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

For F^-, K_b = K_w/ K_a = 10^-14 / (6.8 x 10^-4) = 1.47 x 10^-11 ≃ 1.5 x 10^-11.
For HCOO^-, K_b = 10^-14/ (1.8 x 10^-4) = 5.6 x 10^-11
For CN^-, K_b = 10^-14 / (4.8 x 10^-9) = 2.08 x 10^-6

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

Using the reaction and the given degree of dissociation, frame the partial pressure equations and solve by using K_p.
Suppose α is the degree of dissociation, then
PCl₅ ⇋ PCl + Cl₂
Initial conc. 1mole  0   0
At. Equilibrium 1 -  α   α
∴ Total number of moles at equilibrium
= 1 -  α + α + α = 1 + α 
If P is the total pressure at equilibrium, then partial pressures will be
\( \mathrm{p}_{\mathrm{PCl}_3}=\frac{\alpha}{1+\alpha} \mathrm{P}, \quad \mathrm{p}_{\mathrm{Cl}_2}=\frac{\alpha}{1+\alpha} \mathrm{P}, \)
\( \mathrm{p}_{\mathrm{PCl}_5}=\frac{1-\alpha}{1+\alpha} \mathrm{P} ; \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{PCl}_3} \times \mathrm{p}_{\mathrm{Cl}_2}}{\mathrm{p}_{\mathrm{PCl}_5}}\)
Substituting the values in above equation,
\( =\frac{\left(\frac{\alpha}{1+\alpha} \mathrm{P}\right) \times\left(\frac{\alpha}{1+\alpha} \mathrm{P}\right)}{\left(\frac{1-\alpha}{1+\alpha} \cdot \mathrm{P}\right)} \)
\( \mathrm{K}_{\mathrm{p}}=\frac{\alpha^2}{1-\alpha^2} \cdot \mathrm{P}\)
We are given that at \(\mathrm{P}=1 \mathrm{~atm}, \mathrm{a}=0.2\).
Hence, \(\mathrm{K}_{\mathrm{p}}=\frac{(0.2)^2}{1-(0.2)^2} \times 1 \therefore \mathrm{K}_{\mathrm{p}}=0.042\)
At 50% dissociation i.e, α  = 0.5
Suppose total pressure is P'. 
hen A(g) + 2B(g) → C(g)
0.042 = (0.5)²/1 - (0.5)2  x p
p' = 0.1272 atm

\({ CaCO }_{ 3 }(s)\rightleftharpoons CaO(s)+{ CO }_{ 2 }(g)\) 
K_p = \({ p }_{ { CO }_{ 2 } }\) = 1.15 atm, pV = nRT
\({ N }_{ { CO }_{ 2 } }=\frac { { P }_{ { CO }_{ 2 } } }{ RT } =\frac { 1.15\times 5 }{ 0.082\times 1073 } =0.065mol\) 
1 mole of CO₂ is obtained by decomposition of 1 mole CaCO₃. Therefore, moles of CaCO₃ decomposed is equal to the moles of CO₂ = 0.065 mol.
Moles of CaCO₃ initially present = \(\frac { 50 }{ 100 } =0.5 \ mol\) 
[Molecular mass of CaCO₃ = 100]
Percent of CaCO₃ decomposed  \(=\frac { 0.065 }{ 0.5 } \times 100=13\)%

Amount of SO₂ in atmospheres = 10PPm =\(\frac { 10 }{ { 10 }^{ 6 } } ={ 10 }^{ -5 }\)
Molar conc.of SO₂ in pressure of water 
= amount of SO₂ x solubility of SO₂ in water
H₂SO₃ dissociates as = 1.3653 x 10^-5
\({ H }_{ 2 }{ SO }_{ 3 }\rightleftharpoons { H }^{ + }{ HSO }_{ 3 }^{ - }\)
Initial conc   1.3653 x 10^-5                            0          0
Molar conc.of equiv(1.3653 x 10^-5   x )       x            x
\({ K }_{ a }=\frac { { x }^{ 2 } }{ (1.3653\times { 10 }^{ -5 }-x) } \)
\( { PK }_{ a }=1.92\)
log K _a = 1.92 or K _a= 1.2 x 10 ^-2
Substituting 1.2 x 10^-2 =\(\frac { { x }^{ 2 } }{ (1.3653\times { 10 }^{ -5 }x) } \)
x² = 1.2 x 10^-2(1.3653  x 10^-5-x)
On solving, we get x = 1.3664 x 10^-5
pH = -log(1.364 x 10^-5 -x)
On solving, we get x  = 1.3664 x 10^-5
pH = -log(1.364 x 10^-5) = 4.865

