Question

Match the List I with List II:

List I	List II
(A) Degree of ionization of weak electrolytes increases on	(i) Common ion
(B) pH of water decreases on	(ii) $pH=4$
(C) The solution has hydronium ion concentration of $0.0001$ mol/litre	(iii) Heating
(D) The addition of $NaOH$ to $Ca{\left(OH\right)}_2$ solution precipitates $Ca{\left(OH\right)}_2$	(iv) dilution

• A. A-(iv) B-(iii) C-(i) D-(ii)
• B. A-(iii) B-(iv) C-(ii) D-(i)
• C. A-(iv) B-(iii) C-(ii) D-(i)
• D. A-(iv) B-(ii) C-(ii) D-(i)

Answer 1

The correct option is C A-(iv) B-(iii) C-(ii) D-(i)

(A) On dilution, the dielectric constant value increases, therefore, ionization of weak electrolyte increases as a result degree of ionization $(\propto )$ also increases

$\left(A\right)-\left(iv\right)$

(B) On increasing temperature the tendency of H-bonding formation decreases means more $H^{+}$ ions generated therefore $pH$ value decreases.

$\left(B\right)-\left(iii\right)$

(C) Hydronium on means $H_3{O}^{+}$ that is given

$H_3{O}^{+}=0.0001mLmol/L={10}^{-4}m$

$pH=-\log \left(H_3{O}^{+}\right)=-\log \left({10}^{-4}\right)$

$pH=4$

$C)-(ii)$

(D) Both $NaOH$ and $Ca(OH{)}_2$, contains hydroxyl group

$\left(O{H}^{-}\right)$ that is common in both therefore shows common ion effect.

$\left(D\right)-\left(i\right)$

Hence, option C is correct.