Question

Column I		Column II
A	$10^{-2} M \left( NH _4\right)_2 SO _4$ solution	p	Cationic hydrolysis
B	$10^{-2} M HCl$ solution	q	Anionic hydrolysis
C	$10^{-2} M NH _3$ solution	r	$pH$ charges by one unit when diluted to one - tenth of its concentration
D	$10^{-2} MCH _3 COONH _4$ solution	s	$pH >7$ at $25^{\circ} C$

• A. A - (s,t), B - (r), C - (p,q), D - (p,t)
• B. A - (p,t), B - (s,t), C - (r), D - (p,q)
• C. A - (s,t), B - (p,q), C - (p,t), D - (r)
• D. A - (r), B - (q,t), C - (s,t), D - (p,q)

Answer 1

Answer: (B)

$(a \rightarrow p, t)$ only $\overset{\oplus}{N}H_4$ will get hydrolysed (cation hydrolysis) $pH <7$
$pH =\frac{1}{2}\left( p K_w- p K_b-\log C\right)$
$( b \rightarrow s , t )$ It is a $S_A \cdot pH =2$,
New conc $=\frac{10^{-2}}{10}= 10^{-3}, pH =3$
$( c \rightarrow r )$ It is a $W_B$, so $pH >7$
$( d \rightarrow p , q )$ Salt of $W_A / W_B$ both $CH _3 COO ^{\ominus}$ and $\overset{\oplus}{N}H _4$ will hydrolysed so both cationic and anionic hydrolysis