Question

How many mL of 1.73 M HNO3 will be needed to neutralize 185 mL of 2.89 M Mg(OH)2. mL

Answer 1

Balanced equation between Mg(OH)2 and HNO3

Mg(OH)2 + 2 HNO3 ------> Mg(NO3)2 + 2 H2O

number of moles of Mg(OH)2 = molarity * volume of solution in L

number of moles of Mg(OH)2 = 2.89*0.185 = 0.535 mole

from the balanced equation we can say that

1 mole of Mg(OH)2 requires 2 mole of HNO3 so

0.535 mole of Mg(OH)2 will require 1.07 mole of HNO3

molarity of HNO3 = number of moles of HNO3 / volume of solution in L

1.73 = 1.07 / volume of solution in L

volume of solution in L = 1.07 / 1.73 = 0.6185 L

1 L = 1000 mL

0.6185 L = 618.5 mL

Therefore, the volume of HNO3 needed will be 618.5 ml