Chapter 17, Problem 60PS

What is the solubility, in milligrams per milliliter, of BaF₂, (a) in pure water and (b) in water containing 5.0 mg/mL KF?

Interpretation Introduction

Interpretation:

The solubility of Barium fluoride ${\text{BaF}}_2$ salt in milligrams per milliliter unit has to be calculated in pure water.

Concept introduction:

The solubility of a salt is defined as the maximum amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant $K_{\text{sp}}$ is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

For example, general salt ${\text{A}}_x{\text{B}}_y$ when dissolved in water dissociates as,

  ${\text{A}}_x{\text{B}}_y\left(\text{s}\right)\rightleftharpoons x{\text{A}}^{y+}\left(\text{aq}\right)+y{\text{B}}^{-x}\left(\text{aq}\right)$

The expression for $K_{\text{sp}}$ of a salt is,

$K_{\text{sp}}={\left[{\text{A}}^{y+}\right]}^x{\left[{\text{B}}^{-x}\right]}^y$ (1)

The ICE table (1) for salt ${\text{A}}_x{\text{B}}_y$, which relates the equilibrium concentration of ions in the solution is given as follows,

$\begin{array}{cccccc}\text{Equation}& {\text{A}}_x{\text{B}}_y& \rightleftharpoons & x{\text{A}}^{y+}& +& y{\text{B}}^{-x}\\ \text{Initial}\left(\text{M}\right)& & & 0& & 0\\ \text{Change}\left(\text{M}\right)& & & +xs& & +ys\\ \text{Equilibrium}\left(\text{M}\right)& & & xs& & ys\end{array}$

From the table,

$\begin{array}{l}\left[{\text{A}}^{y+}\right]=xs\\ \left[{\text{B}}^{-x}\right]=ys\end{array}$

Substitute $xs$ for $\left[{\text{A}}^{y+}\right]$ and $ys$ for $\left[{\text{B}}^{-x}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{sp}}={\left(xs\right)}^x{\left(ys\right)}^y\\ =x^x{y}^y{\left(s\right)}^{x+y}\end{array}$

Rearrange for $s$.

$s={\left(\frac{K_{\text{sp}}}{x^x{y}^y}\right)}^{1/\left(x+y\right)}$

Here,

• $x$ is the coefficient of the cation ${\text{A}}^{+y}$.
• $y$ is the coefficient of the anion ${\text{B}}^{-x}$.
• $s$ is the molar solubility.

The value of $K_{\text{sp}}$ is calculated by using molar solubility of the salt.

Explanation of Solution

The solubility of the salt ${\text{BaF}}_2$ in pure water is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of $K_{\text{sp}}$.

The value of solubility product constant,$K_{\text{sp}}$ of ${\text{BaF}}_2$ is $1.8\times {10}^{-7}$.

The balanced chemical reaction for the dissolution of ${\text{BaF}}_2$ in water is,

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

The ICE table(2) is as follows,

$\begin{array}{cccccc}\text{Equation}& {\text{BaF}}_2\left(\text{s}\right)& \rightleftharpoons & {\text{Ba}}^{2+}\left(\text{aq}\right)& +& {\text{2F}}^{-}\left(\text{aq}\right)\\ \text{Initial }\left(\text{M}\right)& & & 0& & 0\\ \text{Change }\left(\text{M}\right)& & & +s& & +2s\\ \text{Equilibrium }\left(\text{M}\right)& & & \text{ s}& & 2s\end{array}$

The $K_{\text{sp}}$ expression for ${\text{BaF}}_2$ is,

$K_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]{\left[{\text{F}}^{-1}\right]}^2$ (2)

From the table,

$\begin{array}{c}\left[{\text{Ba}}^{2+}\right]=s\\ \left[{\text{F}}^{-1}\right]=2s\end{array}$

Substitute $s$ for $\left[{\text{Ba}}^{2+}\right]$ and $2s$ for $\left[{\text{F}}^{-1}\right]$ in equation (2).

$K_{\text{sp}}=\left(s\right){\left(2s\right)}^2$

Here,

• $K_{\text{sp}}$ is solubility product constant
• $s$ is the molar solubility of the salt $\text{AgI}$.

$K_{\text{sp}}=4s^3$

Rearrange for $s$.

$s=\sqrt[3]{\frac{K_{\text{sp}}}{4}}$

Substitute $1.8\times {10}^{-7}$ for $K_{\text{sp}}$.

$\begin{array}{c}s=\sqrt[3]{\frac{1.8\times {10}^{-7}}{4}}\\ =3.56\times {10}^{-3}\text{}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Calculate the solubility of ${\text{BaF}}_2$ in pure water in milligrams per milliliter unit.

Multiply the molar solubility of ${\text{BaF}}_2$ by the molar mass of ${\text{BaF}}_2$ to calculate the solubility in grams per liter.

${\text{ s}}^{\text{'}}=\left(s\frac{\text{mol}}{\text{L}}\right)\left(\text{M}\frac{\text{g}}{\text{mol}}\right)$ (3)

Here,

•  $s^{\text{'}}$ is the solubility of ${\text{BaF}}_2$ in grams per liter.
• $s$ is the solubility of ${\text{BaF}}_2$ in moles per liter.
• $\text{M}$ is the molar mass of ${\text{BaF}}_2$.

