We have taken a saturated solultion of `AgBr, K_(SP)` is `12 xx 10^(-4)`. If `10^(-7) M` of `AgNO_(3)` are added to `1 L` of this solution, find conductivity (specific conductance) of this solution in term of `10^(-7) Sm^(-1)` units.
Given, `lambda_((Ag^(+)))^(@) = 6 xx 10^(-3) Sm^(2) mol^(-1)`,
`lambda_((Br^(-)))^(@) = 8 xx 10^(-3) Sm^(2) mol^(-1)`,
`lambda_((NO_(3)^(-)))^@ = 7 xx 10^(-3) Sm^(2) mol^(-1)`.
Neglect the contribution of solvent.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript and derive the specific conductance of the solution containing AgBr and AgNO3.

### Step 1: Understand the Dissociation of Compounds
AgBr dissociates in solution as follows:
\[ \text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^- \]

AgNO3 dissociates as:
\[ \text{AgNO}_3 \rightleftharpoons \text{Ag}^+ + \text{NO}_3^- \]

### Step 2: Define the Variables
Let:
- \( S \) = solubility of AgBr in mol/L
- \( S' = 10^{-7} \) M (the concentration of AgNO3 added)

### Step 3: Apply the Common Ion Effect
The concentration of Ag\(^+\) ions in the solution after adding AgNO3 will be:
\[ [\text{Ag}^+] = S + S' = S + 10^{-7} \]

The concentration of Br\(^-\) ions will remain as:
\[ [\text{Br}^-] = S \]

### Step 4: Use the Solubility Product Constant (Ksp)
The Ksp expression for AgBr is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
Given \( K_{sp} = 12 \times 10^{-4} \):
\[ 12 \times 10^{-4} = (S + 10^{-7}) \cdot S \]

### Step 5: Solve for S
Expanding the equation:
\[ 12 \times 10^{-4} = S^2 + 10^{-7}S \]

Rearranging gives:
\[ S^2 + 10^{-7}S - 12 \times 10^{-4} = 0 \]

Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 10^{-7} \), \( c = -12 \times 10^{-4} \)

Calculating the discriminant:
\[ b^2 - 4ac = (10^{-7})^2 - 4(1)(-12 \times 10^{-4}) = 10^{-14} + 48 \times 10^{-4} \]
\[ = 48 \times 10^{-4} + 10^{-14} \approx 48 \times 10^{-4} \]

Now substituting into the quadratic formula:
\[ S = \frac{-10^{-7} \pm \sqrt{48 \times 10^{-4}}}{2} \]
Calculating the square root:
\[ \sqrt{48 \times 10^{-4}} = 6.928 \times 10^{-2} \]

Thus:
\[ S = \frac{-10^{-7} + 6.928 \times 10^{-2}}{2} \approx 3.464 \times 10^{-2} \]

### Step 6: Calculate Concentrations
- Concentration of Br\(^-\):
\[ [\text{Br}^-] = S \approx 3.464 \times 10^{-2} \, \text{mol/L} \]

- Concentration of Ag\(^+\):
\[ [\text{Ag}^+] = S + S' \approx 3.464 \times 10^{-2} + 10^{-7} \approx 3.464 \times 10^{-2} \, \text{mol/L} \]

- Concentration of NO3\(^-\):
\[ [\text{NO}_3^-] = S' = 10^{-7} \, \text{mol/L} \]

### Step 7: Calculate Specific Conductivities
Using the provided molar conductivities:
- For Ag\(^+\):
\[ \lambda_{\text{Ag}^+} = 6 \times 10^{-3} \, \text{S m}^2/\text{mol} \]
\[ \text{Specific Conductivity of Ag}^+ = \lambda_{\text{Ag}^+} \times [\text{Ag}^+] = 6 \times 10^{-3} \times 3.464 \times 10^{-2} \approx 2.08 \times 10^{-4} \, \text{S/m} \]

- For Br\(^-\):
\[ \lambda_{\text{Br}^-} = 8 \times 10^{-3} \, \text{S m}^2/\text{mol} \]
\[ \text{Specific Conductivity of Br}^- = \lambda_{\text{Br}^-} \times [\text{Br}^-] = 8 \times 10^{-3} \times 3.464 \times 10^{-2} \approx 2.77 \times 10^{-4} \, \text{S/m} \]

- For NO3\(^-\):
\[ \lambda_{\text{NO}_3^-} = 7 \times 10^{-3} \, \text{S m}^2/\text{mol} \]
\[ \text{Specific Conductivity of NO}_3^- = \lambda_{\text{NO}_3^-} \times [\text{NO}_3^-] = 7 \times 10^{-3} \times 10^{-7} \approx 7 \times 10^{-10} \, \text{S/m} \]

### Step 8: Total Specific Conductivity
Adding the specific conductivities:
\[ \text{Total Specific Conductivity} = 2.08 \times 10^{-4} + 2.77 \times 10^{-4} + 7 \times 10^{-10} \approx 5.85 \times 10^{-4} \, \text{S/m} \]

### Step 9: Convert to \( 10^{-7} \, \text{S/m} \)
To express in terms of \( 10^{-7} \, \text{S/m} \):
\[ \text{Total Specific Conductivity} \approx 58.5 \times 10^{-7} \, \text{S/m} \]

### Final Answer
The specific conductance of the solution is approximately:
\[ \boxed{58.5} \]