Question

Calculate the volume of 0.15M hydrochloric acid required to neutralize 125mL of 0.175 M Mg(OH)2.

Answer 1

Let us first calculate the number of moles of Mg(OH)2 present in 125 mL 0.175 M solution. As the solution is 0.175 M, it means that 0.175 moles of Mg(OH)2 is present in 1000 mL. Therefore we can consider

1000 mL = 0.175 moles

For 125 mL = (125 x 0.175) / 1000

= 0.02187 moles

The balanced chemical equation for the reaction of HCl and Mg(OH)2 is

2 HCl + Mg(OH)2 = MgCl2 + 2 H2O

It is clear from the balanced chemical equation that 1 mole of Mg(OH)2 consumes 2 moles of HCl. Therefore

2 moles of Mg(OH)2 = 1 moles of HCl

For,. 0.02187 moles of Mg(OH)2 = 0.02187 / 2 moles of HCl

= 0.01093 moles of HCl

As the solution if HCl is 0.15 M, it means that 0.15 moles of HCl is present in 1000 mL of solution. Therefore,

0.15 moles of HCl = 1000 mL

For, 0.01093 moles of HCl = (0.01093 x 1000) / 0.15

= 72.91 mL of HCl solution

Therefore answer is 72.91 mL of 0.15 M HCl solution will be required.