Question

If 27.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 32.40 mL of 0.185 M HCl, what is the concentration of the Ca(OH)2 solution? Show all of the work needed to solve this problem. (2 points) Ca(OH)2 + 2HCl yields 2H2 O + CaCl2

Answer 1

Given information : Volume of HCl = 32.40 mL

`32.40 mL* (1 L)/(1000 mL)`

Volume of HCl = 0.0324 L

Concentration of HCl = 0.185 M or 0.185 mol/L (M = mol/L)

Volume of Ca(OH)2 = 27.0 mL

`27.0 mL* (1 L)/(1000 mL)`

Volume of Ca(OH)2 = 0.027 L

We need to find the concentration of Ca(OH)2.

To find the concentration of Ca(OH)2 we need moles and volume of Ca(OH)2.

`Concentration (Molarity) = ((Moles of Ca(OH)2))/((Volume of Ca(OH)2))`

Moles of Ca(OH)2 can be calculated using stoichiometry and volume of Ca(OH)2 is already given to us.

Step 1 : Find the moles of HCl using its given volume and concentration.

`Moles = Concentration * Volume in L`

`Moles = 0.185(mol)/(1L)* 0.0324 L`

Moles of HCl = 0.005994 mol HCl

Step 2 : We need to find moles of Ca(OH)2 using mol of HCl with the help of mole ratio.

Mole ratio are the coefficient present in front of the compound in a balanced equation.

Mole ratio of Ca(OH)2 : HCl = 1:2 ( 1 coefficient of Ca(OH)2 and 2 coefficient of HCl)

`(0.005994 mol HCl)* ((1 mol Ca(OH)2))/((2 mol HCl))`

Moles of Ca(OH)2 = 0.002997 mol Ca(OH)2

Step 3 : Find the concentration of Ca(OH)2 using its moles and volume.

`Concentration (Molarity) = ((Moles of Ca(OH)2))/((Volume of Ca(OH)2))`

Moles of Ca(OH)2 = 0.002997 mol and volume of Ca(OH)2 = 0.027 L

`Concentration (Molarity) = ((0.002997 mol))/((0.027 L))`

Concentration of Ca(OH)2 = 0.111 mol/L or 0.111 M