Question

How many milliliters of 0.2 M Mg(OH)₂ (aq) are required to titrate 25 ml 0.3 M HC₃H₅O₂(aq)? (Hint: What is the net reaction?)

Answer 1

The balanced equation is

Mg(OH)₂ + 2 HC₃H₅O₂ ------> Mg(C₃H₅O₂)₂ + 2 H₂O

Number of moles of HC₃H₅O₂ = molarity* volume of solution in L

Number of moles of HC₃H₅O₂ = 0.3 * 0.025 = 0.0075 mole

From the balanced equation we can say that

2 mole of HC₃H₅O₂ requires 1 mole of Mg(OH)2 so

0.0075 mole of HC₃H₅O₂ will require

= 0.0075 mole of HC₃H₅O₂ *(1 mole of Mg(OH)2 / 2 mole of HC₃H₅O₂ )

= 0.00375 mole of Mg(OH)2

Molarity of Mg(OH)2 = number of moles of Mg(OH)2 / volume of solution in L

0.2 = 0.00375 / volume of solution in L

volume of solution in L = 0.00375 / 0.2 = 0.0188 L

1 L = 1000 mL

0.0188 L = 18.8 mL

Therefore, the volume of solution required would be 18.8 mL