Question

How many milliliters of 0.0708 M Ca( OH)₂would be required to exactly neutralize 151 mL of 0.231 M HCl?

Answer 1

The balanced equation is

2 HCl + Ca(OH)2 ------> CaCl2 + 2 H2O

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

Number of moles of HCl = 0.231 * 0.151 = 0.0349 mol

From the balanced equation we can say that

2 mole of HCl requires 1 mole of Ca(OH)2 so

0.0349 mole of HCl will require

= 0.0349 mole of HCl *(1 mole of Ca(OH)2 / 2 mole of HCl)

= 0.0175 mole of Ca(OH)2

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

0.0708 = 0.0175 / volume of solution in L

volume of solution in L = 0.0175 / 0.0708 = 0.247 L

1L = 1000 mL

0.247 L = 247 mL

Therefore. the volume of Ca(OH)2 required would be 247 mL