(A) . $A$ dilute aqueous solution of $NH_3$ exists as $NH_4OH$,which dissociates as: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$. On adding solid ammoni

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$pH$ decreases when solid ammonium chloride is added to a dilute aqueous solution of $NH_3$

(A) . $A$ dilute aqueous solution of $NH_3$ exists as $NH_4OH$,which dissociates as: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$. On adding solid ammonium chloride $(NH_4Cl)$,it dissociates completely: $NH_4Cl \rightarrow NH_4^+ + Cl^-$. Due to the common ion effect of $NH_4^+$,the equilibrium shifts to the left,decreasing the concentration of $OH^-$. Since $pOH = -\log[OH^-]$,a decrease in $[OH^-]$ increases $pOH$,which in turn decreases the $pH$ $(pH = 14 - pOH)$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Common ion effect

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Among the following,the correct statement is

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A
$pH$ decreases when solid ammonium chloride is added to a dilute aqueous solution of $NH_3$
B
$pH$ decreases when solid sodium acetate is added to a dilute aqueous solution of acetic acid
C
$pH$ decreases when solid $NaCl$ is added to a dilute aqueous solution of $NaOH$
D
$pH$ decreases when solid sodium oxalate is added to a dilute aqueous solution of oxalic acid

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6-2.Equilibrium-II (Ionic Equilibrium)

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Common ion effect

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The following equilibrium exists in an aqueous solution of hydrogen sulphide: $H_2S \rightleftharpoons H^{+} + HS^{-}$. If dilute $HCl$ is added to an aqueous solution of $H_2S$ without any change in temperature,what happens?

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Statement $A$: Addition of $NH_4OH$ in the presence of excess $NH_4Cl$ to an aqueous solution of $BaCl_2$ results in the precipitation of $Ba(OH)_2$.
Reason $R$: $Ba(OH)_2$ is insoluble in water.

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Assertion : $Sb(III)$ is not precipitated as sulphide when $H_2S$ is passed through its alkaline solution.
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DifficultAIIMS 2004

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At $298 \ K$ the molar solubility of $Cd(OH)_2$ in $0.1 \ M \ KOH$ solution is $x \times 10^{-y}$. The values of $x$ and $y$ are respectively (at $298 \ K, \ K_{sp}$ of $Cd(OH)_2 = 2.5 \times 10^{-14}$)

MediumTS EAMCET 2023

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In the presence of $HCl$,$H_2S$ results in the precipitation of Group-$2$ elements but not Group-$4$ elements during qualitative analysis. It is due to

MediumKCET 2015

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Among the following,the correct statement is

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The following equilibrium exists in an aqueous solution of hydrogen sulphide: $H_2S \rightleftharpoons H^{+} + HS^{-}$. If dilute $HCl$ is added to an aqueous solution of $H_2S$ without any change in temperature,what happens?

Statement $A$: Addition of $NH_4OH$ in the presence of excess $NH_4Cl$ to an aqueous solution of $BaCl_2$ results in the precipitation of $Ba(OH)_2$.Reason $R$: $Ba(OH)_2$ is insoluble in water.

Assertion : $Sb(III)$ is not precipitated as sulphide when $H_2S$ is passed through its alkaline solution.Reason : The concentration of $S^{2-}$ ion in alkaline medium is inadequate for precipitation.

At $298 \ K$ the molar solubility of $Cd(OH)_2$ in $0.1 \ M \ KOH$ solution is $x \times 10^{-y}$. The values of $x$ and $y$ are respectively (at $298 \ K, \ K_{sp}$ of $Cd(OH)_2 = 2.5 \times 10^{-14}$)

In the presence of $HCl$,$H_2S$ results in the precipitation of Group-$2$ elements but not Group-$4$ elements during qualitative analysis. It is due to

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Statement $A$: Addition of $NH_4OH$ in the presence of excess $NH_4Cl$ to an aqueous solution of $BaCl_2$ results in the precipitation of $Ba(OH)_2$.
Reason $R$: $Ba(OH)_2$ is insoluble in water.

Assertion : $Sb(III)$ is not precipitated as sulphide when $H_2S$ is passed through its alkaline solution.
Reason : The concentration of $S^{2-}$ ion in alkaline medium is inadequate for precipitation.