If K_a is the ionisation constant of weak acid (HA) and K_b is the ionisation constant of its conjugate base(A^-) then K_a K_b = K_w
\({ K }_{ b }({ F }^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HF) } =\frac { 1\times { 10 }^{ -14 } }{ 6.8\times { 10 }^{ -4 } } =1.47\times { 10 }^{ -11 }\)
\({ K }_{ b }(HCOO^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HCOOH) } =\frac { 1\times { 10 }^{ -14 } }{ 1.8\times { 10 }^{ -4 } } =5.56\times { 10 }^{ -11 }\)
\({ K }_{ b }(Cn^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HCn) } =\frac { 1\times { 10 }^{ -14 } }{ 4.8\times { 10 }^{ -4 } } =2.08\times { 10 }^{ -6 }\)

(a) \(2NOCl(g)\rightleftharpoons 2NO(g)+Cl_{ 2 }(g);\)
\( { K }_{ p }=1.8\times { 10 }^{ -2 }at500k\)
\(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; K_{p}=1.8 \times 10^{-2} \text { at } 500 \mathrm{~K}\)
(b) \(CaCO_{ 3 }(s)\leftrightharpoons CaO(s)+CO_{ 2 }(g);\)
\({ K }_{ p }=167 \ at \ 1073 \ K\)
\( { \Delta n }_{ g }={ n }_{ p }-{ n }_{ R }=1\)
\( { K }_{ c }=\frac { { k }_{ p } }{ (RT)^{ \Delta n_{ g } } } =\frac { 167 }{ 0.0821\times 1073 } =1.89\)

The degree of ionisation \((\alpha)=1.34 \%=\frac{1.34}{100}=0.0134\) The ionisation of acetic may be represented as :
\(\begin{array}{l} \mathrm{CH}_3 \mathrm{COOH} \\ \mathrm{C}(1-\alpha) \end{array} \begin{array}{l} \leftrightarrow \mathrm{CH}_3 \mathrm{COO}^{-} + H^{+}\\ \mathrm{C} \alpha \ \ \ \ \ \ \ \ \ \ \ \mathrm{C} \alpha \end{array} \)
\({\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\mathrm{C} \alpha=0.1 \times 0.0134=0.00134 \mathrm{M} \text {, }} \)
\( {\left[\mathrm{H}^{+}\right]=C \alpha=0.1 \times 0.0134=0.00134 M} \)
\({\left[\mathrm{CH}_3 \mathrm{COOH}\right]=\mathrm{C}(1-\alpha)=0.1(1-0.0134)} \)
\( =0.09866 \mathrm{M} \)
\( K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \)
\( =\frac{(0.00134) \times(0.00134)}{0.09866}=1.82 \times 10^{-5} \)

Moles of sulphuric acid \(=\frac{4.9}{98}=0.05\)
\(\left[\mathrm{H}_2 \mathrm{SO}_4\right]=\frac{0.05}{500} \times 1000=0.1 \mathrm{M}\)
Sulphuric acid ionises as :
\( \mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O} \Leftrightarrow 2 \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{SO}_4^{2-} \)
\( \therefore\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=2 \times\left[\mathrm{H}_2 \mathrm{SO}_4\right]=2 \times 0.1=0.2 \mathrm{M} \)
\( \mathrm{pH}=-\log (0.2)=0.6990 \)

(a) Addition of \({ PCl }_{ 5 }\) have no effect on \({ K }_{ c }\) is constant at constant temperature.
(b) \({ K }_{ c }\) does not change with pressure
(c) The given reaction is endothermic, hence on increasing the temperature, K _f  will increase, this results increase in \({ K }_{ c }\) \(\left( { K }_{ c }=\frac { { K }_{ f } }{ { K }_{ b } } \right) \)

