For any sparingly soluble salt `[M(NH_(3))_(4)Br_(2)] H_(2)PO_(2)`
Given: `lambda_(M(NH_(3))_(4)Br_(2)^(+))^(@) = 400 Sm^(2) - mol^(-1)`.
`lambda_(H_(2)PO_(2)^(-))^(@) = 100 S-m^(2) -mol^(1)`
Specific resistance of saturated `[M(NH_(3))_(4)Br_(2)] H_(2)PO_(2)` solution is `200 Omega-cm`
If solubility product constant of the above salt is `10^(-x)`. What will be the value of `x`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the dissociation of the salt
The sparingly soluble salt \([M(NH_3)_4Br_2]H_2PO_2\) dissociates in solution as follows:
\[ [M(NH_3)_4Br_2]H_2PO_2 \rightleftharpoons [M(NH_3)_4Br_2]^+ + H_2PO_2^- \]
This means that for every mole of the salt that dissolves, we get one mole of \([M(NH_3)_4Br_2]^+\) and one mole of \(H_2PO_2^-\).

### Step 2: Define solubility (s)
Let the solubility of the salt be \(s\) mol/L. Therefore, at equilibrium:
- Concentration of \([M(NH_3)_4Br_2]^+\) = \(s\)
- Concentration of \(H_2PO_2^-\) = \(s\)

### Step 3: Write the solubility product expression
The solubility product constant \(K_{sp}\) is given by:
\[ K_{sp} = [M(NH_3)_4Br_2^+] [H_2PO_2^-] = s \cdot s = s^2 \]
We also know that \(K_{sp} = 10^{-x}\). Therefore, we have:
\[ 10^{-x} = s^2 \tag{1} \]

### Step 4: Calculate the molar conductivity at infinite dilution
The molar conductivity at infinite dilution \(\lambda_{M(NH_3)_4Br_2}^\infty\) is the sum of the molar conductivities of the ions:
\[ \lambda_{M(NH_3)_4Br_2}^\infty = \lambda_{[M(NH_3)_4Br_2]^+} + \lambda_{H_2PO_2^-} \]
Given:
\[ \lambda_{[M(NH_3)_4Br_2]^+} = 400 \, S \cdot m^2 \cdot mol^{-1} \]
\[ \lambda_{H_2PO_2^-} = 100 \, S \cdot m^2 \cdot mol^{-1} \]
Thus,
\[ \lambda_{M(NH_3)_4Br_2}^\infty = 400 + 100 = 500 \, S \cdot m^2 \cdot mol^{-1} \]

### Step 5: Calculate the specific conductivity (k)
The specific resistance (or resistivity) is given as \(200 \, \Omega \cdot cm\). The specific conductivity \(k\) is the reciprocal of the specific resistance:
\[ k = \frac{1}{\text{Specific Resistance}} = \frac{1}{200 \, \Omega \cdot cm} = 0.005 \, S \cdot cm^{-1} \]
To convert \(S \cdot cm^{-1}\) to \(S \cdot m^{-1}\), we multiply by \(100\):
\[ k = 0.005 \times 100 = 0.5 \, S \cdot m^{-1} \]

### Step 6: Calculate solubility (s)
Using the formula:
\[ s = \frac{k \cdot 1000}{\lambda_{M(NH_3)_4Br_2}^\infty} \]
Substituting the values:
\[ s = \frac{0.5 \cdot 1000}{500} = \frac{500}{500} = 1 \, mol/L \]

### Step 7: Substitute s back into the Ksp expression
From equation (1):
\[ 10^{-x} = s^2 = (1)^2 = 1 \]
This can be rewritten as:
\[ 10^{-x} = 10^{0} \implies -x = 0 \implies x = 0 \]

### Final Answer
The value of \(x\) is \(0\).