If 150 ml of Al(OH)3 of unknown concentration with a methyl red indicator presen

Question

If 150 ml of Al(OH)3 of unknown concentration with a methyl red indicator present is titrated with [0.535 M] HCl, the yellow solution turns red after 32.8 ml of the HCl is added. What is the molarity of the Al(OH)3 solution?

Explanation / Answer

Al(OH)3 + 3 HCl ======> AlCl3 + 3 H2O M1 = molarity of Al(OH)3 = ? : V1 = volume of Al(OH)3 = 150mL : n1 = co-efficient of Al(OH)3 in balanced reaction = 1 M2 = Molarity of HCl = 0.535 M : V2 = volume of HCl = 32.8 mL : n2 = co-efficient of HCl in balanced reaction =3 according to titration formula (M1V1/n1) =(M2V2/n2) M1*150/1 = (0.535*32.8)/3=5.85 M1 = 5.85/150 = 0.039 = 0.04 M Al(OH)3

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