The solubility product constant of `Ag_(2)CrO_(4)` and `AgBr` are `1.1xx10^(-12)` and `5.0xx10^(-13)` respectively. Calculate the ratio of the molarities of their saturated solutions.

Explanation

To calculate the ratio of the molarities of the saturated solutions of `Ag2CrO4` and `AgBr`, we will follow these steps:

### Step 1: Write the dissociation equations and expressions for Ksp

1. **For `Ag2CrO4`:**
- Dissociation:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2 Ag^+ (aq) + CrO_4^{2-} (aq) \]
- Let the solubility of `Ag2CrO4` be \( S_1 \). Therefore, the concentration of \( Ag^+ \) will be \( 2S_1 \) and the concentration of \( CrO_4^{2-} \) will be \( S_1 \).
- The expression for the solubility product constant \( K_{sp} \) is:
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2S_1)^2 (S_1) = 4S_1^3 \]

2. **For `AgBr`:**
- Dissociation:
\[ AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq) \]
- Let the solubility of `AgBr` be \( S_2 \). Therefore, the concentration of both \( Ag^+ \) and \( Br^- \) will be \( S_2 \).
- The expression for the solubility product constant \( K_{sp} \) is:
\[ K_{sp} = [Ag^+] [Br^-] = S_2^2 \]

### Step 2: Substitute the given Ksp values and solve for S1 and S2

1. **For `Ag2CrO4`:**
- Given \( K_{sp} = 1.1 \times 10^{-12} \):
\[ 1.1 \times 10^{-12} = 4S_1^3 \]
- Rearranging gives:
\[ S_1^3 = \frac{1.1 \times 10^{-12}}{4} \]
\[ S_1^3 = 2.75 \times 10^{-13} \]
- Taking the cube root:
\[ S_1 = \left(2.75 \times 10^{-13}\right)^{1/3} \approx 0.65 \times 10^{-4} \, \text{M} \]

2. **For `AgBr`:**
- Given \( K_{sp} = 5.0 \times 10^{-13} \):
\[ 5.0 \times 10^{-13} = S_2^2 \]
- Rearranging gives:
\[ S_2 = \sqrt{5.0 \times 10^{-13}} \approx 7.07 \times 10^{-7} \, \text{M} \]

### Step 3: Calculate the ratio of the solubilities

Now we need to find the ratio of the solubilities \( \frac{S_1}{S_2} \):
\[ \frac{S_1}{S_2} = \frac{0.65 \times 10^{-4}}{7.07 \times 10^{-7}} \approx 91.9 \]

### Final Answer:
The ratio of the molarities of their saturated solutions is approximately **91.9**.

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