Chapter 15, Problem 76QAP

Interpretation Introduction

(a)

Interpretation:

The value of $\left[\text{OH}\right]$ at which $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ start to precipitates needs to be determined.

Concept introduction:

The equilibrium reaction of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Mg}}^{2+}\left(\text{aq}\right)+2{\text{OH}}^{-}$.

Solubility product expression for above reaction will be

$K_{sp}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)=\left[{\text{Mg}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$

Answer to Problem 76QAP

At $\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}$, $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ will precipitate.

Explanation of Solution

The equilibrium reaction of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Mg}}^{2+}\left(\text{aq}\right)+2{\text{OH}}^{-}$.

The solubility product expression for above reaction will be

$K_{sp}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)=\left[{\text{Mg}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$   ....... (1)

$\begin{array}{c}{K}_{sp}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)=6\times {10}^{-12}\\ \left[{\text{Mg}}^{2+}\right]=0.056\end{array}$

Above $\left[{\text{Mg}}^{2+}\right]$ and $K_{sp}$ values put in equation (1).

$\begin{array}{c}6\times {10}^{-12}=0.056\times {\left[{\text{OH}}^{-}\right]}^2\\ {\left[{\text{OH}}^{-}\right]}^2=\frac{6\times {10}^{-12}}{0.056}\\ \left[{\text{OH}}^{-}\right]=\sqrt{\frac{6\times {10}^{-12}}{0.056}}\\ \left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}\end{array}$

Hence, at $\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}$, $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ will precipitate.

Interpretation Introduction

(b)

Interpretation:

At the above OH concentration $\left(\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}\right)$, whether any of the other ions precipitate or not needs to be determined.

Concept introduction:

If, $Q<K_{sp}$ precipitation does not take place.

If, $Q>K_{sp}$, precipitation takes place.

If, $Q=K_{sp}$, solution is saturated solution.

Q is ionic product and $K_{sp}$ is solubility product.

Answer to Problem 76QAP

The ions which precipitate at $\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}$ are ${\text{Al}}^{\text{+3}}$ and ${\text{Fe}}^{\text{+3}}$.

Explanation of Solution

The concentration of hydroxide ion is as follows:

$\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{M}$

The precipitation of sodium ions will not take place with OH ion.

The ionic product ( Q) of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is calculated as follows:

$Q=\left[{\text{Ca}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$   ....... (2)

$\left[{\text{Ca}}^{2+}\right]=0.01$

From equation (2), ionic product of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ will be

$\begin{array}{c}Q=\left[{\text{Ca}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2\\ =0.01\times {\left(1\times {10}^{-5}\right)}^2\\ =1\times {10}^{-12}\end{array}$

$K_{sp}\left(\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\right)=4\times {10}^{-6}$

As, $Q<K_{sp}$, so precipitation of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ will not takes place.

From equation (2), ionic product ( Q) of $\text{Al}{\left(\text{OH}\right)}_3$ will be

$Q=\left[{\text{Al}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3$   ....... (3)

Or,

$\left[{\text{Al}}^{\text{3+}}\right]=4\times {10}^{-7}$

From equation (3), ionic product is

$\begin{array}{c}Q=\left[{\text{Al}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3\\ =4\times {10}^{-7}\times {\left(1\times {10}^{-5}\right)}^3\\ =4\times {10}^{-22}\end{array}$

Or,

$K_{sp}\left(\text{Al}{\left(\text{OH}\right)}_3\right)=2\times {10}^{-31}$

As, $Q>K_{sp}$, so precipitation of $\text{Al}{\left(\text{OH}\right)}_3$ takes place.

The ionic product ( Q) of $\text{Fe}{\left(\text{OH}\right)}_3$ is calculated as follows:

$Q=\left[{\text{Fe}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3$   ....... (4)

Or,

$\left[{\text{Fe}}^{\text{3+}}\right]=2\times {10}^{-7}$

From equation (4), ionic product is

$\begin{array}{c}Q=\left[{\text{Fe}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3\\ =2\times {10}^{-7}\times {\left(1\times {10}^{-5}\right)}^3\\ =2\times {10}^{-22}\end{array}$

Or,

$K_{sp}\left(\text{Fe}{\left(\text{OH}\right)}_3\right)=3\times {10}^{-39}$

As, $Q>K_{sp}$, so precipitation of $\text{Fe}{\left(\text{OH}\right)}_3$ takes place.

