Calculate the solubility product of `Ag_2 CrO_4` at `25^@C` if the concentration of `Ag^+` ions is `1.5 xx 10^(-4)` mole/litre in a saturated solution of `Ag_2CrO_4` at `25^@C`

Explanation

To calculate the solubility product (Ksp) of \( Ag_2CrO_4 \) at \( 25^\circ C \), given that the concentration of \( Ag^+ \) ions is \( 1.5 \times 10^{-4} \) moles/litre, we can follow these steps:

### Step 1: Write the Dissolution Reaction
The dissolution of \( Ag_2CrO_4 \) in water can be represented by the following equation:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

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