To determine the maximum concentration of silver ions ([tex]\( Ag^+ \)[/tex]) in a solution that is [tex]\( 0.025 \, \text{M} \)[/tex] in carbonate ([tex]\( CO_3^{2-} \)[/tex]), we need to consider the solubility product constant ([tex]\( K_{sp} \)[/tex]) of silver carbonate ([tex]\( Ag_2CO_3 \)[/tex]), which is [tex]\( 8.1 \times 10^{-12} \)[/tex].
The dissolution of [tex]\( Ag_2CO_3 \)[/tex] in water can be represented by the following equation:
[tex]\[ Ag_2CO_3 \leftrightarrow 2Ag^+ + CO_3^{2-} \][/tex]
The [tex]\( K_{sp} \)[/tex] expression for this equilibrium is given by:
[tex]\[ K_{sp} = [Ag^+]^2 [CO_3^{2-}] \][/tex]
Given:
[tex]\[ K_{sp} = 8.1 \times 10^{-12} \][/tex]
[tex]\[ [CO_3^{2-}] = 0.025 \, M \][/tex]
We need to find the concentration of [tex]\( Ag^+ \)[/tex], which we will denote as [tex]\( x \)[/tex]. This leads to the following equation:
[tex]\[ 8.1 \times 10^{-12} = [Ag^+]^2 (0.025) \][/tex]
Rearranging to solve for [tex]\( [Ag^+] \)[/tex], we get:
[tex]\[ [Ag^+]^2 = \frac{8.1 \times 10^{-12}}{0.025} \][/tex]
Let's calculate this step by step:
1. Compute the denominator:
[tex]\[ 0.025 = 2.5 \times 10^{-2} \][/tex]
2. Now divide the [tex]\( K_{sp} \)[/tex] by this value:
[tex]\[ \frac{8.1 \times 10^{-12}}{2.5 \times 10^{-2}} = \frac{8.1 \times 10^{-12}}{2.5} \times 10^{2} = 3.24 \times 10^{-10} \][/tex]
3. Finally, take the square root of the result to find [tex]\( [Ag^+] \)[/tex]:
[tex]\[ [Ag^+] = \sqrt{3.24 \times 10^{-10}} \][/tex]
4. Calculating the square root:
[tex]\[ \sqrt{3.24} \approx 1.8 \][/tex]
[tex]\[ \sqrt{10^{-10}} = 10^{-5} \][/tex]
[tex]\[ [Ag^+] = 1.8 \times 10^{-5} \, M \][/tex]
Thus, the maximum concentration of silver ions ([tex]\( Ag^+ \)[/tex]) in this solution is [tex]\( 1.8 \times 10^{-5} \, M \)[/tex].
The correct option is:
c. 1
