Ionic product of water `(K_(w)` is `10^(-14))` at `25^(@)C`. What is the dissociation constant of water and auto protonation constant of water?

Explanation

To solve the question regarding the dissociation constant of water and the auto protonation constant of water at 25°C, we will follow these steps:

### Step 1: Understand the Dissociation of Water
The dissociation of water can be represented by the equation:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
The ionic product of water (\(K_w\)) at 25°C is given as \(10^{-14}\).

### Step 2: Write the Expression for the Dissociation Constant (\(K_d\))
The dissociation constant (\(K_d\)) for the reaction can be expressed as:
\[ K_d = \frac{[H^+][OH^-]}{[H_2O]} \]
Since the concentration of water remains relatively constant, we can substitute \(K_w\) into the equation:
\[ K_d = \frac{K_w}{[H_2O]} \]

### Step 3: Calculate the Molar Concentration of Water
The molarity of pure water at 25°C can be calculated using its density. Assuming the density of water is \(1 \, \text{g/mL}\):
- 1 liter of water weighs approximately 1000 grams.
- The molar mass of water is approximately \(18 \, \text{g/mol}\).

Thus, the molarity of water (\([H_2O]\)) is:
\[ [H_2O] = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol/L} \]

### Step 4: Substitute Values into the \(K_d\) Expression
Now we can substitute the values into the \(K_d\) equation:
\[ K_d = \frac{10^{-14}}{55.56} \]

### Step 5: Calculate \(K_d\)
Perform the calculation:
\[ K_d \approx \frac{10^{-14}}{55.56} \approx 1.8 \times 10^{-16} \]

### Step 6: Determine the Auto Protonation Constant (\(K_{ap}\))
The auto protonation of water can be represented as:
\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]
The expression for the auto protonation constant (\(K_{ap}\)) is:
\[ K_{ap} = \frac{[H_3O^+][OH^-]}{[H_2O]^2} \]
Since \([H_3O^+][OH^-] = K_w\), we can write:
\[ K_{ap} = \frac{K_w}{[H_2O]^2} \]

### Step 7: Substitute Values into the \(K_{ap}\) Expression
Substituting the known values:
\[ K_{ap} = \frac{10^{-14}}{(55.56)^2} \]

### Step 8: Calculate \(K_{ap}\)
Perform the calculation:
\[ K_{ap} \approx \frac{10^{-14}}{3086.49} \approx 3.24 \times 10^{-17} \]

### Final Answers
- The dissociation constant of water (\(K_d\)) is approximately \(1.8 \times 10^{-16}\).
- The auto protonation constant of water (\(K_{ap}\)) is approximately \(3.24 \times 10^{-18}\).