`EMF` of the cell
`|Ag|AgNO_(3)(0.1M)||K Br(1 N),AgBr(s)|Ag is -0.6V` at `298K`
`AgNO_(3) ` is `80%` and `KBr` is `60%` dissociated.
Calculate `a.` Solubility and
`b. K_(sp)` of `AgBr` at `298 K`.

Explanation

To solve the problem, we need to calculate the solubility and the solubility product constant (Ksp) of AgBr at 298 K using the given information about the electrochemical cell and the dissociation percentages of AgNO3 and KBr.

### Step-by-Step Solution:

**Step 1: Understand the Cell Reaction**
The cell reaction can be represented as follows:
- At the anode: \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)
- At the cathode: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)
...