Fe=0.01;//concentration of Fe3+ ions in the solution//
Ksp=3.8*10^-38;//Solubility product of Fe(OH)3//
OH=(Ksp/Fe)^0.333;//Concentration of OH- ions in the solution//
printf('Concentration of OH- ions in the solution=OH=%f=1.561*10^-12',OH);
printf('\nAt this PH,Fe(OH)3 starts precipitating and precipitation is complete (Fe3+)=10^-6M\n(10^-6)(OH-)^3 = 3.8*10^-38 ');
printf('\nupon solving this we get (OH-)=3.362*10^-11\nPOH=10.48 or PH=3.52');
printf('\nAt this PH the precipitation of Fe(OH)3 is almost complete.');