Calculation of K_p 
PCl₅(g) ⇌ PCl₃(g) + Cl₂ (g)
     1             0               0
    (1- α)       α              α
Total no. of moles in the equilibrium mixture = 1 - α + α + α
= (1 + α) mol.
Let the total pressure of equilibrium mixture = p atm
Partial pressure of PCl₅
\(p_{ { PCI }_{ 5 } }\ =\ \frac { 1-\alpha }{ 1+\alpha } \times p\ atm\)
Partial pressure of PCI₃ = \(\ \frac { \alpha }{ 1+\alpha } \times p\ atm\)
Partial pressure of CI₂
\(p_{ { CI }_{ 2 } }\ =\ \frac { \alpha }{ 1+\alpha } \times p\ atm\)
K_p = \(\frac { { p }_{ { PCI }_{ 3 } }\times { p }_{ { CI }_{ 2 } } }{ { p }_{ PCI_{ s } } } \)
= \(\frac { \left( \frac { \alpha }{ 1+\alpha } p\ atm \right) \left( \frac { \alpha }{ 1+\alpha } p\ atm \right) }{ \frac { 1-\alpha }{ 1+\alpha } p\ atm } =\frac { { \alpha }^{ 2 }p }{ 1-\alpha ^{ 2 } } atm\)
P = 4 atm and α = 10% = \(\frac { 10 }{ 100 } \) = 0.1
K_p = \(\frac { (0.1)\times (0.1)\times (4\ atm) }{ 1-(0.1{ ) }^{ 2 } } \)
= \(\frac { 0.04 }{ 0.99 } \) = 0.04 atm.
Calculation of P under new condition
α = 0.2, K_p = 0.04 atm
K_p = \(\frac { { \alpha }^{ 2 }p }{ 1-\alpha ^{ 2 } } \)or \(\frac { { K_{ p }(1-\alpha }^{ 2 }) }{ \alpha ^{ 2 } } \)
= \(\frac { (0.04atm)[(1-(0.2)^{ 2 }] }{ (0.2{ ) }^{ 2 } } \)
= \(\frac { 0.04atm\times 0.96 }{ 0.04 } \)
= 0.96 atm.

(a) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}(\mathrm{S}) \rightleftharpoons \mathrm{C}_5 \mathrm{H}_5 \mathrm{COO}^{-}+\mathrm{Ag}^{+}: \mathrm{K}_1=\mathrm{K}_{ \mathrm{sp}}\)
\( \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-4}+\mathrm{H}^{+}=\mathrm{C}_6 \mathrm{H}_2 \mathrm{COOH}^{-} \quad \mathrm{K}_2=\frac{1}{K_a} \)
\(\mathrm{C}_4 \mathrm{H}_5 \mathrm{COOAg}(5)^{+}-\mathrm{H}^{+} \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{Ag}^{+} ; K_3=\frac{K_{sp}}{K_a}\)
\(K_3=\frac{\left.\left[\mathrm{C}_{\mathrm{s}} \mathrm{H}_5 \mathrm{COOH}\right] \mathrm{OAg}^{+}\right] \mathrm{s} . \mathrm{s}}{\left[\mathrm{H}^{+}\right]}=\frac{s^2}{\left[\mathrm{H}^{+}\right]}=\frac{K_{\mathrm{sp}}}{\mathrm{K}_n}\)
Where, s is the solubility of \(\mathrm{C}_{\mathrm{e}} \mathrm{H}_2 \mathrm{COOAg}\), In a buffer of \(\mathrm{pH}=3.19\)
\(\log \left[\mathrm{H}^*\right]=-3.19=\overline{4} .81\)
\( {\left[\mathrm{H}^{+}\right]=\operatorname{antilog} 4 . \mathrm{a} 1=6.46 \times 10^{-4}} \)
\( \frac{\mathrm{s}^2}{\left[\mathrm{H}^{+}\right]}=\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{K}_{\mathrm{a}}} \text { or } \mathrm{s}^2=\frac{\mathrm{K}_{\mathrm{sp}} \times\left[\mathrm{H}^*\right]}{\mathrm{K}_{\mathrm{a}}}\)
\(s=\sqrt{\frac{2.5 \times 10^{-13} \times 6.46 \times 10^{-4}}{6.46 \times 10^{-5}}}\)
\(s=\sqrt{2.5 \times 10^{-13} \times 10}\)
\(\mathrm{s}=1.6 \times 10^{-6} \mathrm{M}(\text { in buffer })\)
In aqueous solution, solubility of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}\);
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]\left[\mathrm{Ag}^{+}\right]=\mathrm{s} . \mathrm{s}=\mathrm{s}^2\)
\(s=\sqrt{K_{s p}}=\sqrt{2.5 \times 10^{-13}}=5 \times 10^{-7} \mathrm{M}\)
\(\frac{\mathrm{S}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}\right) \text { in buffer }}{\mathrm{S}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}\right) \text { in aqueous solutipo } \times 10^{-7}}=3.2\)
C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.

