The `K_(sp)` of `Ag_(2)CrO_(4),AgCl,AgBr` and AgI are respectively, `1.1xx10^(-12),1.8xx10^(-10),5.0xx10^(-13),8.3xx10^(-17)`. Which one of the following salts will precipitate last if `AgNO_(3)` solution is added to the solution containing equal moles of NaCl,NaBr,NaI and `Na_(2)CrO_(4)` ?

Options

A. Agl
B. AgCl
C. AgBr
D. `Ag_(2)CrO_(4)`

Explanation

To determine which salt will precipitate last when AgNO₃ is added to a solution containing equal moles of NaCl, NaBr, NaI, and Na₂CrO₄, we need to analyze the solubility products (Ksp) of the salts formed with AgNO₃. The salts in question are Ag₂CrO₄, AgCl, AgBr, and AgI, with their respective Ksp values given as follows:

- Ksp (Ag₂CrO₄) = 1.1 x 10^(-12)
- Ksp (AgCl) = 1.8 x 10^(-10)
- Ksp (AgBr) = 5.0 x 10^(-13)
- Ksp (AgI) = 8.3 x 10^(-17)

### Step 1: Write the dissociation equations and Ksp expressions
1. **Ag₂CrO₄**:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]
\[ Ksp = [Ag^+]^2 [CrO_4^{2-}] \]

2. **AgCl**:
\[ AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) \]
\[ Ksp = [Ag^+][Cl^-] \]

3. **AgBr**:
\[ AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq) \]
\[ Ksp = [Ag^+][Br^-] \]

4. **AgI**:
\[ AgI (s) \rightleftharpoons Ag^+ (aq) + I^- (aq) \]
\[ Ksp = [Ag^+][I^-] \]

### Step 2: Calculate the solubility (S) of each salt
Using the Ksp values, we can calculate the solubility of each salt.

1. **For Ag₂CrO₄**:
\[ Ksp = 4S^3 \implies S = \sqrt[3]{\frac{Ksp}{4}} = \sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \approx 0.65 \times 10^{-4} \text{ M} \]

2. **For AgCl**:
\[ Ksp = S^2 \implies S = \sqrt{Ksp} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \text{ M} \]

3. **For AgBr**:
\[ Ksp = S^2 \implies S = \sqrt{Ksp} = \sqrt{5.0 \times 10^{-13}} \approx 0.71 \times 10^{-6} \text{ M} \]

4. **For AgI**:
\[ Ksp = S^2 \implies S = \sqrt{Ksp} = \sqrt{8.3 \times 10^{-17}} \approx 0.9 \times 10^{-8} \text{ M} \]

### Step 3: Compare the solubilities
Now we compare the calculated solubilities:
- Ag₂CrO₄: \(0.65 \times 10^{-4}\) M
- AgCl: \(1.34 \times 10^{-5}\) M
- AgBr: \(0.71 \times 10^{-6}\) M
- AgI: \(0.9 \times 10^{-8}\) M

### Step 4: Determine which salt precipitates last
The salt with the highest solubility will precipitate last. From our calculations:
- Ag₂CrO₄ has the highest solubility.

### Conclusion
Thus, the salt that will precipitate last when AgNO₃ is added to the solution is **Ag₂CrO₄** (Option D).

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