Chapter 15, Problem 15.80SP

Interpretation Introduction

Interpretation:

Amount in grams of $\text{CaO}$ needed to dissolve in sufficient water to make 1.00 L of a solution with a pH of 10.50 should be determined.

Concept introduction:

Molarity is defined as mass of solute per liter of solution
$\text{Molarity =}\frac{\text{Number of moles of solute }}{\text{Volume of solution in liter }}$

An acid is defined as the species that increases the concentration of ${\text{H}}^{\text{+}}\left(\text{aq}\right)$ , while a base increases the concentration of ${\text{H}}^{-}\left(\text{aq}\right)$.
$\begin{array}{l}\text{pH of solution is defined as }\\ \text{pH = }-\log \left[{\text{H}}^{\text{+}}\right]\\ \left[H_3{O}^{+}\right]\text{ or }\left[{\text{H}}^{\text{+}}\right]={10}^{-\text{pH}\text{.}}\\ \text{pOH=-log}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}\end{array}$

For any aqueous solution at ${25}^{\circ }C$

$\begin{array}{l}\text{pH +pOH = 14}\\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]={\text{K}}_{\text{w}}{\text{ , where K}}_w={10}^{-14}\end{array}$

$\begin{array}{l}{\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2\left(\text{conservation of amount of solute }\right)\\ {\text{M}}_1:{\text{ initial molarity , V}}_1:\text{ initial value valume of solution}\\ {\text{M}}_2:{\text{ final molarity , V}}_2:\text{ Final volume of solution }\end{array}$

$\text{M = }\frac{{\text{M}}_1{\text{V}}_1+{\text{M}}_2{\text{V}}_2}{{\text{V}}_1+{\text{V}}_2}\text{ M: molarity of solution}$

$\begin{array}{l}{\text{M}}_2:\text{ molarity of second solution }\\ {\text{V}}_2:\text{ volume of second solution}\end{array}$

$\begin{array}{l}{\text{M}}_1:\text{ Molarity of first salution }\\ {\text{V}}_1:\text{ volume of first solution }\end{array}$

Reaction of lime:
$\text{CaO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

The dissociation of calcium hydroxide

$\begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_2{\text{+H}}_2\text{O}\to {\text{ Ca}}^{++}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \text{pH + pOH = 14}\\ \left[{\text{OH}}^{-}\right]={\text{ 10}}^{-p\text{OH}}\end{array}$

Mass of solution in a solution $=\text{ Molarity }\times \text{volume }\left(L\right)$