The first step of the dissociation of H₂S is 
\({ H }_{ 2 }S\rightleftharpoons { H }^{ \oplus }{ +HS }^{ \ominus }\)
\({ K }_{ 1 }=\frac { \left[ { H }^{ \oplus } \right] \left[ { HS }^{ \ominus } \right] }{ { H }_{ 2 }S } =4\times { 10 }^{ -3 }\)
\(\left[ { H }^{ \oplus } \right] =C\alpha ,\left[ { HS }^{ \ominus } \right] =C\alpha ,\left[ { H }_{ 2 }S \right] =C(1-\alpha )\)
\(4\times { 10 }^{ -3 }=\frac { C\alpha .C\alpha }{ C(1-\alpha ) } =\frac { { C }\alpha ^{ 2 } }{ (1-\alpha ) }\)
\(4\times { 10 }^{ -3 }=\frac { 0.1\times { \alpha ^{ 2 } } }{ (1-\alpha ) } (1-\alpha \ shpold \ not \ be \ negleced)\)
\(\alpha =0.18,\)
\(\left[ { H }^{ \oplus } \right] =C\alpha =0.1\times 0.18=0.018M\)
\(\left[ { HS }^{ \ominus } \right] =C\alpha =0.1\times 0.18=0.018M\)
\( \left[ { H }_{ 2 }S \right] =C(1-\alpha )=0.1(1-0.18)=0.082M\)
Now, \(\left[ { HS }^{ \ominus } \right] \) further dissociates to \(\left[ { H }^{ \oplus } \right] \) and S^2-
\(\underset { \underset { (1-{ \alpha }_{ 1 }) }{ 1 } }{ { HS }^{ \ominus } } \quad \rightleftharpoons \quad \underset { \underset { { \alpha }_{ 1 } }{ 1 } }{ { H }^{ \oplus } } +\quad \underset { \underset { { \alpha }_{ 1 } }{ 0 } }{ { S }^{ 2- } } \)
\({ K }_{ 2 }=1\times { 10 }^{ -5 }=\frac { \left[ { H }^{ \oplus } \right] \left[ { S }^{ 2- } \right] }{ \left[ { HS }^{ \ominus } \right] } \)
Because \(\left[ { H }^{ \oplus } \right] \)already in solution = 0.018 and thus, dissociation of \({ HS }^{ \ominus }\) further suppresses due to common ion effect and \(1-\alpha \approx 1\)
\(1\times { 10 }^{ -5 }=\frac { 0.018\times { C }_{ 1 }{ \alpha }_{ 1 } }{ { C }_{ 1 }{ (1-\alpha }_{ 1 }) } =0.018\times { \alpha }_{ 1 }\)
\( { \alpha }_{ 1 }\quad =\frac { 1\times { 10 }^{ -5 } }{ 0.018 } =5.55\times { 10 }^{ -4 }\)
\(\left[ { S }^{ 2- } \right] ={ C }_{ 1 }{ \alpha }_{ 1 }=0.018\times 5.55\times { 10 }^{ -4 }=0.099\times { 10 }^{ -4 }\)

We know that, \({ K }_{ p }={ K }_{ c }{ \left( RT \right) }^{ \triangle ng }\)
For the above reaction,
\(\triangle ng=\left( 2+1 \right) -2=1\)
\({ K }_{ p }=3.75\times { 10 }^{ -6 }\left( 0.0831\times 1069 \right) \)
K_p = 0.033

Let \( \mathrm{ VmL}\) of \(\mathrm{NH}_4 \mathrm{OH}\) be mixed with \(\mathrm{NH}_4 \mathrm{CI}\) to have a buffer of \(p H \ 8.65\). The total volume after mixing becomes (V+30) m L.
m mole of \(\mathrm{NH}_4 \mathrm{OH}=0.3 \times \mathrm{V}\)
\( \therefore\left[\mathrm{NH}_4 \mathrm{OH}\right]=\frac{0.3 \times \mathrm{V}}{(\mathrm{V}+30)} \)
\( \mathrm{m} \text { mole of } \mathrm{NH}_4 \mathrm{CI}=0.2 \times 30 \)
\( \therefore\left[\mathrm{NH}_4 \mathrm{CI}\right]=\frac{0.2 \times 30}{(V+30)}\)
Also pOH of buffer mixture is given by:
\( p O H=p K_b=\log \frac{[\text { Salt }]}{[\text { Base }]} \text { Itbgt or } 14-8.65=4.74+\log \)
\( \frac{(0.2 \times 30) /(V+30)}{(0.3 \times V) /(V+30)},(p O H=n 14-p H) \)
\( 0.61=\log \frac{6}{0.3 \times V} \)
\(\therefore V=4.91 m L\)
Similary calculate,
\(\left(14-10=4.74+\log \left(\frac{0.2 \times 30 /\left(V_1+30\right)}{0.3 \times V_1 /\left(V_1+30\right)}\right)\right)\)
For \(p H=10, V=109.9 m L\)

When catalyst is added, the state of equilibrium is not distributed but equilibrium is attained quickly. This is because the catalyst increases the rate of forward and backward reaction to the same extent.