## Chemistry and Chemical Reactivity (9th Edition)

$FeCO_3$ is more soluble in $0.1M$ Correct answer: $(a)$
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. ------- 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: $[Cl^{-}] = 0.1$ $1.8 \times 10^{-10}= (0.1)^ 1 \times ( 1S)^ 1$ $1.8 \times 10^{-10}= 0.1 \times ( 1S)^ 1$ $\frac{1.8 \times 10^{-10}}{0.1} = ( 1S)^ 1$ $1.8 \times 10^{-9} = S$