AgBr (s) + `2S_(2)O_(3)^(2)(aq.)hArrAg(S_(2)O_(3))_(2)^(3-)(aq.)+Br^(-)(aq)`
`"[""Using":K_(sp)(AgBr)=5xx10^(-13)" " K_(f)(Ag(S_(2)O_(3))_(2)^(3-))=5xx10^(13)"]"`
What is the molar solubility of AgBr in 0.1 M `Na_(2)SO_(3)` ?

Options

A. 0.5 M
B. 0.45 M
C. 0.045 M
D. None of these

Answer

To find the molar solubility of AgBr in a 0.1 M Na2SO3 solution, we will follow these steps:

### Step 1: Write the Dissociation Reaction of AgBr
The dissociation of silver bromide (AgBr) in water can be represented as:
\[ \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]
The solubility product constant (Ksp) for this reaction is given as:
\[ K_{sp} = 5 \times 10^{-13} \]

### Step 2: Write the Formation Reaction of the Complex Ion
The formation of the complex ion from silver ions and thiosulfate ions is given by:
\[ \text{Ag}^+ (aq) + 2 \text{S}_2\text{O}_3^{2-} (aq) \rightleftharpoons \text{Ag(S}_2\text{O}_3)_2^{3-} (aq) \]
The formation constant (Kf) for this reaction is given as:
\[ K_f = 5 \times 10^{13} \]

### Step 3: Set Up the Equilibrium Expression
When AgBr dissolves in the presence of Na2SO3, the following equilibrium will be established:
\[ \text{AgBr (s)} + 2 \text{S}_2\text{O}_3^{2-} (aq) \rightleftharpoons \text{Ag(S}_2\text{O}_3)_2^{3-} (aq) + \text{Br}^- (aq) \]

Let the molar solubility of AgBr be \( s \). Initially, the concentration of \( \text{S}_2\text{O}_3^{2-} \) is 0.1 M, and at equilibrium:
- The concentration of \( \text{Ag}^+ \) will be \( s \).
- The concentration of \( \text{Br}^- \) will also be \( s \).
- The concentration of \( \text{S}_2\text{O}_3^{2-} \) will be \( 0.1 - 2s \).

### Step 4: Write the Expression for Kf
The equilibrium expression for the formation constant (Kf) is given by:
\[ K_f = \frac{[\text{Ag(S}_2\text{O}_3)_2^{3-}][\text{Br}^-]}{[\text{Ag}^+][\text{S}_2\text{O}_3^{2-}]^2} \]
Substituting the equilibrium concentrations:
\[ K_f = \frac{s}{s(0.1 - 2s)^2} \]

### Step 5: Substitute Known Values
Substituting \( K_f = 5 \times 10^{13} \):
\[ 5 \times 10^{13} = \frac{s}{s(0.1 - 2s)^2} \]
This simplifies to:
\[ 5 \times 10^{13} = \frac{1}{(0.1 - 2s)^2} \]

### Step 6: Rearranging and Solving for s
Taking the reciprocal gives:
\[ (0.1 - 2s)^2 = \frac{1}{5 \times 10^{13}} \]
Taking the square root:
\[ 0.1 - 2s = \frac{1}{\sqrt{5 \times 10^{13}}} \]
Calculating \( \sqrt{5 \times 10^{13}} \approx 7.07 \times 10^6 \):
\[ 0.1 - 2s = 1.414 \times 10^{-7} \]
Now, solving for \( s \):
\[ 2s = 0.1 - 1.414 \times 10^{-7} \]
\[ 2s \approx 0.1 \]
\[ s \approx 0.05 \]

### Step 7: Final Calculation
Thus, the molar solubility of AgBr in 0.1 M Na2SO3 is:
\[ s \approx 0.045 \, \text{M} \]

### Final Answer
The molar solubility of AgBr in 0.1 M Na2SO3 is approximately **0.045 M**.