
An aqueous solution contains the ions as $Hg_{2}^{2+},H{{g}^{2+}},P{{b}^{2+}}$,and $C{{d}^{2+}}$. The addition of dilute $HCl(6N)$ precipitates:
A. $H{{g}_{2}}C{{l}_{2}}$ only
B. $PbC{{l}_{2}}$ only
C. $PbC{{l}_{2}}$ and $HgC{{l}_{2}}$
D. $H{{g}_{2}}C{{l}_{2}}$ and $PbC{{l}_{2}}$
Answer
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Hint: An ionic solid is rarely soluble in a solution of its ions which are in equilibrium. The solubility product $({{k}_{sp}})$ of the solid is the equilibrium constant. If a solution has a small value of $({{k}_{sp}})$, it means there aren’t many hydrated ions in a saturated solution.
Complete Step by Step Solution:
Let’s start by defining $HCl(6N)$: Muriatic acid, which is an aqueous solution of hydrogen chloride, is another name for hydrochloric acid. It has an unmistakably strong odour and is colourless. It's also categorised as a strong acid. In the digestive tracts of the majority of animal species, including humans, it is a part of the stomach acid.
Let's now examine the ions$Hg_{2}^{2+},H{{g}^{2+}},P{{b}^{2+}}$ and $C{{d}^{2+}}$ that an aqueous solution.
The group $I$or the silver group comprises of $H{g^{2 + }},P{b^{2 + }}$and $A{g^{2 + }}$ions, and the cations of the group $I$elements have very low solubility products and hence are precipitated as chloride by the addition of dilute $HCl$.
In the absence of group $I$, the test for the group $II$is conducted then we noticed that $C{u^{2 + }},P{b^{2 + }}$in group $IIA$gives black precipitation. This means that when dilute $HCl$is present, the group $II$ radicals precipitate as sulphides and cations of group $II$have low solubility products.
$Hg_2^{2 + } + 2HCl \to H{g_2}C{l_2} \downarrow + 2{H^ + } \\$
$P{b^{2 + }} + 2HCl \to PbC{l_2} \downarrow + 2{H^ + } \\$
Therefore, $Hg_2^{2 + }$and $P{b^{2 + }}$ have been placed in first group of basic radicals as their$({k_{sp}})$ are less than the $II$group ions $H{g^{2 + }}$and $C{d^{2 + }}$. So, only these will precipitate.
Thus, the correct option is: (D) $H{g_2}C{l_2}$and $PbC{l_2}$.
Note: It should be noted that an ion with a basic character that comes from a base is referred to as a basic radical. It's a positive-charged chemical species. Small amounts of inorganic salt could be present. The hydroxyl ion from a base can be removed to create a basic radical.
Complete Step by Step Solution:
Let’s start by defining $HCl(6N)$: Muriatic acid, which is an aqueous solution of hydrogen chloride, is another name for hydrochloric acid. It has an unmistakably strong odour and is colourless. It's also categorised as a strong acid. In the digestive tracts of the majority of animal species, including humans, it is a part of the stomach acid.
Let's now examine the ions$Hg_{2}^{2+},H{{g}^{2+}},P{{b}^{2+}}$ and $C{{d}^{2+}}$ that an aqueous solution.
The group $I$or the silver group comprises of $H{g^{2 + }},P{b^{2 + }}$and $A{g^{2 + }}$ions, and the cations of the group $I$elements have very low solubility products and hence are precipitated as chloride by the addition of dilute $HCl$.
In the absence of group $I$, the test for the group $II$is conducted then we noticed that $C{u^{2 + }},P{b^{2 + }}$in group $IIA$gives black precipitation. This means that when dilute $HCl$is present, the group $II$ radicals precipitate as sulphides and cations of group $II$have low solubility products.
$Hg_2^{2 + } + 2HCl \to H{g_2}C{l_2} \downarrow + 2{H^ + } \\$
$P{b^{2 + }} + 2HCl \to PbC{l_2} \downarrow + 2{H^ + } \\$
Therefore, $Hg_2^{2 + }$and $P{b^{2 + }}$ have been placed in first group of basic radicals as their$({k_{sp}})$ are less than the $II$group ions $H{g^{2 + }}$and $C{d^{2 + }}$. So, only these will precipitate.
Thus, the correct option is: (D) $H{g_2}C{l_2}$and $PbC{l_2}$.
Note: It should be noted that an ion with a basic character that comes from a base is referred to as a basic radical. It's a positive-charged chemical species. Small amounts of inorganic salt could be present. The hydroxyl ion from a base can be removed to create a basic radical.
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