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 idetification and separation of basic radicals
Potussium chromate is slowly aded toa solution containing 0.20M Ag `NO_(3)`, and 0.20M `Ba(NO_(3))_(2)`. Describe what happensif the `K_(sp)` for `Ag_(2),CrO_(4)`, is `1.1 xx 10^(-12)` and the `K_(sp)` of `BaCiO_(4)`, is `1.2 xx 10^(-10)`,

Options

A. The `Ag_(2),CrO_(4)`, pecipitates irst out of solution and then `BaCrO_(4)`, preciptates
B. The `BaCrO_(4)`, pecipitates first out of solution and then `Ag_(2),CrO_(4)`, preciptates
C. Both `Ag_(2),CrO_(4)`, and `BaCrO_(4)`, precipitate simultaneously out of solution
D. Neither `Ag_(2),CrO_(4)`, nor `BaCrO_(4)`, precipitates

Answer

To solve the problem, we need to analyze the solubility products (Ksp) of the two compounds formed when potassium chromate is added to a solution containing silver nitrate (AgNO3) and barium nitrate (Ba(NO3)2).

We are given:
- Ksp of Ag2CrO4 = 1.1 × 10^(-12)
- Ksp of BaCrO4 = 1.2 × 10^(-10)
- Concentration of AgNO3 = 0.20 M
- Concentration of Ba(NO3)2 = 0.20 M

### Step 1: Calculate the concentration of CrO4^2- ions at which Ag2CrO4 will start to precipitate.

Using the formula for Ksp:
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \]

Rearranging gives:
\[ [CrO_4^{2-}] = \frac{K_{sp}}{[Ag^+]^2} \]

Substituting the values:
\[ [CrO_4^{2-}] = \frac{1.1 \times 10^{-12}}{(0.20)^2} \]
\[ [CrO_4^{2-}] = \frac{1.1 \times 10^{-12}}{0.04} \]
\[ [CrO_4^{2-}] = 2.75 \times 10^{-11} \, M \]

### Step 2: Calculate the concentration of CrO4^2- ions at which BaCrO4 will start to precipitate.

Using the formula for Ksp:
\[ K_{sp} = [Ba^{2+}][CrO_4^{2-}] \]

Rearranging gives:
\[ [CrO_4^{2-}] = \frac{K_{sp}}{[Ba^{2+}]} \]

Substituting the values:
\[ [CrO_4^{2-}] = \frac{1.2 \times 10^{-10}}{0.20} \]
\[ [CrO_4^{2-}] = 6.0 \times 10^{-10} \, M \]

### Step 3: Compare the concentrations of CrO4^2- ions at which precipitation occurs.

From the calculations:
- For Ag2CrO4, precipitation occurs at [CrO4^2-] = 2.75 × 10^(-11) M.
- For BaCrO4, precipitation occurs at [CrO4^2-] = 6.0 × 10^(-10) M.

Since 2.75 × 10^(-11) M < 6.0 × 10^(-10) M, Ag2CrO4 will precipitate first.

### Conclusion:
When potassium chromate is added to the solution, Ag2CrO4 will precipitate before BaCrO4.