Acid A with pK_a = 1.5 is strongest acid , Lower the value of pK_a stronger will be the acid.

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
\({ K }'_{ c }=\frac { 1 }{ { K }_{ c } } =\frac { 1 }{ 6.3\times { 10 }^{ 14 } } =1.587\times { 10 }^{ -15 }\)

\(pH=\cfrac { 1 }{ 2 } p{ k }_{ w }+\cfrac { 1 }{ 2 } p{ k }_{ a }-\cfrac { 1 }{ 2 } p{ K }_{ b }\)
\(=\cfrac { 1 }{ 2 } (14)+\cfrac { 1 }{ 2 } (4.75)-\cfrac { 1 }{ 2 } (4.74)=7\)

\({ 3O }_{ 2 }(g)\rightleftharpoons { 2O }_{ 3 }(g)\quad \)
\({ K }_{ c }=\frac { \left[ { O }_{ 3 } \right] ^{ 2 } }{ \left[ { O }_{ 2 } \right] ^{ 3 } } \)
or \({ O }_{ 3 }=\sqrt { { K }_{ c }\left[ { O }_{ 2 } \right] } ^{ 3 }=\sqrt { (2.0\times { 10 }^{ -50 })(1.6\times { 10 }^{ -2 })^{ 3 } } \)
\(\left[ { O }_{ 3 } \right] =2.86\times { 10 }^{ -28 }mol{ L }^{ -1 }\)

Precipitation takes place when ionic product exceeds the solubility product.

\(\mathrm{Ag}_2 \mathrm{CrO}_4(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cro}_4^{2-}(a q)\)
\(K_{\text {sp }}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{Cro}_4^{2-}\right]=(2 s)^2(s)=4 s^3\)

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

It will remain the same because rate of inflow is equal to rate of outflow. The state is called of 'equilibrium'.

\(pH={ pK }^{ a }+\log { \frac { \left[ Salt \right] }{ \left[ Acid \right] } } =-\log { \left( 1.75\times { 10 }^{ -5 } \right) } +\log { \frac { 0.15 }{ 0.10 } } \quad \)
= (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.

The equilibrium constant for the reaction, 
\({ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)\) can be written as,
\( { K }_{ c }=\frac { \left[ { NH }_{ 3 }(g) \right] ^{ 2 } }{ \left[ { N }_{ 2 }(g) \right] \left[ { H }_{ 2 }(g) \right] ^{ 3 } } \)
\({ K }_{ c }=\frac { \left( 1.2\times { 10 }^{ -2 } \right) ^{ 2 } }{ \left( 1.5\times { 10 }^{ -2 } \right) \left( 3.0\times { 10 }^{ -2 } \right) ^{ 3 } } \)
= 0.106 × 10⁴ = 1.06 × 10³

\(2NaI{ O }_{ 3 }+Cu(CIo_{ 3 }{ ) }_{ 2 }\longrightarrow 2NaCl{ O }_{ 3 }+Cu(I{ O }_{ 3 }{ 0 }_{ 2 }\)
\([C{ u }^{ 2+ }{ ] }_{ mix }=\cfrac { 0.002 }{ 2 } =0.001M\)
\( [I{ O }_{ 3 }^{ - }{ ] }_{ mix }=\cfrac { 0.002 }{ 2 } =0.001M\)
\(Cu(I{ O })_{ 3 }\rightleftharpoons C{ u }^{ 2+ }+2I{ o }_{ 3 }^{ - }\)
\( [C{ u }^{ 2+ }{ ] }[I{ O }_{ 3 }^{ - }{ ] }=(1.0\times { 10 }^{ -3 })\times (1.0\times { 10 }^{ -3 }{ ) }^{ 2 }=1\times { 10 }^{ -9 }\)
\(Ionic\quad product<{ K }_{ sp }\)

On mixing 50ml of 0.01 M NaCl and 50 ml of 0.01 M AGNO3, the total volume become 100 ml
Conc.of NaCl in 100 ml = \(\cfrac { 0.01\times 50 }{ 100 } =0.005M\)
Conc.Of AgN\(O_{3}\)in 100 ml = \(\cfrac { 0.01\times 50 }{ 100 } =0.005M\)
\(NaCl(aq)={ Na }^{ + }(aq)+{ Cl }^{ - }(aq)\)
\( [{ Cl }^{ - }]=[NacL]=0.005N\)
\( [Ag^{ + }]=[AgN{ O }_{ 3 }]=0.005M\)
\(Ionic \ product \ of \ [A{ g }^{ + }]=[{ Cl }^{ - }]=0.0005\times 0.0005=2.5\times { 10 }\)
Since ionic product is greater than its solubility product, precipitation will occur.

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

\({ CaSO }_{ 4 }\rightleftharpoons { Ca }^{ 2+ }+{ SO }_{ 4 }^{ 2- };{ K }_{ sp }=9.1\times { 10 }^{ -6 }\)
     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

\(Millimoles \ of \ acid, \ { H }_{ 2 }{ SO }_{ 4 }={ M }_{ 1 }{ V }_{ 1 }=2\times 0.01\times 10=0.2\)
\( Millimoles \ of \ base,\ Ca({ OH }_{ 2 })={ M }_{ 2 }{ V }_{ 2 }\)
\(=2\times 0.01\times 10=0.2\)
\( \therefore { M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 } \ hence \ solution \ is \ neutral.\)
\( \therefore pH=7\)

The answer is given in the following Table :