Interpretation Introduction

(c)

Interpretation:

The percentage of precipitated ions needs to be calculated, if enough ${\text{OH}}^{-}$ is added to precipitate $\text{50\%}$ of ${\text{Mg}}^{\text{2+}}$ ions.

Concept introduction:

The equilibrium reaction of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Mg}}^{2+}\left(\text{aq}\right)+2{\text{OH}}^{-}$.

The expression of solubility productfor above reaction is represented as follows:

$K_{sp}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)=\left[{\text{Mg}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$

The equilibrium reaction of $\text{Al}{\left(\text{OH}\right)}_3$ is

$\text{Al}{\left(\text{OH}\right)}_3\left(\text{s}\right)\rightleftharpoons {\text{Al}}^{3+}\left(\text{aq}\right)+3{\text{OH}}^{-}$.

The expression of solubility productfor above reaction is

$K_{sp}\left(\text{Al}{\left(\text{OH}\right)}_{\text{2}}\right)=\left[{\text{Al}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3$

The equilibrium reaction of $\text{Fe}{\left(\text{OH}\right)}_3$ is

$\text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Fe}}^{3+}\left(\text{aq}\right)+3{\text{OH}}^{-}$.

The expression of solubility productfor above reaction is

$K_{sp}\left(\text{Fe}{\left(\text{OH}\right)}_3\right)=\left[{\text{Fe}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3$

Answer to Problem 76QAP

So, under mentioned conditions, $100\%$ of ${\text{Fe}}^{\text{3+}}$ and ${\text{Al}}^{\text{3+}}$ ions precipitate.

Explanation of Solution

Calculation of $\left[{\text{OH}}^{-}\right]$, if enough OH- is added to precipitate $\text{50\%}$ of ${\text{Mg}}^{\text{2+}}$ ions is represented as follows:

$\begin{array}{c}{K}_{sp}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)=6\times {10}^{-12}\\ \left[{\text{Mg}}^{2+}\right]=0.056\end{array}$

Or,

$\begin{array}{c}{K}_{\text{sp}}\left({\text{MgOH}}_{\text{2}}\right)\text{=}\left[\text{50\%}\text{of}{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2\\ 6\times {10}^{-12}=\left[\frac{50}{100}\times 0.056\right]{\left[{\text{OH}}^{-}\right]}^2\\ 6\times {10}^{-12}=0.028\times {\left[{\text{OH}}^{-}\right]}^2\\ {\left[{\text{OH}}^{-}\right]}^2=\frac{6\times {10}^{-12}}{0.028}\end{array}$

Also,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\frac{6\times {10}^{-12}}{0.028}}\\ \left[{\text{OH}}^{-}\right]=1.464\times {10}^{-5}\text{M}\end{array}$

Hence, to precipitate $\text{50\%}$ of ${\text{Mg}}^{\text{2+}}$ ions, required OH- concentration is $1.464\times {10}^{-5}\text{M}$.

Use above $\left[{\text{OH}}^{-}\right]$ to calculate final concentration of $\left[{\text{Al}}^{\text{3+}}\right]$ and $\left[{\text{Fe}}^{\text{3+}}\right]$.

$\begin{array}{c}{K}_{\text{sp}}\left(\text{Al}{\left(\text{OH}\right)}_3\right)\text{=}\left[{\text{Al}}^{\text{3+}}\right]{\left[{\text{OH}}^{-}\right]}^3\\ 2\times {10}^{-31}=\left[{\text{Al}}^{\text{3+}}\right]{\left[1.464\times {10}^{-5}\right]}^3\\ \left[{\text{Al}}^{\text{3+}}\right]=\frac{2\times {10}^{-31}}{3.138\times {10}^{-15}}\\ =6.37\times {10}^{-17}\text{M}\end{array}$

