for 4biii)
the reaction
AlCl3 + 3NaOH --> Al(OH)3 + 3NaCl
Al(OH)3 is insoluble (precipitated), however, when all AlCl3 is reacted, the excess NaOH that is added reacts with Al(OH)3
Al(OH)4- ions are formed in the presence of high concentrations of hydroxide: this is soluble. Thus all the precipitate dissolves and the height of the precipitate becomes 0. for the graph, just draw a rough graph where it continues till the maximum point but falles to 0 when excess is added.