Substitute $3.56\times {10}^{-3}\text{}\text{mol}\cdot {\text{L}}^{-1}$ for $s$ and $175.3\text{g}\cdot {\text{mol}}^{-\text{1}}$ for $\text{M}$ in equation (3).

The solubility of ${\text{BaF}}_2$ in grams per liter,

$\begin{array}{c}{s}^{\text{'}}=\left(3.56\times {10}^{-3}\text{}\text{}\text{}\text{}\frac{\text{mol}}{\text{L}}\right)\left(175.3\text{}\frac{\text{g}}{\text{mol}}\right)\\ =0.624\text{g}\cdot {\text{L}}^{-1}\end{array}$

Solubility in milligrams per millilitre is,

$\begin{array}{c}{s}^{\text{'}}=0.624\text{g}\cdot {\text{L}}^{-1}\\ =\left(0.624\frac{\text{g}}{\text{L}}\right)\left(\frac{1000\text{}\text{}\text{ mg}}{1\text{g}}\right)\left(\frac{1\text{}\text{}\text{ L}}{1000\text{mL}}\right)\\ =0.624mg\cdot m{L}^{-1}\end{array}$

The solubility of ${\text{BaF}}_2$ in pure water is $0.624mg\cdot m{L}^{-1}$.

Interpretation Introduction

Interpretation:

The solubility of salt ${\text{BaF}}_2$ has to be calculated in water containing $5.0\text{}\text{}\text{ mg}\cdot {\text{mL}}^{-1}$ $\text{KF}$ and also compared it with the solubility value in pure water.

Concept introduction:

The solubility of a salt decreases if in the solution one of the ions common to the dissolved ion of salt is already present before the dissolution of salt due to common ion effect. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium.

For example, general salt ${\text{A}}_x{\text{B}}_y$ when dissolved in water dissociates as,

  ${\text{A}}_x{\text{B}}_y\left(\text{s}\right)\rightleftharpoons x{\text{A}}^{y+}\left(\text{aq}\right)+y{\text{B}}^{-x}\left(\text{aq}\right)$

The expression for $K_{\text{sp}}$ of a salt is,

$K_{\text{sp}}={\left[{\text{A}}^{y+}\right]}^x{\left[{\text{B}}^{-x}\right]}^y$ (4)

For example, if anion ${\text{B}}^{-x}$ is already present in the solution before equilibrium, therefore a modified ICE table (3) is used to give the concentration relationships between ions.

$\begin{array}{cccccc}\text{Equation}& {\text{A}}_x{\text{B}}_y\left(\text{s}\right)& =& x{\text{A}}^{y+}\left(\text{aq}\right)& +& y{\text{B}}^{-x}\left(\text{aq}\right)\\ \text{Initial}\left(\text{M}\right)& & & 0& & s^{\text{'}}\\ \text{Change}\left(\text{M}\right)& & & +xs& & +ys\\ \text{Equilibrium}\left(\text{M}\right)& & & xs& & s^{\text{'}}+ys\end{array}$

Here,

• $s^{\text{'}}$ is the initial concentration of the anion ${\text{B}}^{-x}$ (common ion coming from strong electrolyte) present in the solution before the dissociation of a weak salt.

 From the ICE table (3),

 $\begin{array}{l}\left[{\text{A}}^{y+}\right]=xs\\ \left[{\text{B}}^{-x}\right]=s^{\text{'}}+ys\end{array}$

 Substitute $xs$ for $\left[{\text{A}}^{y+}\right]$ and $s^{\text{'}}+ys$ for $\left[{\text{B}}^{-x}\right]$ in equation (4).

$K_{\text{sp}}={\left(xs\right)}^x{\left(s^{\text{'}}+ys\right)}^y$

The value of $s$ is very small in comparison to the value of $s^{\text{'}}$. Therefore it can be neglected and the expression of $K_{\text{sp}}$ changes to,

$K_{\text{sp}}={\left(xs\right)}^x{\left(s^{\text{'}}\right)}^y$

Explanation of Solution

The solubility of the salt ${\text{BaF}}_2$ in the presence of $5.0\text{}\text{}\text{ mg}\cdot {\text{mL}}^{-1}$$\text{KF}$ is calculated below.

Given:

The mass of $\text{KF}$ dissolved in per milliliter of water is $5.0\text{ mg}$.

The balanced chemical reaction  for the dissolution of ${\text{BaF}}_2$ in water is,

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

$\text{KF}$ is a strong salt and undergoes complete dissociation to give potassium ions and fluoride ions in the solution and therefore the concentration of both the potassium and fluoride ions is equal to the initial concentration of $\text{KF}$.

