Question

If 30.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 40.0 mL of 0.175 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

Concentration of Ca(OH)₂:

0.117 M.

Step-by-step explanation:

How many moles of HCl is consumed?

Note the unit of concentration: moles per liter solution.

`c(\text{HCl}) = 0.175\;\text{M} = 0.175\;\text{mol}\cdot\textbf{L}^(-1)`.

Convert milliliters to liters.

`V(\text{HCl})=40.0\;\text{mL} = 0.0400\;\text{L}`.

`n(\text{HCl}) = c(\text{HCl})\cdot V(\text{HCl})= 0.175\;\text{mol}\cdot\text{L}^(-1) * 0.0400\;\text{L}= 7.00* 10^(-3)\;\text{mol}`.

How many moles of NaOH in the solution?

Refer to the equation. The coefficient in front of Ca(OH)₂ is 1. The coefficient in front of HCl is 2. In other words, it takes two moles of HCl to neutralize one mole of Ca(OH)₂. That
`7.00* 10^(-3)\;\text{mol}` of HCl will neutralize only half that much Ca(OH)₂.

`\displaystyle n(\text{Ca}(\text{OH})_2)=(1)/(2)\;n(\text{HCl}) = 3.50* 10^(-3)\;\text{mol}`.

What's the concentration of the Ca(OH)₂ solution?

Concentration is the number of moles of solute per unit volume.

`\displaystyle c(\text{Ca}(\text{OH})_2) = \frac{n(\text{Ca}(\text{OH})_2)}{V(\text{Ca}(\text{OH})_2)} = \frac{3.50* 10^(-3)\;\text{mol}}{0.0300\;\text{L}}=0.117\;\text{mol}\cdot\text{L}^(-1)`.