) is the equilibrium constant for the dissociation of water into hydrogen ions (H⁺) and hydroxide ions (OH⁻).

Chemical Equation: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

Expression for $K_w$: $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]$

2. Value of Kw

At 25°C (298 K), the value of $K_w$ is: $K_w = 1.0 \times 10^{-14}$

Temperature Dependence:

• The value of $K_w$ increases with temperature because water ionizes more at higher temperatures.
• For example, at 50°C, $K_w$ is approximately $5.5 \times 10^{-14}$.

3. Relationship with pH and pOH

pH and pOH Relationship:

• The product of the concentrations of hydrogen ions and hydroxide ions in water is constant at a given temperature.
• Therefore, the relationship between pH and pOH is given by: $\text{pH} + \text{pOH} = 14$

Example Calculation:

• If the pH of a solution is 4, then: $\text{pH}=\mathrm{4 }$${\text{[H}}^{+}]=10^{-4}\text{ M }$
• Using the relationship $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]:$$[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}^{+}]}=\frac{1.0\times 10^{-14}}{10^{-4}}=1.0\times 10^{-10}\text{ M}$
• Therefore, the pOH is: $\text{pOH} = -\log [\text{OH}^-] = -\log (1.0 \times 10^{-10}) = 10$

Sample Question and Answers

Question 1

At 318 K, the ionic product of water is $4 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$, while at 298 K, the ionic product of water ($K_w$) is $1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$. What can be deduced from this change in $K_w$ regarding the dissociation of water at the higher temperature?

Solution:

Step 1: Understanding the Ionic Product of Water

The ionic product of water ($K_w$) is the equilibrium constant for the self-ionization of water:

$$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$$

The value of $K_w$ is given by:

$$K_w = [\text{H}^+][\text{OH}^-]$$

At 298 K (25°C), the ionic product of water is $1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$, and at 318 K (45°C), it is $4 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$.

Step 2: Effect of Temperature on $K_w$

The dissociation of water is an endothermic process, meaning that as temperature increases, the equilibrium shifts to favor more dissociation of water into $\text{H}^+$ and $\text{OH}^-$.

• At 298 K: $K_w = 1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$
• At 318 K: $K_w = 4 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$

Since $K_w$ increases with temperature, we can deduce that more water molecules dissociate into $\text{H}^+$ and $\text{OH}^-$ at higher temperatures.

Step 3: Implications of the Increased $K_w$

At 318 K, the ionic product of water is four times larger than at 298 K. This implies:

1. Higher concentration of $\text{H}^+$ and $\text{OH}^-$ ions in the solution at higher temperatures.
2. The pH of pure water at 318 K will be lower than at 298 K because the concentration of $\text{H}^+$ ions is higher.

Step 4: Calculation of $\text{H}^+$ and $\text{OH}^-$ Concentrations at 318 K

We know:

$$K_w = [\text{H}^+][\text{OH}^-]$$

Since water dissociates in equal amounts of $\text{H}^+$ and $\text{OH}^-$, we have:

$$[\text{H}^+] = [\text{OH}^-]$$

Thus, at 318 K:

$$K_w = [\text{H}^+]^2$$

Substitute the value of $K_w$:

$$4 \times 10^{-14} = [\text{H}^+]^2$$

Taking the square root of both sides:

$$[{\text{H}}^{+}]=[{\text{OH}}^{-}]=\sqrt{4\times 10^{-14}}=2\times 10^{-7}\text{mol}{\text{dm}}^{-3}$$

At 318 K, the concentration of both $\text{H}^+$ and $\text{OH}^-$ is $2 \times 10^{-7} \, \text{mol} \, \text{dm}^{-3}$.

Step 5: Calculation of pH at 318 K

pH is defined as:

$$\text{pH} = -\log [\text{H}^+]$$

At 318 K:

$$\text{pH} = -\log (2 \times 10^{-7}) = 6.70$$

So, the pH of pure water at 318 K is 6.70, which is lower than the pH of 7 at 298 K.

At 318 K, the ionic product of water increases to $4 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$, indicating greater dissociation of water at higher temperatures. This results in a lower pH (6.70) compared to the pH of 7 at 298 K. The increase in $K_w$ demonstrates that water dissociates more at elevated temperatures, producing higher concentrations of hydrogen and hydroxide ions.

Question 2

Calculate the hydroxide ion concentration $[\text{OH}^-]$ at 25°C in a solution whose pH is 5.78. (Given: $K_w = 1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$ at 25°C).

