What is the solubility product of `Ag_2CrO_4,` if 0.0166 g of the salt dissolves in 500 mL of water at `18^@C?`

Answer

To find the solubility product (Ksp) of \( Ag_2CrO_4 \), we will follow these steps:

### Step 1: Calculate the molar mass of \( Ag_2CrO_4 \)

The molar mass can be calculated using the atomic masses of silver (Ag), chromium (Cr), and oxygen (O):
- Atomic mass of Ag = 107.87 g/mol
- Atomic mass of Cr = 51.996 g/mol
- Atomic mass of O = 16.00 g/mol

The molar mass of \( Ag_2CrO_4 \) is calculated as follows:
\[ \text{Molar mass of } Ag_2CrO_4 = 2 \times \text{(mass of Ag)} + \text{(mass of Cr)} + 4 \times \text{(mass of O)} \]
\[ = 2 \times 107.87 + 51.996 + 4 \times 16.00 \]
\[ = 215.74 + 51.996 + 64.00 = 331.736 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of \( Ag_2CrO_4 \)

Using the mass of \( Ag_2CrO_4 \) that dissolves:
\[ \text{Mass of } Ag_2CrO_4 = 0.0166 \, \text{g} \]

The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ = \frac{0.0166 \, \text{g}}{331.736 \, \text{g/mol}} \approx 0.0000500 \, \text{mol} \]

### Step 3: Calculate the concentration of \( Ag_2CrO_4 \)

The concentration in a 500 mL solution is given by:
\[ \text{Concentration} = \frac{\text{number of moles}}{\text{volume in liters}} \]
\[ = \frac{0.0000500 \, \text{mol}}{0.500 \, \text{L}} = 0.000100 \, \text{mol/L} = 10^{-4} \, \text{mol/L} \]

### Step 4: Write the dissociation equation

The dissociation of \( Ag_2CrO_4 \) in water is:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

### Step 5: Determine the concentration of ions

From the dissociation, we see that:
- For every 1 mole of \( Ag_2CrO_4 \), 2 moles of \( Ag^+ \) and 1 mole of \( CrO_4^{2-} \) are produced.

Let \( S \) be the solubility of \( Ag_2CrO_4 \) in mol/L:
\[ \text{Concentration of } Ag^+ = 2S \]
\[ \text{Concentration of } CrO_4^{2-} = S \]

Given \( S = 10^{-4} \, \text{mol/L} \):
\[ \text{Concentration of } Ag^+ = 2 \times 10^{-4} \, \text{mol/L} \]
\[ \text{Concentration of } CrO_4^{2-} = 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate the solubility product (Ksp)

The solubility product \( Ksp \) is given by:
\[ Ksp = [Ag^+]^2 \times [CrO_4^{2-}] \]
Substituting the concentrations:
\[ Ksp = (2S)^2 \times S = (2 \times 10^{-4})^2 \times (10^{-4}) \]
\[ = 4 \times 10^{-8} \times 10^{-4} = 4 \times 10^{-12} \]

### Final Answer

Thus, the solubility product \( Ksp \) of \( Ag_2CrO_4 \) is:
\[ Ksp = 4 \times 10^{-12} \]

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