The `[Ag^(+)]` ion in a saturated solution of `Ag_(2)CrO_(4)` at `25^(@)C is 1.5xx10^(-4)M`. Determine `K_(SP)` of `Ag_(2)CrO_(4)` at `25^(@)C`.

Answer

To determine the solubility product constant (`K_sp`) of `Ag_(2)CrO_(4)` at `25°C`, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of `Ag_(2)CrO_(4)` in water can be represented as:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

### Step 2: Define solubility
Let the solubility of `Ag_(2)CrO_(4)` be denoted as `S` (in mol/L). According to the dissociation equation:
- The concentration of `Ag^+` ions will be `2S`.
- The concentration of `CrO_4^{2-}` ions will be `S`.

### Step 3: Substitute the given concentration
We are given that the concentration of `Ag^+` ions in a saturated solution is `1.5 × 10^{-4} M`. Therefore:
\[ 2S = 1.5 × 10^{-4} \]

### Step 4: Solve for S
To find `S`, we can rearrange the equation:
\[ S = \frac{1.5 × 10^{-4}}{2} = 0.75 × 10^{-4} = 7.5 × 10^{-5} M \]

### Step 5: Write the expression for K_sp
The solubility product constant (`K_sp`) can be expressed as:
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \]
Substituting the concentrations in terms of `S`:
\[ K_{sp} = (2S)^2 (S) \]

### Step 6: Substitute S into the K_sp expression
Now substituting `S = 7.5 × 10^{-5}` into the `K_sp` expression:
\[ K_{sp} = (2(7.5 × 10^{-5}))^2 (7.5 × 10^{-5}) \]
\[ = (1.5 × 10^{-4})^2 (7.5 × 10^{-5}) \]

### Step 7: Calculate K_sp
Calculating the square:
\[ (1.5 × 10^{-4})^2 = 2.25 × 10^{-8} \]
Now multiplying by `7.5 × 10^{-5}`:
\[ K_{sp} = 2.25 × 10^{-8} × 7.5 × 10^{-5} \]
\[ = 1.6875 × 10^{-12} \]

### Step 8: Final result
Thus, the solubility product constant `K_sp` of `Ag_(2)CrO_(4)` at `25°C` is:
\[ K_{sp} = 1.69 × 10^{-12} \]