Calculate the simultaneous solubility of AgSCN and AgBr. ` K_(sp) ` for AgSCN and AgBr are` 1 xx 10 ^(-12) and 5 xx 10 ^(-13)` respectively.

Explanation

To calculate the simultaneous solubility of AgSCN and AgBr, we will follow these steps:

### Step 1: Write the dissociation equations
For AgSCN:
\[ \text{AgSCN (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{SCN}^- (aq) \]

For AgBr:
\[ \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]

### Step 2: Define the solubility variables
Let:
- \( A \) = solubility of AgSCN in moles per liter
- \( B \) = solubility of AgBr in moles per liter

From the dissociation equations, we can express the concentrations of the ions:
- For AgSCN:
- \([ \text{Ag}^+ ] = A\)
- \([ \text{SCN}^- ] = A\)

- For AgBr:
- \([ \text{Ag}^+ ] = B\)
- \([ \text{Br}^- ] = B\)

### Step 3: Write the expressions for \( K_{sp} \)
The solubility product \( K_{sp} \) expressions for both salts are:
1. For AgSCN:
\[ K_{sp} = [ \text{Ag}^+ ][ \text{SCN}^- ] = A(A) = A^2 \]
Given \( K_{sp} = 1 \times 10^{-12} \):
\[ A^2 = 1 \times 10^{-12} \]
(Equation 1)

2. For AgBr:
\[ K_{sp} = [ \text{Ag}^+ ][ \text{Br}^- ] = B(A + B) \]
Given \( K_{sp} = 5 \times 10^{-13} \):
\[ 5 \times 10^{-13} = B(A + B) \]
(Equation 2)

### Step 4: Substitute and simplify
From Equation 1, we have:
\[ A = \sqrt{1 \times 10^{-12}} = 1 \times 10^{-6} \text{ moles per liter} \]

Now substitute \( A \) into Equation 2:
\[ 5 \times 10^{-13} = B(1 \times 10^{-6} + B) \]

### Step 5: Solve for \( B \)
Rearranging gives:
\[ 5 \times 10^{-13} = B \times 1 \times 10^{-6} + B^2 \]
This can be rearranged to:
\[ B^2 + 1 \times 10^{-6}B - 5 \times 10^{-13} = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 1, b = 1 \times 10^{-6}, c = -5 \times 10^{-13} \):
\[ B = \frac{-1 \times 10^{-6} \pm \sqrt{(1 \times 10^{-6})^2 - 4 \times 1 \times -5 \times 10^{-13}}}{2 \times 1} \]

Calculating the discriminant:
\[ (1 \times 10^{-6})^2 + 20 \times 10^{-13} = 1 \times 10^{-12} + 20 \times 10^{-13} = 21 \times 10^{-13} = 2.1 \times 10^{-12} \]

Now substituting back:
\[ B = \frac{-1 \times 10^{-6} \pm \sqrt{2.1 \times 10^{-12}}}{2} \]

Calculating \( \sqrt{2.1 \times 10^{-12}} \):
\[ \sqrt{2.1} \approx 1.45 \]
Thus:
\[ B = \frac{-1 \times 10^{-6} \pm 1.45 \times 10^{-6}}{2} \]

Taking the positive root:
\[ B = \frac{0.45 \times 10^{-6}}{2} = 0.225 \times 10^{-6} = 2.25 \times 10^{-7} \text{ moles per liter} \]

### Step 7: Calculate \( A \)
Now, substitute \( B \) back into the equation for \( A \):
\[ A = 1 \times 10^{-6} \text{ moles per liter} \]

### Final Answer
The simultaneous solubility of AgSCN and AgBr is:
- \( A = 1 \times 10^{-6} \) moles per liter for AgSCN
- \( B = 2.25 \times 10^{-7} \) moles per liter for AgBr