\[\begin{matrix} NaOH & + & C{{H}_{3}}COOH & \rightleftharpoons \\ 20\times 0.2 & {} & 50\times 0.2 & {} \\ 4 & {} & 10 & {} \\ 0 & {} & 6 & {} \\ \end{matrix}\,\,\,\,\,\begin{matrix} C{{H}_{3}}COONa & + & {{H}_{2}}O & {} \\ 0 & {} & 0 & {} \\ 0 & {} & 0 & {} \\ 4 & {} & 4 & {} \\ \end{matrix}\]
$$Conc. = \frac{{{\text{Millimoles}}}}{{{\text{Total Volume}}}}$$
$$\left[ {C{H_3}COOH} \right] = \frac{6}{{70}};$$     $$\left[ {C{H_3}COONa} \right] = \frac{4}{{70}}$$
$$pH = - \log \,1.8 \times {10^{ - 5}} + \log \frac{{\frac{4}{{70}}}}{{\frac{6}{{70}}}},$$       $$pH = 4.56$$

The characteristics of the given solutions are :
$$NaCl\,\, - $$   neutral solution
$$N{H_4}Cl\,\, - $$   slightly acidic due to the following reaction
$$NH_4^ + + {H_2}O \rightleftharpoons N{H_4}OH + {H^ + }$$
$$NaCN\,\, - $$   slightly alkaline due to the following reaction
$$C{N^ - } + {H_2}O \rightleftharpoons HCN + O{H^ - }$$
$$HCl\,\, - $$   highly acidic
The $$pH$$  of the solution will follow the order highly acidic < slightly acidic < neutral < slightly alkaline i.e. $$HCl < N{H_4}Cl < NaCl < NaCN$$

$$\left[ {{H^ + }} \right] = \frac{{\left( {25 \times 0.3 - 25 \times 0.1} \right)}}{{100}}meq = 0.05M$$

$$\eqalign{ & {\text{In}}\,\,{\text{equation,}}\,\,\,\,\,\,\,X \rightleftharpoons \,\,\,\,\,Y\,\, + Z \cr & {\text{Initial moles}}\,\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{At equil}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\,\left( {1 - \alpha } \right)\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\alpha \cr & {\text{where,}}\,\alpha = {\text{degree of dissociation}} \cr & {\text{Total number of moles}} \cr & \,\,\,\,\,\,\,\, = 1 - \alpha + \alpha + \alpha \cr & \,\,\,\,\,\,\,\, = \left( {1 + \alpha } \right) \cr & {p_X} = \left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)\,{p_1} \cr & {p_Y} = \left( {\frac{\alpha }{{1 + \alpha }}} \right)\,{p_1} \cr & {p_Z} = \frac{\alpha }{{\left( {1 - \alpha } \right)}}\,{p_1} \cr & {K_{{p_1}}} = \frac{{\left[ {{p_Y}} \right]\left[ {{p_Z}} \right]}}{{\left[ {{p_X}} \right]}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{\left( {\frac{\alpha }{{1 + \alpha }}} \right){p_1} \times \left( {\frac{\alpha }{{1 + \alpha }}} \right)\,{p_1}}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)\,{p_1}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{{{\left( {\frac{\alpha }{{1 + \alpha }}} \right)}^2}{p_1}}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)}}\,\,\,...{\text{(i)}} \cr & {\text{For}}\,{\text{equation,}}\,\,\,{\text{A}}\,\,\,\,\,\,\,\,\,\, \rightleftharpoons \,\,\,\,{\text{2B}} \cr & {\text{Initial moles}}\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}} \cr & {\text{At equil}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\alpha \cr} $$
Total number of moles at equilibrium $$ = \left( {1 + \alpha } \right)$$
$$\eqalign{ & {p_B} = \left( {\frac{{2\alpha }}{{1 + \alpha }}} \right)\,{p_2} \cr & {P_A} = \left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)\,{p_2} \cr & {K_{{p_2}}} = \frac{{{{\left[ {{p_B}} \right]}^2}}}{{\left[ {{p_A}} \right]}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{{{\left[ {\left( {\frac{{2\alpha }}{{1 + \alpha }}} \right)\,{p_2}} \right]}^2}}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)\,{p_2}}} \cr & {K_{{p_2}}} = \frac{{{{\left( {\frac{{2\alpha }}{{1 + \alpha }}} \right)}^2}{p_2}}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right)}}\,\,\,...{\text{(ii)}} \cr & {\text{Eq}}{\text{. (i) divide by Eq, (ii)}} \cr & \frac{{{K_{{p_1}}}}}{{{K_{{p_2}}}}} = \frac{{{\alpha ^2} \times {p_1}}}{{4{\alpha ^2} \times {p_2}}} \cr & \,\,\,\,\,\frac{9}{1} = \frac{{{p_1}}}{{4{p_2}}} \cr & \,\,\,\frac{{{p_1}}}{{{p_2}}} = \frac{{36}}{1} = 36:1 \cr} $$

$$\eqalign{ & HA \rightleftharpoons {H^ + } + {A^ - } \cr & {\text{At equilibrium}}\,\left[ {{H^ + } = {A^ - }} \right] \cr & {K_a} = \frac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}} \cr & \,\,\,\,\,\,\,\, = \frac{{{{\left[ {{H^ + }} \right]}^2}}}{{\left[ {HA} \right]}} \cr & \left[ {{H^ + }} \right] = \sqrt {{K_a}\left[ {HA} \right]} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1 \times {{10}^{ - 5}} \times 0.1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1 \times {{10}^{ - 6}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 3}} \cr & \alpha = \frac{{{\text{Actual ionisation}}}}{{{\text{Molar concentration}}}} \cr & \,\,\,\,\, = \frac{{{{10}^{ - 3}}}}{{0.1}} \cr & \,\,\,\,\, = {10^{ - 2}} \cr & \% \,\,{\text{of acid dissociated}} \cr & = {10^{ - 2}} \times 1.00 \cr & = 1\% = 100\% \cr} $$

Blood is an example of buffer solution, which contains serum protein, so its $$pH$$  does not change appreciably by adding small amount of an acid or a base to it.

$$\eqalign{ & {A_x}{B_y} \rightleftharpoons x{A^{y + }} + y{B^{x - }} \cr & {K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y} \cr} $$

$$B{F_3}$$  is an electron deficient compound and hence is a Lewis acid.

$$\eqalign{ & pH = \frac{1}{2}\left[ {p{K_w} + p{K_a} - p{K_b}} \right] \cr & pH = 7 + \frac{1}{2}\left( {5.76 - 5.25} \right) = 7.255 \cr} $$

NOTE: A buffer is a solution of weak acid and its salt with strong base and vice versa.
$$HCl$$  is strong acid and $$NaCl$$  is its salt with strong base.
$$pH$$ is less than 7 due to $$HCl.$$