
\[{K_{sp}}\] of $CuS$ , $A{g_2}S$ and $HgS$ are ${10^{ - 31}}$ ,${10^{ - 44}}$ and ${10^{ - 54}}$ $mo{l^2}$ $litr{e^{ - 2}}$ respectively. Select the correct order for their solubility in water.
1. $A{g_2}S$>$HgS$>$CuS$
2. $HgS$>$CuS$>$A{g_2}S$
3. $HgS$>$A{g_2}S$>$CuS$
4. $A{g_2}S$>$CuS$>$HgS$
Answer
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Hint: The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol \[{K_{sp}}\]. It is found out by using the solubility product of the components given.
Complete Step by step answer:
Equilibrium reaction for dissolution reaction of $CuS$ is:
$CuS \rightleftharpoons [C{u^{ + 2}}] + [{S^{ - 2}}]$
So, \[{K_{sp}}\]= $[C{u^{ + 2}}][{S^{ - 2}}]$ = ${10^{ - 31}}$
Hence, we get ${S^2}$ =${10^{ - 31}}$
Therefore S= 3.16 $ \times {10^{ - 16}}$
Equilibrium reaction for dissolution reaction of $A{g_2}S$ is:
$A{g_2}S \rightleftharpoons 2A{g^ + } + {S^{ - 2}}$
So, \[{K_{sp}}\]=${[A{g^ + }]^2}[{S^{ - 2}}]$ = ${10^{ - 44}}$
Hence, we get ${(2S)^2}(S)$ = ${10^{ - 44}}$
Therefore, 4${S^3}$ = ${10^{ - 44}}$
So, S= 1.357$ \times {10^{ - 15}}$
Equilibrium reaction for dissolution of $HgS$ is:
$HgS \rightleftharpoons H{g^{ + 2}} + {S^{ - 2}}$
So, \[{K_{sp}}\]=$[H{g^{ + 2}}][{S^{ - 2}}]$ =${10^{ - 54}}$
Hence, we get ${S^2}$= ${10^{ - 54}}$
Therefore, S= $1 \times {10^{ - 27}}$
Therefore solubility of $CuS$ , $A{g_2}S$ and $HgS$ are 3.16$ \times {10^{ - 16}}$ , 1.357$ \times {10^{ - 15}}$and$1 \times {10^{ - 27}}$. This implies that the highest solubility is that of $A{g_2}S$ and lower than this is of $CuS$. And lowest solubility is that of $HgS$ according to the values found. So the correct order according to the solubility in water is $A{g_2}S$>$CuS$>$HgS$.
So the correct option is 4th.
Note: The solubility product is a kind of equilibrium constant and its value depends on temperature. \[{K_{sp}}\] Usually increases with an increase in temperature due to increased solubility. It should be kept in mind that solubility products represent the extent of dissolution of any compound.
Complete Step by step answer:
Equilibrium reaction for dissolution reaction of $CuS$ is:
$CuS \rightleftharpoons [C{u^{ + 2}}] + [{S^{ - 2}}]$
So, \[{K_{sp}}\]= $[C{u^{ + 2}}][{S^{ - 2}}]$ = ${10^{ - 31}}$
Hence, we get ${S^2}$ =${10^{ - 31}}$
Therefore S= 3.16 $ \times {10^{ - 16}}$
Equilibrium reaction for dissolution reaction of $A{g_2}S$ is:
$A{g_2}S \rightleftharpoons 2A{g^ + } + {S^{ - 2}}$
So, \[{K_{sp}}\]=${[A{g^ + }]^2}[{S^{ - 2}}]$ = ${10^{ - 44}}$
Hence, we get ${(2S)^2}(S)$ = ${10^{ - 44}}$
Therefore, 4${S^3}$ = ${10^{ - 44}}$
So, S= 1.357$ \times {10^{ - 15}}$
Equilibrium reaction for dissolution of $HgS$ is:
$HgS \rightleftharpoons H{g^{ + 2}} + {S^{ - 2}}$
So, \[{K_{sp}}\]=$[H{g^{ + 2}}][{S^{ - 2}}]$ =${10^{ - 54}}$
Hence, we get ${S^2}$= ${10^{ - 54}}$
Therefore, S= $1 \times {10^{ - 27}}$
Therefore solubility of $CuS$ , $A{g_2}S$ and $HgS$ are 3.16$ \times {10^{ - 16}}$ , 1.357$ \times {10^{ - 15}}$and$1 \times {10^{ - 27}}$. This implies that the highest solubility is that of $A{g_2}S$ and lower than this is of $CuS$. And lowest solubility is that of $HgS$ according to the values found. So the correct order according to the solubility in water is $A{g_2}S$>$CuS$>$HgS$.
So the correct option is 4th.
Note: The solubility product is a kind of equilibrium constant and its value depends on temperature. \[{K_{sp}}\] Usually increases with an increase in temperature due to increased solubility. It should be kept in mind that solubility products represent the extent of dissolution of any compound.
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