The solubility product (`K_(sp)`) of `Ca(OH)_(2)` at `25^@` is `4.42xx10^(-5)`. A 500 mL of saturated solution of `Ca(OH)_(2)` is mixed with equal volume of 0.4 M NaOH. How much `Ca(OH)_(2)` in milligrams is precipitated?
At equilibrium, the solution contains 0.0358 mole of `K_(2)CO_(3)`. Assuming the degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be equal, calculate the solubility product of `Ag_(2)CO_(3)`.

Explanation

To solve the problem of how much `Ca(OH)₂` precipitates when a saturated solution is mixed with `NaOH`, we will follow these steps:

### Step 1: Determine the solubility of `Ca(OH)₂`
The solubility product (`Ksp`) of `Ca(OH)₂` is given as `4.42 x 10^(-5)`. The dissociation of `Ca(OH)₂` in water can be represented as:
\[ Ca(OH)₂ \rightleftharpoons Ca^{2+} + 2OH^{-} \]

Let the solubility of `Ca(OH)₂` be `S`. Then at equilibrium:
- The concentration of `Ca^{2+}` ions = `S`
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