Specific conductance of pure water at `25^(@)C` is `0.58 xx 10^(-7)` mho `cm^(-1)`. Calculate ionic product of wter `(K_(w))` if ionic conductances of `H^(+)` and `OH^(-)`ions at infinite are `350` and `198` mho `cm^(2)` respectively at `25^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between specific conductance and ionic conductance
The specific conductance (κ) of pure water is given as \(0.58 \times 10^{-7} \, \text{mho cm}^{-1}\). The ionic conductance of the ions in water can be expressed as:
\[ \kappa = \lambda_{H^+} [H^+] + \lambda_{OH^-} [OH^-] \]
where \(\lambda_{H^+}\) and \(\lambda_{OH^-}\) are the ionic conductances of \(H^+\) and \(OH^-\) ions, respectively.

### Step 2: Use the ionic conductances
The ionic conductances at infinite dilution are given as:
- \(\lambda_{H^+} = 350 \, \text{mho cm}^2\)
- \(\lambda_{OH^-} = 198 \, \text{mho cm}^2\)

### Step 3: Set up the equation
Since in pure water at equilibrium, the concentration of \(H^+\) ions is equal to the concentration of \(OH^-\) ions, we can denote this concentration as \(s\). Therefore, we can rewrite the equation as:
\[ \kappa = \lambda_{H^+} s + \lambda_{OH^-} s = s (\lambda_{H^+} + \lambda_{OH^-}) \]
Substituting the values:
\[ \kappa = s (350 + 198) = s \times 548 \]

### Step 4: Solve for \(s\)
We can now substitute the value of \(\kappa\) into the equation:
\[ 0.58 \times 10^{-7} = s \times 548 \]
To find \(s\), we rearrange the equation:
\[ s = \frac{0.58 \times 10^{-7}}{548} \]

### Step 5: Calculate \(s\)
Calculating \(s\):
\[ s = \frac{0.58 \times 10^{-7}}{548} \approx 1.06 \times 10^{-9} \, \text{mol/L} \]

### Step 6: Calculate the ionic product of water \(K_w\)
The ionic product of water \(K_w\) is given by:
\[ K_w = [H^+][OH^-] = s^2 \]
Substituting the value of \(s\):
\[ K_w = (1.06 \times 10^{-9})^2 = 1.12 \times 10^{-18} \, \text{mol}^2/\text{L}^2 \]

### Step 7: Final result
Thus, the ionic product of water \(K_w\) at \(25^\circ C\) is approximately:
\[ K_w \approx 1.12 \times 10^{-18} \, \text{mol}^2/\text{L}^2 \]

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