We have taken a saturated solution of `AgBr`, whose `K_(sp)` is `12xx10^(-14).` If `10^(-7)M` of `AgNO_(3)` are added to `1L` of this solution, find the conductivity `(` specific conductance `)` of the solution in terms of `10^(-7)S m^(-1)` units.
Given `:`
`lambda^(@)._((Ag^(+)))=6xx10^(-3)S m^(2) mol^(-1)`
`lambda^(@)._((Br^(-)))=8xx10^(-3)S m^(2)mol^(-1)`
`lambda^(@)._((NO_(3)^(-)))=7xx10^(-3)S m^(2) mol^(-1)`

Explanation

To solve the problem, we need to calculate the specific conductance of a saturated solution of AgBr after adding AgNO3. We will follow these steps:

### Step 1: Determine the concentrations of ions in the solution
1. **Saturated Solution of AgBr**: The solubility product (Ksp) of AgBr is given as \( K_{sp} = 12 \times 10^{-14} \).
- Let the solubility of AgBr be \( s \). Therefore, the concentration of \( \text{Ag}^+ \) ions is \( s \) and the concentration of \( \text{Br}^- \) ions is also \( s \).
- The expression for Ksp is:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] = s^2 \]
- Thus, we have:
\[ s^2 = 12 \times 10^{-14} \implies s = \sqrt{12 \times 10^{-14}} = 3.464 \times 10^{-7} \, \text{M} \]

2. **Adding AgNO3**: When \( 10^{-7} \, \text{M} \) of AgNO3 is added, it dissociates completely:
- \( \text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- \)
- The concentration of \( \text{Ag}^+ \) ions from AgNO3 is \( 10^{-7} \, \text{M} \).
- Therefore, the total concentration of \( \text{Ag}^+ \) ions in solution becomes:
\[ [\text{Ag}^+] = s + 10^{-7} = 3.464 \times 10^{-7} + 10^{-7} = 4.464 \times 10^{-7} \, \text{M} \]
- The concentration of \( \text{Br}^- \) ions remains \( s = 3.464 \times 10^{-7} \, \text{M} \).
- The concentration of \( \text{NO}_3^- \) ions is \( 10^{-7} \, \text{M} \).

### Step 2: Calculate the molar conductivities
1. **Molar Conductivity Values**:
- \( \lambda^{\circ}_{\text{Ag}^+} = 6 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \lambda^{\circ}_{\text{Br}^-} = 8 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \lambda^{\circ}_{\text{NO}_3^-} = 7 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)

### Step 3: Calculate the specific conductance (κ)
1. **Specific Conductance for Each Ion**:
- For \( \text{Ag}^+ \):
\[ \kappa_{\text{Ag}^+} = \lambda^{\circ}_{\text{Ag}^+} \times [\text{Ag}^+] = 6 \times 10^{-3} \times 4.464 \times 10^{-7} = 2.664 \times 10^{-9} \, \text{S m}^{-1} \]
- For \( \text{Br}^- \):
\[ \kappa_{\text{Br}^-} = \lambda^{\circ}_{\text{Br}^-} \times [\text{Br}^-] = 8 \times 10^{-3} \times 3.464 \times 10^{-7} = 2.7712 \times 10^{-9} \, \text{S m}^{-1} \]
- For \( \text{NO}_3^- \):
\[ \kappa_{\text{NO}_3^-} = \lambda^{\circ}_{\text{NO}_3^-} \times [\text{NO}_3^-] = 7 \times 10^{-3} \times 10^{-7} = 7 \times 10^{-10} \, \text{S m}^{-1} \]

### Step 4: Total Conductivity
1. **Total Conductivity**:
\[ \kappa_{\text{total}} = \kappa_{\text{Ag}^+} + \kappa_{\text{Br}^-} + \kappa_{\text{NO}_3^-} \]
\[ \kappa_{\text{total}} = 2.664 \times 10^{-9} + 2.7712 \times 10^{-9} + 7 \times 10^{-10} = 5.2052 \times 10^{-9} \, \text{S m}^{-1} \]

### Step 5: Convert to \( 10^{-7} \, \text{S m}^{-1} \)
1. **Expressing in terms of \( 10^{-7} \, \text{S m}^{-1} \)**:
\[ \kappa_{\text{total}} = 5.2052 \times 10^{-9} \, \text{S m}^{-1} = 52.052 \times 10^{-7} \, \text{S m}^{-1} \]
- Rounding gives us \( \approx 52 \times 10^{-7} \, \text{S m}^{-1} \).

### Final Answer
The specific conductance of the solution is approximately **52 \( \times 10^{-7} \, \text{S m}^{-1} \)**.