$${C{H_3}COOK}$$   is a salt of weak acid $$\left( {C{H_3}COOK} \right)$$   and strong base $$\left( {KOH} \right)$$
$$FeC{l_3}$$  is a salt of weak base $$\left[ {Fe{{\left( {OH} \right)}_3}} \right]$$   and strong acid $$\left( {HCl} \right)$$  .
$$Pb{\left( {C{H_3}COO} \right)_2}$$    is a salt of weak base $$Pb{\left( {OH} \right)_2}$$  and weak acid $$\left( {C{H_3}OOH} \right)$$
$$Al{\left( {CN} \right)_3}$$  is a salt of weak base $$\left[ {Al{{\left( {OH} \right)}_3}} \right]$$  and weak acid $$\left( {HCN} \right).$$

TIPS/Formulae :
The pH of the solution at the equivalence point will be greater than 7 due to salt hydrolysis. So an indicator giving colour in basic medium will be suitable.
Phenolphthalein is a good indicator if the base is strong because strong base immediately changes the $$pH$$ at end point.

The strength of oxyacids can also be decided with the help of the oxidation number of central atom. Higher the oxidation number of central atom, more acidic is the oxyacid.
$$\eqalign{ & \mathop {{H_2}S{O_4}}\limits^{ + 6} ,\mathop {HCl{O_3}}\limits^{ + 5} ,\mathop {HCl{O_4}}\limits^{ + 7} ,\mathop {{H_2}S{O_3}}\limits^{ + 4} \cr & {\text{Order of acidic nature}} \cr & HCl{O_4} > {H_2}S{O_4} > HCl{O_3} > {H_2}S{O_3} \cr} $$
Since, in $$HCl{O_4},$$  oxidation number of $$Cl$$  is highest, so, $$HCl{O_4}$$  is the strongest acid among the given acids.

$$\eqalign{ & Mg{\left( {OH} \right)_2} \rightleftharpoons M{g^{ + + }} + 2O{H^ - } \cr & {K_{sp}} = \left[ {M{g^{ + + }}} \right]{\left[ {O{H^ - }} \right]^2} \cr & 1.0 \times {10^{ - 11}} = {10^{ - 3}} \times {\left[ {O{H^ - }} \right]^2} \cr & \left[ {O{H^ - }} \right] = \sqrt {\frac{{{{10}^{ - 11}}}}{{{{10}^{ - 3}}}}} = {10^{ - 4}} \cr & \therefore \,pOH = 4\,\,\,\,\,\,\therefore \,\,pH + pOH = 14\,\,\,\,\therefore pH = 10 \cr} $$

Any species which can accept a proton is Bronsted base while which can give a proton is Bronsted acid.

For precipitation, ionic product > solubility product. In all other options, ionic product is less than solubility product.

$$PbS{O_4}\left( s \right) \rightleftharpoons $$   $$P{b^{2 + }}\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\left( {{K_{sp}}} \right)...\left( {\text{i}} \right)$$
$$HSO_4^ - \left( {aq} \right) \rightleftharpoons $$   $${H^ + }\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\,\left( {{K_a}} \right)...\left( {{\text{ii}}} \right)$$
$${\text{Subtracting equation (ii) from (i), then}}$$
$$PbS{O_4}\left( s \right) + {H^ + }\left( {aq} \right) \rightleftharpoons $$      $$P{b^{2 + }}\left( {aq} \right) + HSO_4^ - \left( {aq} \right)\left( {{K_{eq}}} \right)$$
$$\eqalign{ & \therefore \,\,{K_{eq}} = \frac{{{K_{sp}}}}{{{K_a}}} \cr & = \frac{{1.8 \times {{10}^{ - 8}}}}{{1.0 \times {{10}^{ - 2}}}} \cr & = 1.8 \times {10^{ - 6}} \cr} $$

Conjugate acid of $$SO_4^{2 - }$$  is $$HSO_4^ - .$$

NOTE : In case of alkaline earth hydroxides solubility increases on moving down the group.
$$Be{\left( {OH} \right)_2}$$  has lowest solubility and hence lowest solubility product. [ $$Be$$ at tip of the group ]

In $$0.1\,M\,NaCl,$$   the solubility of $$AgCl$$  is minimum due to the phenomenon of common ion effect.