Question

1. the solubility of silver sulfide is $8.0 \times 10^{-17} m$. determine the $k_{sp}$ of this salt.

a. $64 \times 10^{-51}$
b. $16.0 \times 10^{-17}8.0 \times 10^{-17}$
c. $16.0 \times 10^{-17}^28.0 \times 10^{-17}$
d. $8.0 \times 10^{-17}4.0 \times 10^{-17}$
e. $4.0 \times 10^{-17}^28.0 \times 10^{-17}$

Explanation:

Step1: Write the dissociation equation

Silver sulfide ($\ce{Ag2S}$) dissociates as $\ce{Ag2S(s) <=> 2Ag+(aq) + S^{2-}(aq)}$. Let the solubility of $\ce{Ag2S}$ be $s = 8.0\times10^{-17}\ M$. Then, $[\ce{Ag+}] = 2s$ and $[\ce{S^{2-}}] = s$.

Step2: Calculate ion concentrations

Given $s = 8.0\times10^{-17}\ M$, so $[\ce{Ag+}] = 2\times8.0\times10^{-17}=16.0\times10^{-17}\ M$ and $[\ce{S^{2-}}] = 8.0\times10^{-17}\ M$.

Step3: Write the $K_{sp}$ expression

For $\ce{Ag2S}$, $K_{sp} = [\ce{Ag+}]^2[\ce{S^{2-}}]$. Substituting the concentrations, we get $K_{sp} = [16.0\times10^{-17}]^2[8.0\times10^{-17}]$.

Answer:

c. $[16.0\times 10^{-17}]^2 [8.0\times 10^{-17}]$