`Ag^(+)` ion concentration in saturated solution of `Ag_(3)PO_(4) (K_(sp)=2.7xx10^(-19))` :

Options

A. `10^(-5) M`
B. `2.7xx10^(-4) M`
C. `3xx10^(-5) M`
D. `10^(-4) M`

Answer

To find the concentration of `Ag^+` ions in a saturated solution of `Ag₃PO₄`, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of `Ag₃PO₄` in water can be represented as:
\[ Ag_3PO_4 (s) \rightleftharpoons 3Ag^+ (aq) + PO_4^{3-} (aq) \]

### Step 2: Define solubility
Let the solubility of `Ag₃PO₄` be \( s \) mol/L. This means that in a saturated solution:
- The concentration of `Ag^+` ions will be \( 3s \) (since 3 moles of `Ag^+` are produced for every mole of `Ag₃PO₄` that dissolves).
- The concentration of `PO₄^{3-}` ions will be \( s \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation of `Ag₃PO₄` can be expressed as:
\[ K_{sp} = [Ag^+]^3 [PO_4^{3-}] \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (3s)^3 \cdot s = 27s^4 \]

### Step 4: Substitute the value of \( K_{sp} \)
Given that \( K_{sp} = 2.7 \times 10^{-19} \):
\[ 27s^4 = 2.7 \times 10^{-19} \]

### Step 5: Solve for \( s \)
Now, we can solve for \( s \):
\[ s^4 = \frac{2.7 \times 10^{-19}}{27} \]
Calculating the right side:
\[ s^4 = 1.0 \times 10^{-20} \]

### Step 6: Take the fourth root
To find \( s \), we take the fourth root:
\[ s = (1.0 \times 10^{-20})^{1/4} \]
Calculating this gives:
\[ s = 10^{-5} \]

### Step 7: Calculate the concentration of \( Ag^+ \)
Now, we can find the concentration of `Ag^+`:
\[ [Ag^+] = 3s = 3 \times 10^{-5} \, \text{mol/L} \]

### Final Answer
Thus, the concentration of `Ag^+` ions in the saturated solution of `Ag₃PO₄` is:
\[ [Ag^+] = 3 \times 10^{-5} \, \text{mol/L} \]