Solution:

Step 1: Calculate the $[\text{H}^+]$ concentration from pH

The pH of a solution is related to the hydrogen ion concentration $[\text{H}^+]$ by the following formula:

$$\text{pH}=-\log [{\text{H}}^{+}]$$

Rearranging this formula to solve for $[\text{H}^+]$:

$$[{\text{H}}^{+}]=10^{-\text{pH}}$$

Substitute the given pH value (5.78):

$$[{\text{H}}^{+}]=10^{-5.78}$$

Calculating:

$$[{\text{H}}^{+}]\approx 1.66\times 10^{-6}\text{mol}{\text{dm}}^{-3}$$

Step 2: Use the ionic product of water ($K_w$) to find $[\text{OH}^-]$

The ionic product of water is expressed as:

$$K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]$$

Given $K_w = 1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$ at 25°C, we can rearrange the equation to solve for $[\text{OH}^-]$:

$$[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}^{+}]<span\; class="vlist"><span\; class="mord"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span></span></span></span><span\; class="vlist-s"></span>}$$

Substitute the known values:

$$[{\text{OH}}^{-}]=\frac{1\times 10^{-14}}{1.66\times 10^{-\mathrm{6}}}$$

Calculating:

$$[{\text{OH}}^{-}]\approx 6.02\times 10^{-9}\text{mol}{\text{dm}}^{-3}$$

The hydroxide ion concentration $[\text{OH}^-]$ at 25°C in the solution with a pH of 5.78 is approximately $6.02\times 10^{-9}\text{mol}{\text{dm}}^{-3}$

Question 3

(i) Calculate the concentrations of $[ \text{H}_3\text{O}^+ ]$, $[ \text{NO}_3^- ]$, and $[ \text{OH}^- ]$ in a 0.009 M HNO₃ solution.
(ii) If 100 cm³ of 0.009M HNO₃ is mixed with 100 cm³ of 0.01 M NaOH solution, will the final solution be acidic or basic?
(iii) Justify your answer to (ii) by calculating the $[ \text{H}_3\text{O}^+ ]$ of the mixture.
Given: $K_w = 1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6}$.

Solution:

(i) Concentrations in 0.009 M HNO₃

Step 1: $\text{HNO}_3$ dissociates completely in water:

For every mole of $\text{HNO}_3$, 1 mole of $\text{H}_3\text{O}^+$ and 1 mole of $\text{NO}_3^-$ are produced.

Thus:

Step 2: Use $K_w$ to calculate $[ \text{OH}^- ]$.

Since:

Substitute $[ \text{H}_3\text{O}^+ ] = 0.009 \, \text{M}$:

Thus, in 0.009 M HNO₃:

(ii) Will the Final Solution Be Acidic or Basic?

We are mixing 100 cm³ of 0.009 M HNO₃ with 100 cm³ of 0.01 M NaOH. To determine if the final solution is acidic or basic, we need to compare the moles of acid and base present.

Step 1: Calculate the moles of HNO₃.

Step 2: Calculate the moles of NaOH.

Step 3: Compare the moles of HNO₃ and NaOH.

• Moles of NaOH = $1 \times 10^{-3} \, \text{mol}$
  
• Moles of HNO₃ = $9 \times 10^{-4} \, \text{mol}$

Since the moles of NaOH exceed the moles of HNO₃, the final solution will be basic.

(iii) Justify by Calculating $[ \text{H}_3\text{O}^+ ]$

To calculate $[ \text{H}_3\text{O}^+ ]$, we need to determine the excess moles of NaOH after neutralization.

Step 1: Calculate the excess moles of NaOH.

Step 2: Determine the concentration of $[ \text{OH}^- ]$ in the final solution.

The total volume of the mixture is $100 \, \text{cm}^3 + 100 \, \text{cm}^3 = 200 \, \text{cm}^3 = 0.200 \, \text{dm}^3$.

Step 3: Use $K_w$ to calculate $[ \text{H}_3\text{O}^+ ]$.

We know:

Substitute $[\text{OH}^-] = 5 \times 10^{-4} \, \text{M}$:

Thus, $[ \text{H}_3\text{O}^+ ] \approx 2 \times 10^{-11} \, \text{M}$, confirming that the solution is basic.

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Final Answers:

• (i) In 0.009 M HNO₃: $[{\text{H}}_3{\text{O}}^{+}]=0.009\text{M, }$$[{\text{NO}}_3^{-}]=0.009\text{M, and }$$[ \text{OH}^-] \approx 1.11 \times 10^{-12} \, \text{M}$.
• (ii) The final solution will be basic.
• (iii) The $[{\text{H}}_3{\text{O}}^{+}]\; of\; the\; final\; solution\; is$$2 \times 10^{-11} \, \text{M}$, confirming its basic nature.

4. Implications and Applications

Acid-Base Reactions:

• The ionic product of water is essential in calculating the pH and pOH of solutions, especially for dilute acids and bases.
• It helps in understanding the behavior of weak acids and bases in aqueous solutions.

Buffer Solutions:

• Knowledge of $K_w$ is used in buffer calculations to maintain a stable pH.

Water Quality:

• Monitoring $K_w$ helps in analyzing the pH and overall water quality in various environmental and industrial contexts.

Summary

The ionic product of water ($K_w$) is a key parameter in aqueous chemistry that reflects the equilibrium between hydrogen ions and hydroxide ions in water. At 25°C, $K_w$ is $1.0 \times 10^{-14}$, and it plays a crucial role in determining the pH and pOH of solutions.

External References

• Acids and Bases - Ionic Product of Water (A-Level Chemistry) - Study Mind
• Khan Academy - Acid-Base Equilibria
• 21.8: Ion-Product of Water - Chemistry LibreTexts

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