  $\text{KF}\left(\text{s}\right)\to {\text{ K}}^{+}\left(\text{aq}\right){\text{+ F}}^{-}\left(\text{aq}\right)$

The concentration of potassium fluoride ($\text{KF}$) is calculated as follows,

$\left[\text{KF}\right]=\frac{\text{w}}{\text{MV }}$ (5)

Here,

• $\text{w}$ is the given mass of potassium fluoride.
• $\text{M}$ is the gram molecular mass of the potassium fluoride.
• $\text{V }$ is the volume of the solution in liter.

The volume of the solution is,

$\begin{array}{c}\text{V =}\left(\text{1}\text{.00 ml}\right)\left(\frac{1.00\text{ L}}{1000\text{ ml}}\right)\\ ={10}^{-3}\text{L}\end{array}$

Gram molecular mass of potassium fluoride is $\text{58}\text{.1 g}\cdot \text{mol}$.

Substitute $5.0\text{ mg}$ for $\text{w}$, $\text{58}\text{.1 g}\cdot \text{mol}$ for $\text{M}$ and ${10}^{-3}\text{ L}$ for $\text{V }$ in equation (5).

$\begin{array}{c}\left[\text{KF}\right]=\frac{\text{5}\text{.0 mg}\left(\frac{1\text{ g}}{1000\text{}\text{ mg}}\right)}{\left(\text{58}\text{.1 g}\cdot \text{mol}\right)\left({\text{10}}^{-3}\text{ L}\right)}\\ =0.086\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The ICE table(4) is given as,

$\begin{array}{cccccc}\text{Equation}& {\text{BaF}}_2\left(\text{s}\right)& \rightleftharpoons & {\text{Ba}}^{2+}\left(\text{aq}\right)& +& 2{\text{F}}^{-}\left(\text{aq}\right)\\ \text{Initial }\left(\text{M}\right)& & & 0& & 0.086\\ \text{Change }\left(\text{M}\right)& & & +s& & +2s\\ \text{Equilibrium }\left(\text{M}\right)& & & \text{ s}& & \left(0.086+2s\right)\end{array}$

Here,

• $0.086$ is the concentration of the fluoride ions coming from strong salt $\text{KF}$.

From the table,

$\begin{array}{c}\left[{\text{Ba}}^{2+}\right]=s\\ \left[{\text{F}}^{-1}\right]=0.086+2s\end{array}$

Substitute $s$ for $\left[{\text{Ba}}^{2+}\right]$ and $\left(0.086+s\right)$ for $\left[{\text{F}}^{-1}\right]$ in equation (2).

$K_{\text{sp}}=\left(s\right){\left(0.086+s\right)}^2$

The value of $s$ is very small in comparison to the value of $0.086$. Therefore it can be neglected and $\left(0.086+s\right)\approx 0.086$.

The expression for $K_{\text{sp}}$ changes as,

$K_{\text{sp}}=\left(s\right){\left(0.086\right)}^2$

Rearrange for $s$.

$s=\frac{K_{\text{sp}}}{{\left(0.086\right)}^2}$

Substitute $1.8\times {10}^{-7}$ for $K_{\text{sp}}$.

$\begin{array}{c}s=\frac{\left(1.8\times {10}^{-7}\right)}{{\left(0.086\right)}^2}\\ =2.43\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Calculate the solubility of ${\text{BaF}}_2$ in the presence of $5.0\text{}\text{}\text{ mg}\cdot {\text{mL}}^{-1}$$\text{KF}$ in milligrams per milliliter unit.

Substitute $2.43\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}$ for $s$ and $175.3\text{g}\cdot {\text{mol}}^{-\text{1}}$ for $\text{M}$ in equation (3).

The solubility of ${\text{BaF}}_2$ in grams per liter,

$\begin{array}{c}{s}^{\text{'}}=\left(2.43\times {10}^{-5}\text{}\text{}\text{}\frac{\text{mol}}{\text{L}}\right)\left(\text{175}\text{.3 }\frac{\text{g}}{\text{mol}}\right)\\ =4.26\text{g}\cdot {\text{L}}^{-1}\end{array}$

Solubility in milligrams per millilitre is,

$\begin{array}{c}{s}^{\text{'}}=4.26\times {10}^{-3}\text{ g}\cdot {\text{L}}^{-1}\\ =\left(4.26\times {10}^{-3}\frac{\text{g}}{\text{L}}\right)\left(\frac{1000\text{}\text{}\text{ mg}}{1\text{g}}\right)\left(\frac{1\text{}\text{}\text{ L}}{1000\text{mL}}\right)\\ =4.26\times 10^{-3}mg\cdot m{L}^{-1}\end{array}$

The solubility of ${\text{BaF}}_2$ in the presence of $5.0\text{}\text{}\text{ mg}\cdot {\text{mL}}^{-1}$ $\text{KF}$ is $4.26\times 10^{-3}mg\cdot m{L}^{-1}$.

The solubility of ${\text{BaF}}_2$ is less in the water containing $5.0\text{}\text{}\text{ mg}\cdot {\text{mL}}^{-1}$$\text{KF}$ than in pure water. It decreases to $4.26\times 10^{-3}mg\cdot m{L}^{-1}$. from $0.624mg\cdot m{L}^{-1}$ due to common ion effect.