Chapter 19, Problem 19.125P

Interpretation Introduction

Interpretation:

$pH$ of a solution of $6.5\times {10}^{-9}molofCa{\left(OH\right)}_2$ in $10.0L$ of water $\left[K_{sp}ofCa{\left(OH\right)}_2=6.5\times {10}^{-6}\right]$ has to be calculated. 

Concept Introduction:

Solubility Product:

The solubility product constant, $K_{sp}$, is the product of concentrations of ions of electrolyte, each raised to the power  of their coefficients in the balanced chemical equation in a saturated solution.

Consider the general dissolution reaction below (in aqueous solutions):

$a{A}_{\left(s\right)}\rightleftarrows c{C}_{\left(aq\right)}+d{D}_{\left(aq\right)}$

The solubility product for the above solid can be written as

$K_{sp}={\left[C\right]}^c{\left[D\right]}^d$

Where,

$\begin{array}{l}\left[C\right]\text{}=ConcentrationofC\\ \\ \left[D\right]=ConcentrationofD\end{array}$

Ionic Product:

The ionic product, $Q_{sp}$ is the product of concentrations of ions of electrolyte, each raised to the power of their coefficients in the balanced chemical equation at any concentration. 

Solubility:

The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.  It is expressed by the symbol S. Unit of solubility is g/L. 

$\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

$\text{pOH}\text{=}\text{-log}\left[O{H}^{-}\right]$

$\text{pH}\text{+}\text{pOH}\text{=}\text{14}$