Question

A 34.62 mL of 0.1510 M Ca(OH)₂ was needed to neutralize 55.7 mL of an HCl solution. What is the concentration of the original HC(aq) solution? please show work

Answer 1

first write the balanced equation

Ca(OH)₂ + 2 HCl -----> CaCl₂ + 2H₂O

from this reaction we can say that one mole of Ca(OH)2 required 2 moles of HCl keep in mind

calculate the no ofmoles of Ca(OH)2 = molarity of Ca(OH)2 x volume of Ca(OH)2 in liters

= 0.151 mol/L x 0.03462 L

= 0.00523 moles

from the balanced equation

= 0.00523 moles required 2 x 0.00523 = 0.0104 moles

now no of moles of HCl =0.0104

volume of HCl = 55.7 mL = 0.0557 L

Molarity = no of moles / volume liters

Molarity = 0.0104 / 0.0557 L

= 0.188 M