(i) Its pH does not change on the addition of small amount of acid or base.
(ii) Its pH does not change on dilution or standing.

Conjugate acid is H₃O⁺ and conjugate base is OH^- .

Common salt (Nacl) is an electrolyte which in the aqueous solution gives Na⁺ and Cl^- ions.Hence, it conducts electricity.Sugar is sucrose (C₁₂H₂₂O₁₁ ) which is a non-electrolyte and does not give ions in the solution.Hence, it does not conduct electricity.

For the reaction 
\(NO(g)+{ O }_{ 3 }(g)\rightleftharpoons NO_{ 2 }(g)+{ O }_{ 2 }(g)\)
\(({ K }_{ c }=6.3\times { 10 }^{ 14 }at\quad 1000K)\)
\({ K }_{ c }=\frac { [{ NO }_{ 2 }][{ O }_{ 2 }] }{ [NO][{ O }_{ 3 }] } =6.3\times { 10 }^{ 14 }\)     ...(i)
For reverse reaction,
\(​​NO_{ 2 }(g)+{ O }_{ 2 }(g)\rightleftharpoons NO(g)+{ O }_{ 3 }(g)\)
\({ K }'_{ c }=\frac { [{ NO }][{ O }_{ 3 }] }{ [NO_{ 2 }][{ O }_{ 2 }] } \)    ...(ii)
From Eqs. (i) and (ii) we, get
\({ K }'_{ c }=\frac { 1 }{ { K }_{ c } } =\frac { 1 }{ 6.3\times { 10 }^{ 14 } } =1.587\times { 10 }^{ -15 }\)

Precipitation takes place when ionic product exceeds the solubility product.

\( A g B r \rightarrow[A g][B r]\)
\( k s p \rightarrow(s) \times(s) \)
\( k s p=s^2 \)
\(s=\sqrt{k s p}=\sqrt{3.3 \times 10^{-3}} \)
\( =5.477 \times 10^{-7} \)
= 5.7 x 10^-7 mol L^-1

k=k_f/k_b . In exothermic reaction, with increase of temperature, k_b increases much more than k_f. Hence, K decreases.

Molar concentration of pure solid or liquid (if in excess) is constant (i.e. independent of the amount present ) . That's why pure liquids and solids can be ignored while the equilibrium constant expression.

(i) K_c = \(\frac { [Ag^{ + }(aq)]^{ 2 } }{ [Cu^{ 2+ }(aq)] } \)
(ii) K_c = \(\frac { [CI_{ 2 }(g)]^{ 2 }[H_{ 2 }O(g)]^{ 2 } }{ [HCI(g)]^{ 4 }[O_{ 2 }(g)] } \)

The equilibrium in the reaction is:
C₂H₆ (g) ⇌ C₂H₄ (g) + H₂(g)
Initial pressure: 4 atm 0 0
Eqn. pressure: (4-p) atm p atm p atm
K_p = \(\frac { p_{ c_{ 2 }H_{ 4 } }\times p_{ H_{ 2 } } }{ p_{ c_{ 2 }H_{ 6 } } } \) or 0.04 = \(\frac { { p }^{ 2 } }{ (4-p) } \)
p² = 0.04 (4-p) or p² + 0.04 p - 0.16 = 0
p = \(\frac { (-0.04)\pm \sqrt { 0.0016-4(-0.16) } }{ 2 } \)
= \(\frac { (-0.04)\pm 0.8 }{ 2 } \)
= \(\frac { 0.76 }{ 2 } \)
= 0.38
Equilibrium pressure or concentration of C₂H₆ = (4 - 0.38) = 3.62 atm.

K_p = K_c (RT)^∆ng or K_c = \(\frac { K_{ p } }{ (RT)^{ \Delta ng } } \) = K_p (RT)^-∆ng
K_p = 2.0 x 10¹⁰ bar^-1 ; R = 0.083 L bar K^-1 mol^-1 ; T = 450 K ; ∆^ng = 2-3 = -1
K_c = (2.0 x 10¹⁰ bar^-1) x [(0.083 L bar K^-1 mol^-1) x (450 K)]^-(-1)
= 7.47 x 10¹¹ mol^-1 
L = 7.47 x 10¹¹ mol^-1

(I). Calculation of hydrolysis constant \(\left(K_h\right)\).
Since sodium acetate is a salt of strong base and weak acid
\(K_h=\frac{K_w}{K_a}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.5 \times 10^{-10}\)
(II). Calculation of degree of hydrolysis (h).
\(K_h=5.5 \times 10^{-10}, C=0.1 M\)
Degree of hydrolysis (h)
\( =\sqrt{\frac{K_h}{C}}=\sqrt{\frac{5.5 \times 10^{-10}}{0.1}}=\left(55 \times 10^{-10}\right)^{1 / 2}  =7.42 \times 10^{-5} \)

