This is the three cases that help to determine the minimum concentration of KOH required for precipitation
Part a) 1.5×10^−2 M K CaCl2
Part b) 2.3×10^−3 M Fe (NO3)2
Part c) 2.0×10^−3 M MgBr2
a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^-
ksp = 1.5*10^-2 + x^2
4.68*10^-6 = 1.5*10^-2 + x^2
x= [KOH] = 0.01766
b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3
Fe (OH)2 <=> Fe^2+ + 2OH^-
ksp = 2.3*10^-3 + x^2
4.87*10^-17 = 2.3*10^-3 + x^2
x= 1.46*10^-7
c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr
Mg (OH) 2 <=> Mg^2+ + 2OH^-
ksp = 2.0*10^-3 + x^2
2.06*10^-13 = 2.0*10^-3 + x^2
x= 1.015*10^-5
