Answer
$FeCO_3$ is more soluble in $0.1M$
Correct answer: $(a)$
Work Step by Step
According to the common ion effect, if a solution has $Cl^-$ ions, the solubility of $AgCl$ will be decreased.
Since its $K_{sp}$ is already lower than the $K_{sp}$ for $FeCO_3$, and its solubility will be decreased, we can affirm that $FeCO_3$ is more soluble.
1. Write the $K_{sp}$ expression:
$ FeCO_3(s) \lt -- \gt 1Fe^{2+}(aq) + 1C{O_3}^{2-}(aq)$
$3.1 \times 10^{-11} = [Fe^{2+}]^ 1[C{O_3}^{2-}]^ 1$
2. Considering a pure solution: $[Fe^{2+}] = 1S$ and $[C{O_3}^{2-}] = 1S$
$3.1 \times 10^{-11}= ( 1S)^ 1 \times ( 1S)^ 1$
$3.1 \times 10^{-11} = S^ 2$
$ \sqrt [ 2] {3.1 \times 10^{-11}} = S$
$5.6 \times 10^{-6} = S$
- This is the molar solubility value for this salt.
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3. Write the $K_{sp}$ expression:
$ AgCl(s) \lt -- \gt 1Cl^{-}(aq) + 1Ag^{+}(aq)$
$1.8 \times 10^{-10} = [Cl^{-}]^ 1[Ag^{+}]^ 1$
$1.8 \times 10^{-10} = (0.1 + S)^ 1( 1S)^ 1$
4. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Cl^{-}] = 0.1$
$1.8 \times 10^{-10}= (0.1)^ 1 \times ( 1S)^ 1$
$1.8 \times 10^{-10}= 0.1 \times ( 1S)^ 1$
$ \frac{1.8 \times 10^{-10}}{0.1} = ( 1S)^ 1$
$1.8 \times 10^{-9} = S$
