Question

A solution is prepared by dissolving 0.148 grams of Al(OH)3 in 2.00 L of water.

Determine the Molarity of this solution. Molarity ____________________________

Determine the Molarity of the [OH-] in solution. Concentration of [OH-] _________________

Determine the solutions pH. pH_________________________________

Determine [H3O+ ] for this solution. [H3O+ ] _______________________

Answer 1

Molarity is the number of moles per litre of the solution. To determine the molarity of the solution,divide the number of moles y volume of solution in litres. Al(OH)3 is an amphoteric compound,both act as base and acid. Molarity(M)   =Moles/volume in litre Moles =mass/molecular mass The molecular mass of   Al(OH)3 =78 g/mol So,moles =(0.148 g)/ (78 g/mol)                  =0.001 mol Molarity =(0.001 mol)/ (2 litre)                =0.0005                 = 5 * 10-4M Molarity becomes the [H30+] concentration of 5 * 10-4M SO [H30+] = 5 * 10-4M pH =-log[H3O+]       = -log [5 * 10-4M ]        =3.30 So,to find out pOH=14 -3.30 =10.7 [OH-] =10-10.7           =1.99 * 10-11 SO,THE [OH-] =1.99 * 10-11M