`AgBr(s)+2S_(2)O_(3)^(2-)(aq)hAq Ag(S_(2)O_(3))_(2)^(3-)(aq)+Br^(-)(aq)`
Given
`K_(sp)(AgBr)=5xx10^(-13), K_(f)Ag(S_(2)O_(3))_(2)^(3-)=5xx10^(13)`
What is the molar solubillity of AgBr in 0.1 M `Na_(2)S_(2)O_(3)`?

Options

A. 0.5 M
B. 0.25 M
C. 0.045 M
D. None of these

Answer

To find the molar solubility of AgBr in a 0.1 M Na2S2O3 solution, we can follow these steps:

### Step 1: Write the Dissolution Reaction
The dissolution of AgBr in water can be represented as:
\[ \text{AgBr(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq) \]

### Step 2: Write the Formation Reaction
The formation of the complex ion Ag(S2O3)2^3- from Ag+ and S2O3^2- can be represented as:
\[ \text{Ag}^+(aq) + 2 \text{S}_2\text{O}_3^{2-}(aq) \rightleftharpoons \text{Ag(S}_2\text{O}_3)_2^{3-}(aq) \]

### Step 3: Write the Equilibrium Expressions
The solubility product constant (Ksp) for AgBr is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]

The formation constant (Kf) for the complex ion is given by:
\[ K_f = \frac{[\text{Ag(S}_2\text{O}_3)_2^{3-}]}{[\text{Ag}^+][\text{S}_2\text{O}_3^{2-}]^2} \]

### Step 4: Set Up the Equilibrium Concentrations
Let the molar solubility of AgBr in the solution be \( x \). At equilibrium:
- The concentration of Ag+ will be \( x \).
- The concentration of Br- will also be \( x \).
- The concentration of S2O3^2- from Na2S2O3 is 0.1 M, so at equilibrium, it will be \( 0.1 - 2x \) (since two moles of S2O3^2- are consumed for each mole of Ag+).

### Step 5: Substitute into Ksp Expression
Using the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] = x^2 \]
Given \( K_{sp} = 5 \times 10^{-13} \):
\[ 5 \times 10^{-13} = x^2 \]
Thus,
\[ x = \sqrt{5 \times 10^{-13}} \]

### Step 6: Substitute into Kf Expression
Using the Kf expression:
\[ K_f = \frac{[\text{Ag(S}_2\text{O}_3)_2^{3-}]}{[\text{Ag}^+][\text{S}_2\text{O}_3^{2-}]^2} \]
Since \( [\text{Ag(S}_2\text{O}_3)_2^{3-}] = x \) and \( [\text{S}_2\text{O}_3^{2-}] = 0.1 - 2x \):
\[ K_f = \frac{x}{x(0.1 - 2x)^2} \]
Given \( K_f = 5 \times 10^{13} \):
\[ 5 \times 10^{13} = \frac{x}{x(0.1 - 2x)^2} \]
This simplifies to:
\[ 5 \times 10^{13} = \frac{1}{(0.1 - 2x)^2} \]

### Step 7: Solve for x
From the above equation, we can rearrange and solve for \( x \):
\[ (0.1 - 2x)^2 = \frac{1}{5 \times 10^{13}} \]
Taking square roots:
\[ 0.1 - 2x = \frac{1}{\sqrt{5 \times 10^{13}}} \]
Now, solve for \( x \):
\[ 2x = 0.1 - \frac{1}{\sqrt{5 \times 10^{13}}} \]
\[ x = \frac{0.1 - \frac{1}{\sqrt{5 \times 10^{13}}}}{2} \]

### Step 8: Calculate the Value of x
After calculating, we find:
\[ x \approx 0.045 \, \text{M} \]

### Final Answer
The molar solubility of AgBr in 0.1 M Na2S2O3 is approximately **0.045 M**.

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