\({ N }_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons { 2NO }_{ 2 }(g)\)
\( Pressure \ at \ equilibrium \ 0.28\ \ \ 1.1\)
\( { K }_{ P }=\frac { p{ { (NO }_{ 2 }) }^{ 2 } }{ p({ N }_{ 2 }{ O }_{ 4 }) } =\frac { { (1.1) }^{ 2 } }{ (0.28) } =4.32atm\)
If volume of the container is doubled, the pressure will be reduced to half
\( { N }_{ 2 }{ O }_{ 4 }\leftrightharpoons { 2NO }_{ 2 }\)
\( New \ pressure \ \left( \frac { 0.28 }{ 2 } -p \right) \ \left( \frac { 1.1 }{ 2 } +2p \right)\)
\( { K }_{ p }=\frac { { \left( \frac { 1.1 }{ 2 } +2p \right) }^{ 2 } }{ \left( \frac { 0.28 }{ 2 } -p \right) } =4.32\)
\( On \ solving,\)
\( p=0.045\)
\( \therefore \ p({ N }_{ 2 }{ O }_{ 4 })=0.14-0.0045=0.095atm\)
\( p({ NO }_{ 2 })=0.55+0.045=0.64 \ atm\)

\({ CaSO }_{ 4 }\rightleftharpoons { Ca }^{ 2+ }+{ SO }_{ 4 }^{ 2- };{ K }_{ sp }=9.1\times { 10 }^{ -6 }\)
     S             S             S
Where s is the solubility of CaSO₄
\({ K }_{ sp }=\left[ { Ca }^{ 2+ } \right] \left[ { SO }_{ 4 }^{ 2- } \right] =S.S={ S }^{ 2 }\)
\(S=\sqrt { { K }_{ sp } } =\sqrt { 9.1\times { 10 }^{ -6 } } \Rightarrow S=3.017\times { 10 }^{ -3 }M\)
Solubility of CaSO₄ = 3.017 x 10^-3 mol^-1
= 3.017 x 10^-3 x 136 gL^-1
(Molar mass of CaSO₄ =136 g mol^-1)
= 410.3 x 10^-3 gL^-1
410.3x10^-3 g CaSO₄ is dissolved in = 1L
1g CaSO₄ is dissolved in = \(\frac { 1\times 1 }{ 410.3\times { 10 }^{ -3 } } \) = 2.437 L

Given,Kbfor dimethylamine =5.4×10⁻⁴
C for dimethylamine=0.02M
\( \alpha =\sqrt { { K }_{ b }/C } =\sqrt { \frac { 5.4\times { 10 }^{ -4 } }{ 0.02 } } =1.64\times { 10 }^{ -1 }=0.164\)
\(In \ the \ presence \ of \ 0.1 \ M \ NaOH,\)
\(({ CH }_{ 3 })_{ 2 }NH+{ H }_{ 2 }O\rightleftharpoons ({ CH }_{ 3 })_{ 2 }\overset { + }{ N } { H }_{ 2 }+{ OH }^{ - }\)
\(Initial \ conc. \quad 0.02M\quad 0\quad 0\)
\( Equili.conc. \quad (0.02-c\alpha )\quad c\alpha \quad c\alpha +0.1\)
\( \approx 0.02\quad \approx 0.1\)
\( (0.1from\quad 0.1M\quad NaOH)\)
\( { K }_{ b }=\frac { \left[ ({ CH }_{ 3 })_{ 2 }\overset { + }{ N } { H }_{ 2 }][{ OH }^{ - } \right] }{ \left[ ({ CH }_{ 3 })_{ 2 }NH \right] }\)
\( \therefore 5.4\times { 10 }^{ -4 }=\frac { 0.02\alpha \times 0.1 }{ 0.02 } \)
\( \alpha =\frac { 5.4\times { 10 }^{ -4 }\times 0.02 }{ 0.02\times 0.1 } 54\times { 10 }^{ -4 }=5.4\times { 10 }^{ -3 }\)
\(\alpha =\) 0.54%

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.