Calculate the dissociation constant of water at `25^(@)C` from the following data. Specific conductance of `H_(2)O = 5.8 xx 10^(-8) mho cm^(-1), lambda_(H^(+))^(oo) = 350.0` and `lambda_(OH^(-))^(oo) = 198.0 mho cm^(2)`

Explanation

To calculate the dissociation constant of water at 25°C, we will follow these steps:

### Step 1: Understand the given data
We have the following data:
- Specific conductance of water, \( \kappa = 5.8 \times 10^{-8} \, \text{mho cm}^{-1} \)
- Conductivity of \( H^+ \) at infinite dilution, \( \lambda_{H^+}^\infty = 350.0 \, \text{mho cm}^2 \)
- Conductivity of \( OH^- \) at infinite dilution, \( \lambda_{OH^-}^\infty = 198.0 \, \text{mho cm}^2 \)

### Step 2: Calculate the total conductivity at infinite dilution
The total conductivity of water at infinite dilution can be calculated using the formula:
\[ \lambda_{H_2O}^\infty = \lambda_{H^+}^\infty + \lambda_{OH^-}^\infty \]
Substituting the values:
\[ \lambda_{H_2O}^\infty = 350.0 + 198.0 = 548.0 \, \text{mho cm}^2 \]

### Step 3: Relate specific conductance to concentration
The specific conductance \( \kappa \) is related to the conductivity and concentration by the formula:
\[ \kappa = \lambda_{H_2O}^\infty \cdot C \]
Where \( C \) is the concentration of ions produced from the dissociation of water.

Rearranging the formula gives:
\[ C = \frac{\kappa}{\lambda_{H_2O}^\infty} \]

### Step 4: Substitute the values to find concentration
Substituting the values we have:
\[ C = \frac{5.8 \times 10^{-8}}{548.0} \]
Calculating this gives:
\[ C \approx 1.06 \times 10^{-10} \, \text{mol/L} \]

### Step 5: Calculate the concentration of \( H^+ \) and \( OH^- \)
Since water dissociates into equal amounts of \( H^+ \) and \( OH^- \):
\[ [H^+] = [OH^-] = C \approx 1.06 \times 10^{-10} \, \text{mol/L} \]

### Step 6: Calculate the concentration of water
The concentration of water can be approximated as:
\[ \text{Concentration of water} = \frac{1000 \, \text{g/L}}{18 \, \text{g/mol}} \approx 55.5 \, \text{mol/L} \]

### Step 7: Calculate the dissociation constant of water
The dissociation constant \( K_w \) is given by:
\[ K_w = [H^+][OH^-] = C^2 \]
Substituting the value of \( C \):
\[ K_w = (1.06 \times 10^{-10})^2 \approx 1.12 \times 10^{-20} \, \text{mol}^2/\text{L}^2 \]

### Step 8: Final calculation of \( K_w \)
To express \( K_w \) in terms of molarity:
\[ K_w = \frac{1.12 \times 10^{-20}}{55.5} \approx 1.8 \times 10^{-16} \, \text{mol/L} \]

### Conclusion
The dissociation constant of water at 25°C is approximately \( K_w = 1.8 \times 10^{-16} \).

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