For a general reaction given below, the value of solubility product can be given us
`{:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):}`
`K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y)`
Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation `[H^(+)]` ion, `[OH^(-)`] ion. It is also useful in qualitative analysis for the identification and separation of basic radicals
Potassium chromate is slowly added to a solution containing `0.20M Ag NO_(3)`, and `0.20M Ba(NO_(3))_(2)`. Describe what happens if the `K_(sp)` for `Ag_(2)CrO_(4)`, is `1.1 xx 10^(-12)` and the `K_(sp)` of `BaCrO_(4)` is `1.2` x`10^(-10)`,

Answer

To solve the problem, we need to analyze the situation when potassium chromate is added to a solution containing silver nitrate and barium nitrate. We will determine which compound precipitates first based on the solubility product constants (Ksp) provided.

### Step-by-Step Solution:

1. **Identify Initial Concentrations**:
- The initial concentration of Ag⁺ ions from AgNO₃ is 0.20 M.
- The initial concentration of Ba²⁺ ions from Ba(NO₃)₂ is also 0.20 M.

2. **Write the Ksp Expressions**:
- For silver chromate (Ag₂CrO₄):
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \]
- For barium chromate (BaCrO₄):
\[ K_{sp} = [Ba^{2+}] [CrO_4^{2-}] \]

3. **Substitute Known Values into Ksp Expressions**:
- For Ag₂CrO₄:
\[ 1.1 \times 10^{-12} = (0.20)^2 [CrO_4^{2-}] \]
\[ [CrO_4^{2-}] = \frac{1.1 \times 10^{-12}}{(0.20)^2} = \frac{1.1 \times 10^{-12}}{0.04} = 2.75 \times 10^{-11} \, \text{M} \]

- For BaCrO₄:
\[ 1.2 \times 10^{-10} = (0.20) [CrO_4^{2-}] \]
\[ [CrO_4^{2-}] = \frac{1.2 \times 10^{-10}}{0.20} = 6.0 \times 10^{-10} \, \text{M} \]

4. **Compare the Chromate Ion Concentrations**:
- The concentration of CrO₄²⁻ required to start precipitation for Ag₂CrO₄ is \(2.75 \times 10^{-11} \, \text{M}\).
- The concentration of CrO₄²⁻ required to start precipitation for BaCrO₄ is \(6.0 \times 10^{-10} \, \text{M}\).

5. **Determine Which Precipitates First**:
- Since \(2.75 \times 10^{-11} \, \text{M}\) (for Ag₂CrO₄) is less than \(6.0 \times 10^{-10} \, \text{M}\) (for BaCrO₄), Ag₂CrO₄ will precipitate first when potassium chromate is added to the solution.

### Conclusion:
When potassium chromate is added to the solution containing 0.20 M AgNO₃ and 0.20 M Ba(NO₃)₂, silver chromate (Ag₂CrO₄) will precipitate first.

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