$($PLEASE DONT FORGET THE GRAPH$)$

Using the following set of reactions, generate a $logAl$ vs$.pH$ diagram depicting the solubility of gibbsite, $Al(OH{)}_3(am).$

$Al(OH{)}_3(am)+3H^{+},$larr$A{l}^{3+},+3H_{2}O,log{K}_{sp,Al(OH)}=10\mathrm{.}38$

$A{l}^{3+}+H_{2}O,$larr$AlO{H}^{2+},+H^{+},log{K}_{1,OH}=-4\mathrm{.}99$

$A{l}^{3+}+2H_{2}O,$larr$Al(OH{)}_2^{+},+2H^{+},log{K}_{2,OH}=-10\mathrm{.}1$

$A{l}^{3+}+3H_{2}O,$larr$Al(OH{)}_3^0(aq),+3H^{+},log{K}_{3,OH}=-16\mathrm{.}0$

$A{l}^{3+}+4H_{2}O,$larr$Al(OH{)}_4^{-},+4H^{+},log{K}_{4,OH}=-23\mathrm{.}0$

Assume this mineral controls the solubility of $Al$ in a coagulation$-$flocculation drinking water plant. If the system is dosed with alum, aluminum sulfate$($s$),$ at a concentration equivalent to $2\mathrm{.}7$mgA$\frac{l}{L}$ and operated at $pH7\mathrm{.}0,$ what is the minimum solubility of $Al$ in equilibrium with gibbsite? Report your answer as $($ mgA$\frac{1}{L}$