Question

A sample of solid calcium hydroxide, Ca(OH)₂, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10^(-2) M HCl requires 36.6 mL of the acid to reach the end point. Ca(OH)₂(aq)+2HCl(aq)⟶CaCl₂(aq)+2H₂O(l)

a) 0.025 g
b) 0.050 g
c) 0.075 g
d) 0.100 g

Answer 1

The molarity of the calcium hydroxide solution is 0.0244 M.

Step-by-step explanation:

To find the molarity of the calcium hydroxide (Ca(OH)2) solution, we can use the equation:

Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

From the equation, we can see that the ratio of moles of Ca(OH)2 to moles of HCl is 1:2. We know that 75.00 mL of the Ca(OH)2 solution requires 36.6 mL of 5.00 × 10-2 M HCl to reach the end point.

Using the equation:

Molarity(M) = Moles of solute / Volume of solution (in liters)

We can calculate the moles of Ca(OH)2 and HCl:

Moles of Ca(OH)2 = (36.6 mL)(5.00 × 10-2 M)(1 L/1000 mL) = 1.83 x 10-3 mol

Moles of HCl = 2 x 1.83 x 10-3 mol = 3.66 x 10-3 mol

Since the moles of Ca(OH)2 is half of the moles of HCl, the molarity of the Ca(OH)2 solution is half of the molarity of the HCl solution:

Molarity of Ca(OH)2 = 1.83 x 10-3 mol / 0.07500 L = 0.0244 M