(i)

(ii) The solubility equilibrium in the saturated solution is
AgCl (s) ⇌ Ag⁺(aq) + CI^-(aq)
The solubility of AgCl is 1.06 x 10^-5 mole per litre.
[Ag⁺(aq)] = 1.06 x 10^-5 mol L^-1
[Cl^-(aq)] 1.06 x 10^-5 mol L^-1
Ksp [Ag⁺(aq)] [Cl^- (aq)]
(1.06 x 10^-5 mol L^-1) x (1.06 x 10^-5 mol L^-1)
= 1.12 x 10^-2 mol²L^-2

\(pH={ pK }^{ a }+\log { \frac { \left[ Salt \right] }{ \left[ Acid \right] } } =-\log { \left( 1.75\times { 10 }^{ -5 } \right) } +\log { \frac { 0.15 }{ 0.10 } } \quad \)
= (5 - 0.2430) + 0.1761 =4.757 + 0.1761 = 4.933
1 cc of 1 M NaOH contains NaOH = 10^-3 mol. This will convent 10^-3 mol of acetic acid into the salt so that salt formed = 10^-3 mol.
Now, [Acid] = 0.10 - 0.001 = 0.099 M
[Salt]  = 0.15 + 0.001 = 0.151 M
pH = 4.757 + \(\log { \frac { 0.151 }{ 0.099 } } \)
= 4.757 + 0.183 =4.940
Increase in pH = 4.940 - 4.933 = 0.007 which is negligible.

Method 1
1) \(\mathrm{CH}_3 \mathrm{COOH} \leftrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \quad \mathrm{K}_3=1.74 \times 10^{-5}\)
2) \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-} \quad K_w=1.0 \times 10^{-14}\)
Since \(\mathrm{Ka} >> \mathrm{K}_2\) :
\(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CHCOO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}\)
c1          0.05                    0               0
              0.05-.05 a          0.5a          0.5a
\( K_a=\frac{(.05 a)(.05 a)}{(.05 a-0.05 a)} \)
\( =\frac{(.05 a \times 0.05 a)}{.05(1-a)} \)
\( =\frac{.05 a^2}{1-a} \)
\( 1.74 \times 10^{-5}=\frac{0.05 a^2}{1-a} \)
\( 1.74 \times 10^{-5}-1.74 \times 10^{-5} a=0.05 a^2 \)
\( 0.05 a^2+1.74 \times 10^{-5} a-1.74 \times 10^{-5} \)
\( D=b^2-4 a c \)
\( =\left(1.74 \times 10^{-5}\right)^2-4(.05)\left(1.74 \times 10^{-5}\right) \)
\( =3.02 \times 10^{-25}+.348 \times 10^{-5}\)
\( a=\sqrt{\frac{K_a}{c}}\)
\( a=\sqrt{\frac{1.74 \times 10^{-5}}{.05}} \)
\( =\sqrt{\frac{34.8 \times 10^{-5} \times 10}{10}} \)
\( =\sqrt{3.48 \times 10^{-6}} \)
\( =\mathrm{CH}_3 \mathrm{COOH} \leftrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \)
\( \frac{a 1.86 \times 10^{-5}}{\left[\mathrm{CH} \mathrm{COO}_3^{-7}\right]} \)
\(=\frac{0.93 \times 10^{-5}}{1000} \)
\( =.00009 \times 1.86 \times 10^{-3}\)
= .000093
Method 2
Degree of dissociation,
\( a=\sqrt{\frac{K_a}{c}} \)
\( C=0.05 \mathrm{M} \)
\( \mathrm{K}_a=1.74 \times {10^{-5}}\)
\({\text { Then, }} \mathrm{a}=\sqrt{\frac{1.74 \times 10^{-5}}{.05}} \mathrm{a}=\sqrt{34.8 \times 10^{-5}}\)
\(\mathrm{a}=\sqrt{3.48} \times 10^{-4}\)
\(\mathrm{a}=1.8610^{-2}\)
\(\mathrm{CH}_3 \mathrm{COOH} \leftrightarrow \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\)
Thus, concentration of \(\mathrm{CH}_3 \mathrm{COO}^{-}=\mathrm{C}\). a
=0.5 ×1.86×10⁻²
=.093 × 10⁻²=.00093M
Since [oAc−]=[H⁺][H⁺]=.00093
=.093×10⁻²pH=−log[H⁺]=−log(.093×10⁻²)
∴pH=3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.