In a saturated solution of the sparingly soluble strong electrolyte `AgIO_3` (molecular mass= 283) , the equilibrium which sets in is
` AgIO_3 (s) hArr Ag^(+) (aq) +IO_3^(-) (aq)`
If the solubility product constant ` K_(sp) " of " AgIO_3` at a given temperature is ` 1.0 xx 10^(-8) , ` what is the mass of `AgIO_3` contained in 100 mL of its saturated solution ?

Options

A. ` 28 . 3 xx 10 ^(-2) g`
B. ` 2.83 xx 10 ^(-3) g `
C. ` 1.0 xx 10 ^(-7) g `
D. ` 1.0 xx 10 ^(-4) g.`

Explanation

To solve the problem, we need to determine the mass of AgIO₃ contained in 100 mL of its saturated solution. We will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of AgIO₃ in water can be represented as:
\[ \text{AgIO}_3 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{IO}_3^- (aq) \]

### Step 2: Define the solubility product constant (Ksp)
The solubility product constant \( K_{sp} \) for this equilibrium can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{IO}_3^-] \]
If we let \( S \) be the solubility of AgIO₃ (in moles per liter), then at equilibrium:
\[ [\text{Ag}^+] = S \]
\[ [\text{IO}_3^-] = S \]
Thus, we can write:
\[ K_{sp} = S \times S = S^2 \]

### Step 3: Substitute the given Ksp value
We are given that \( K_{sp} = 1.0 \times 10^{-8} \). Therefore, we can set up the equation:
\[ S^2 = 1.0 \times 10^{-8} \]

### Step 4: Solve for S
To find \( S \), we take the square root of both sides:
\[ S = \sqrt{1.0 \times 10^{-8}} \]
\[ S = 1.0 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Calculate the moles in 100 mL
Since the solubility \( S \) is in moles per liter, we can find the number of moles in 100 mL (which is 0.1 L):
\[ \text{Moles of AgIO}_3 = S \times \text{Volume in L} \]
\[ \text{Moles of AgIO}_3 = 1.0 \times 10^{-4} \, \text{mol/L} \times 0.1 \, \text{L} \]
\[ \text{Moles of AgIO}_3 = 1.0 \times 10^{-5} \, \text{mol} \]

### Step 6: Calculate the mass of AgIO₃
To find the mass of AgIO₃, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
Given that the molar mass of AgIO₃ is 283 g/mol:
\[ \text{Mass} = 1.0 \times 10^{-5} \, \text{mol} \times 283 \, \text{g/mol} \]
\[ \text{Mass} = 2.83 \times 10^{-3} \, \text{g} \]

### Final Answer
The mass of AgIO₃ contained in 100 mL of its saturated solution is:
\[ \text{Mass} = 2.83 \times 10^{-3} \, \text{g} \]

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