Chapter 17, Problem 50QP

From the solubility data given, calculate the solubility products for the following compounds: $\left(\text{a}\right){\text{ SrF}}_{\text{2}}\text{, 7}{\text{.3 x 10}}^{\text{-2}}\text{g/L, }\left(\text{b}\right){\text{ Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\text{, 6}\text{.7 }\times {\text{10}}^{\text{-3}}\text{g/L}\text{.}$

Interpretation Introduction

Interpretation:

The solubility product constant of ${\text{SrF}}_2\left(s\right)$ and ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$ is to be calculated from the given solubility data.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperature is termed as the solubility of the solute in the solvent at that temperature.

The solubility product of a sparingly-soluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation, after the dissociation of the electrolyte.

The number of moles of the solute dissolved per litre of the solution is called molar solubility.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: ${\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant and $sp$ stands for solubility product.

To convert $\left(\text{g/L}\right)\text{to}\left(\text{mol/L}\right)$, divide it by its molar mass $\left(\text{g}/\text{L}\right)\text{to(mol/L)}=\left(\frac{\text{grams of compound}}{1\text{L solution}}\right)\times \left(\frac{\text{1}}{\text{molar}\text{mass}}\right)$

Answer to Problem 50QP

Solution:

$\left(\text{a}\right)$ The solubility product constant of ${\text{SrF}}_2\left(s\right)$ is $7.8\times {10}^{-4}$

$\left(\text{b}\right)$ The solubility product constant of ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$ is $1.8\times {10}^{-18}$

Explanation of Solution

a) ${\text{SrF}}_2$, $7.3\times {10}^{-2}\text{g}/\text{L}$.

Convert from $\left(\text{g/L}\right)\text{to}\left(\text{mol/L}\right)$ by dividing with molar mass of the ${\text{SrF}}_2$.

$\left(\text{g}/\text{L}\right)\text{to(mol/L)}=\left(\frac{\text{grams of compound}}{1\text{L solution}}\right)\times \left(\frac{\text{1}}{\text{molar}\text{mass}}\right)$

$\begin{array}{l}\left(\text{g}/\text{L}\right)=\left(\frac{7.3\times {10}^{-2}\text{g}{\text{SrF}}_2}{1\text{L solution}}\right)\times \left(\frac{1\text{mol}{\text{SrF}}_2}{125.6\text{g}{\text{SrF}}_2}\right)\\ =\text{5}\text{.8}\times {\text{10}}^{-4}\text{mol}/\text{L}\end{array}$

The equation for the dissociation of ${\text{SrF}}_2$ in water is as follows:

${\text{SrF}}_2\left(s\right)\rightleftharpoons {\text{Sr}}^{2+}\left(aq\right)+{\text{2F}}^{2-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of ${\text{SrF}}_2\left(s\right)$ is as follows:

$\begin{array}{l}{\text{SrF}}_2\left(s\right)\rightleftharpoons {\text{Sr}}^{2+}\left(aq\right)+{\text{2F}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 2s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Sr}}^{\text{2+}}{\text{][F}}^{\text{-}}{\text{]}}^2$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Sr}}^{2+}\right]$ is the concentration of the strontium ion, and $\left[{\text{F}}^{-}\right]$ is theconcentration of the fluorine ion.

Substitute the values of $\left[{\text{Sr}}^{2+}\right]$ and $\left[{\text{F}}^{-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=\left[s\right]{\left[2s\right]}^2\\ =4{\left(s\right)}^3\\ =4{\left(5.8\times {10}^{-4}\right)}^3\\ =7.8\times {10}^{-10}\end{array}$

Hence, the solubility product constant of ${\text{SrF}}_2\left(s\right)$ is $7.8\times {10}^{-10}$.

b) ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$, $6.7\times {10}^{-3}\text{g}/\text{L}$.

Convert from $\left(\text{g/L}\right)\text{to}\left(\text{mol/L}\right)$ by dividing with molar mass of the ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$.

$\left(\text{g}/\text{L}\right)\text{to(mol/L)}=\left(\frac{\text{grams of compound}}{1\text{L solution}}\right)\times \left(\frac{\text{1}}{\text{molar}\text{mass}}\right)$

$\begin{array}{l}\left(\text{g}/\text{L}\right)=\left(\frac{6.7\times {10}^{-3}\text{g}{\text{Ag}}_3{\text{PO}}_4}{1\text{L solution}}\right)\left(\frac{1\text{mol}{\text{Ag}}_3{\text{PO}}_4}{418.7\text{g}{\text{Ag}}_3{\text{PO}}_4}\right)\\ =\text{1}\text{.6}\times {\text{10}}^{-5}\text{mol}/\text{L}\end{array}$

The equation for the dissociation of ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$ in water is as follows:

${\text{Ag}}_3{\text{PO}}_4\left(s\right)\rightleftharpoons 3{\text{Ag}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$ is as follows:

$\begin{array}{l}{\text{Ag}}_3{\text{PO}}_4\left(s\right)\rightleftharpoons 3{\text{Ag}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +3s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& 3s& s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^3{\text{[PO}}_4^{3-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Ag}}^{+}\right]$ is the concentration of the silver ion, and $\left[{\text{PO}}_4^{3-}\right]$ is the concentration of the phosphate ion.

Substitute the values of $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{PO}}_4^{3-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}={\left[3s\right]}^3\left[s\right]\\ =27{\left(s\right)}^4\\ =27{\left(1.6\times {10}^{-5}\right)}^4\\ =1.8\times {10}^{-18}\end{array}$

Hence, the solubility product constant of ${\text{Ag}}_3{\text{PO}}_4\left(s\right)$ is $1.8\times {10}^{-18}$.