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

Answer:

Concentration of Ca(OH)₂:

0.117 M.

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.

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

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

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 $7.00\times 10^{-3}\;\text{mol}$ of HCl will neutralize only half that much Ca(OH)₂.

$\displaystyle n(\text{Ca}(\text{OH})_2)=\frac{1}{2}\;n(\text{HCl}) = 3.50\times 10^{-3}\;\text{mol}$.

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

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

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