Question

What volume of 2.85×10^-3 M HCl is needed to neutralize 11.5 mL of a saturated Ca(OH)2 solution?

A) 10.87 mL
B) 14.04 mL
C) 12.56 mL
D) 16.32 mL

Answer 1

To neutralize the Ca(OH)2 solution, you need approximately 8.06 L of 2.85x10^-3 M HCl.

Step-by-step explanation:

To neutralize the Ca(OH)2 solution, we can use the equation:
Ca(OH)2 + 2HCl → CaCl2 + 2H2O

According to the balanced equation, the ratio between Ca(OH)2 and HCl is 1:2. The Moles of Ca(OH)2 in 11.5 mL (0.0115 L) of the solution can be calculated as:
Moles of Ca(OH)2 = (0.0115 L) * (Molarity of Ca(OH)2)

Molarity of HCl is given as 2.85x10-3 M. To find the volume of HCl solution needed, we can rearrange the formula:
Volume of HCl solution = (Moles of Ca(OH)2) * (2 / Molarity of HCl)

Plugging in the values, we get:
Volume of HCl solution = (0.0115 L) * (2 / 2.85x10-3 M) = 0.0115 L * 701.754386 M = 8.06 L

Therefore, the volume of 2.85x10-3 M HCl needed to neutralize 11.5 mL of the Ca(OH)2 solution is approximately 8.06 L.