$\mathrm { Ag } _ { 2 } \mathrm { S } ( s )$ has a larger molar solubility than CuS even though $\mathrm { Ag } _ { 2 } \mathrm { S }$ has the smaller $K _ { \mathrm { sp } }$ value. Explain how this is possible.

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Quizlet Admin · Accepted Answer

$\bullet$ The solubility product of Ag$_2$S is $\mathrm{ K_{sp} = 1.6 \cdot 10^{-49} }$\
$\bullet$ The solubility product of CuS is $\mathrm{ K_{sp} = 8.5 \cdot 10^{-45} }$\

\subsection*{Ag$_2$S}
Let us calculate the solubility of Ag$_2$S
$$
\mathrm{ Ag_2S(s) ightleftharpoons 2Ag^+ + S^{2-} }
$$
Let us assume that x moles of Ag$_2$S are dissolved.\

\begin{table}[ht]
\begin{tabular}{l|lll}
                                                                         & $\mathrm{ Ag_2S(s) }$ $ightleftharpoons$ & $\mathrm{ 2Ag^+ }$ + & $\mathrm{ S^{2-} }$ \ \hline
\begin{tabular}[c]{@{}l@{}}Initial \ concentration (M)\end{tabular}     & /                                          & 0                    & 0                   \ \cline{2-4} 
\begin{tabular}[c]{@{}l@{}}Change in\ concentration (M)\end{tabular}    & /                                          & +2x                  & +x                  \ \cline{2-4} 
\begin{tabular}[c]{@{}l@{}}Equilibrium \ concentration (M)\end{tabular} & /                                          & 2x                   & x                  
\end{tabular}
\end{table}

Now, let us calculate the value of x
\begin{align*}
\mathrm{ K_{sp} } &= \mathrm{ [Ag^+]^2 \cdot [S^{2-}] }\
1.6 \cdot 10^{-49} &= \mathrm{ (2x)^2 \cdot x }\
&= \mathrm{ 4x^3 }\
\mathrm{ x^3 } &= 4 \cdot 10^{-50}\
\mathrm{ x } &= (4 \cdot 10^{-50})^{1/3}\
\mathrm{ x } &= 3.43 \cdot 10^{-17}
\end{align*}

Therefore, solubility of Ag$_2$S is {\color{blue}$3.43 \cdot 10^{-17}$ }.

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$\mathrm { Ag } _ { 2 } \mathrm { S } ( s )$ has a larger molar solubility than CuS even though $\mathrm { Ag } _ { 2 } \mathrm { S }$ has the smaller $K _ { \mathrm { sp } }$ value. Explain how this is possible.

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$\mathrm { Ag } _ { 2 } \mathrm { S } ( s )$ has a larger molar solubility than CuS even though $\mathrm { Ag } _ { 2 } \mathrm { S }$ has the smaller $K _ { \mathrm { sp } }$ value. Explain how this is possible.