A certain insoluble compound of `M^(2+)`, when shaken with water, provides an `M^(2+)` concentration of `1.0 xx 10^(-4)M`. A ligand is added to the system in a quantify which forms a soluble complex with `M^(2+)` and leaves `1.0 xx 10^(-6)M, M^(2+)` in solution. Will the insoluble compound tend to dissolve? Explain.

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Condition
The insoluble compound of \( M^{2+} \) has a concentration of \( 1.0 \times 10^{-4} \, M \) when shaken with water. This means that the solubility product \( K_{sp} \) can be expressed as:
\[ K_{sp} = [M^{2+}][\text{anion}] \]
Since the concentration of \( M^{2+} \) is \( 1.0 \times 10^{-4} \, M \), we can assume that the concentration of the anion is also at a similar level (let's denote it as \( [A^-] \)). Thus, we can write:
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