What will be the change in acidity if(i) CuSO | vedclass

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(A) $(i)$ In a saturated $(NH_4)_2SO_4$ solution, $NH_4^+$ ions are in equilibrium with $NH_3$ and $H^+$. When $CuSO_4$ is added, $Cu^{2+}$ reacts wit

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increase, increase

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(A) $(i)$ In a saturated $(NH_4)_2SO_4$ solution, $NH_4^+$ ions are in equilibrium with $NH_3$ and $H^+$. When $CuSO_4$ is added, $Cu^{2+}$ reacts with $NH_3$ to form the stable complex $[Cu(NH_3)_4]^{2+}$.$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightarrow [Cu(NH_3)_4]^{2+}_{(aq)}$This reaction consumes $NH_3$, shifting the equilibrium $NH_4^+ \rightleftharpoons NH_3 + H^+$ to the right, which increases the $H^+$ ion concentration and thus increases the acidity.$(ii)$ In anhydrous $HF$, the auto-ionization equilibrium is $2HF \rightleftharpoons H_2F^+ + F^-$.When $SbF_5$ is added, it acts as a strong Lewis acid and reacts with $F^-$ to form the stable $SbF_6^-$ ion $(SbF_5 + F^- \rightarrow SbF_6^-)$.This removes $F^-$ from the solution, shifting the equilibrium to the right, which increases the concentration of $H_2F^+$ (the superacidic species), thereby increasing the acidity.

List-$I$	List-$II$
$A$. $Aq.$ solution of $AlCl_3$	$I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$	$II$. Acidic
$C$. $Aq.$ solution of $KCl$	$III$. Highly conductive
$D$. $Al_2O_3$	$IV$. Strongly basic
	$V$. Amphoteric

What will be the change in acidity if
$(i)$ $CuSO_4$ is added in saturated $(NH_4)_2SO_4$ solution
$(ii)$ $SbF_5$ is added in anhydrous $HF$?

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Match the following:
List-$I$ List-$II$
$A$. $Aq.$ solution of $AlCl_3$ $I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$ $II$. Acidic
$C$. $Aq.$ solution of $KCl$ $III$. Highly conductive
$D$. $Al_2O_3$ $IV$. Strongly basic
$V$. Amphoteric

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What will be the change in acidity if$(i)$ $CuSO_4$ is added in saturated $(NH_4)_2SO_4$ solution$(ii)$ $SbF_5$ is added in anhydrous $HF$?