`HgCrO_(4)` just begins to peripitate when equal volumes of `4 xx 10^(-4)M Hg_(2)(NO_(3))_(2)` and `2 xx 10^(-5)M K_(2)CrO_(4)` are combind. What is the approximate `K_(sp)` value of `Hg_(2)CrO_(4)`?

Options

A. `1 xx 10^(-18) mol L^(-1)`
B. `8 xx 10^(-9)mol L^(-1)`
C. `2xx 10^(-9) mol L^(-1)`
D. `4 xx 10^(-9) mol L^(-1)`

Explanation

To find the approximate \( K_{sp} \) value of \( Hg_2CrO_4 \), we can follow these steps:

### Step 1: Determine the concentrations after mixing
When equal volumes of two solutions are mixed, the concentrations of the ions will be halved.

- For \( Hg_2(NO_3)_2 \):
- Initial concentration = \( 4 \times 10^{-4} \, M \)
- After mixing (halved) = \( \frac{4 \times 10^{-4}}{2} = 2 \times 10^{-4} \, M \)

- For \( K_2CrO_4 \):
- Initial concentration = \( 2 \times 10^{-5} \, M \)
- After mixing (halved) = \( \frac{2 \times 10^{-5}}{2} = 1 \times 10^{-5} \, M \)

### Step 2: Write the dissociation equations
The dissociation of \( Hg_2(NO_3)_2 \) and \( K_2CrO_4 \) can be represented as follows:

- \( Hg_2(NO_3)_2 \) dissociates to give:
\[ Hg_2(NO_3)_2 \rightarrow Hg_2^{2+} + 2 NO_3^{-} \]
From this, we see that 1 mole of \( Hg_2(NO_3)_2 \) produces 1 mole of \( Hg_2^{2+} \).

- \( K_2CrO_4 \) dissociates to give:
\[ K_2CrO_4 \rightarrow 2 K^{+} + CrO_4^{2-} \]
From this, we see that 1 mole of \( K_2CrO_4 \) produces 1 mole of \( CrO_4^{2-} \).

### Step 3: Calculate the concentrations of ions
- The concentration of \( Hg_2^{2+} \) will be the same as that of \( Hg_2(NO_3)_2 \) after mixing:
\[ [Hg_2^{2+}] = 2 \times 10^{-4} \, M \]

- The concentration of \( CrO_4^{2-} \) will be the same as that of \( K_2CrO_4 \) after mixing:
\[ [CrO_4^{2-}] = 1 \times 10^{-5} \, M \]

### Step 4: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( Hg_2CrO_4 \) can be expressed as:
\[ K_{sp} = [Hg_2^{2+}][CrO_4^{2-}] \]

### Step 5: Substitute the concentrations into the \( K_{sp} \) expression
Substituting the values we calculated:
\[ K_{sp} = (2 \times 10^{-4})(1 \times 10^{-5}) \]
\[ K_{sp} = 2 \times 10^{-9} \]

### Conclusion
The approximate \( K_{sp} \) value of \( Hg_2CrO_4 \) is \( 2 \times 10^{-9} \).

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