Hence, final concentration of ${\text{Al}}^{\text{3+}}$ is $6.37\times {10}^{-17}\text{M}$

Percentage (%) of ${\text{Al}}^{\text{3+}}$ ion is calculated as follows:

$\begin{array}{c}\%{\text{Al}}^{\text{3+}}\text{ion=}\frac{\text{Intial}\left({\text{Al}}^{\text{3+}}\right)-\text{Final}\left({\text{Al}}^{\text{3+}}\right)}{\text{Intial}\left({\text{Al}}^{\text{3+}}\right)}\times 100\%\\ =\frac{4\times {10}^{-7}\text{M}-6.37\times {10}^{-17}\text{M}}{4\times {10}^{-7}\text{M}}\times 100\%\end{array}$

Since, $6.37\times {10}^{-17}\text{M}$ is very small quantity, so it can be neglected.

So,

$\begin{array}{c}\%{\text{Al}}^{\text{3+}}\text{ion}=\frac{4\times {10}^{-7}\text{M}}{4\times {10}^{-7}\text{M}}\times 100\%\\ =100\%\end{array}$

Hence, percentage of ${\text{Al}}^{\text{3+}}$ ions in the solution is $100\%$.

Use $\left[{\text{OH}}^{-}\right]=1.464\times {10}^{-5}$ to calculate final concentration of $\left[{\text{Fe}}^{\text{3+}}\right]$.

$\begin{array}{c}K\text{sp }\left(\text{Fe}{\left(\text{OH}\right)}_3\right)\text{=}\left[{\text{Fe}}^{\text{3+}}\right]{\left[{\text{OH}}^{-}\right]}^3\\ 3\times {10}^{-39}=\left[{\text{Fe}}^{\text{3+}}\right]{\left[1.464\times {10}^{-5}\right]}^3\\ \left[{\text{Fe}}^{\text{3+}}\right]=\frac{3\times {10}^{-39}}{3.138\times {10}^{-15}}\\ =9.56\times {10}^{-25}\text{M}\end{array}$

Hence, final concentration of ${\text{Fe}}^{\text{3+}}$ is $9.56\times {10}^{-25}\text{M}$

Percentage (%) of ${\text{Fe}}^{\text{3+}}$ ion is calculated as follows:

$\begin{array}{c}\%{\text{Fe}}^{\text{3+}}\text{ion=}\frac{\text{Intial}\left({\text{Fe}}^{\text{3+}}\right)-\text{Final}\left({\text{Fe}}^{\text{3+}}\right)}{\text{Intial}\left({\text{Fe}}^{\text{3+}}\right)}\times 100\%\\ =\frac{2\times {10}^{-7}\text{M}-9.56\times {10}^{-25}\text{M}}{2\times {10}^{-7}\text{M}}\times 100\%\end{array}$

Since, $9.56\times {10}^{-25}\text{M}$ is very small quantity, so it can be neglected.

So,

$\begin{array}{c}\%{\text{Fe}}^{\text{3+}}\text{ion}=\frac{2\times {10}^{-7}\text{M}}{2\times {10}^{-7}\text{M}}\times 100\%\\ =100\%\end{array}$

Hence, percentage of ${\text{Fe}}^{\text{3+}}$ ions in the solution is $100\%$.

So, under mentioned conditions, $100\%$ of ${\text{Fe}}^{\text{3+}}$ and ${\text{Al}}^{\text{3+}}$ ions precipitate.

Interpretation Introduction

(d)

Interpretation:

The mass of precipitateunder the conditions in(c), in one liter of seawater needs to be determined.

Concept introduction:

The total mass of precipitate in one liter precipitate is the sum of mass of precipitate of every ion that is $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$, $\text{Fe}{\left(\text{OH}\right)}_3$ and $\text{Al}{\left(\text{OH}\right)}_3$.

Answer to Problem 76QAP

Mass of precipitate is $\text{1}\text{.63}\text{g}$ in one lire of seawater.

Explanation of Solution

At concentration of $1.464\times {10}^{-5}\text{M}$ of OH^-, it is estimated that $\text{50\%}$ of ${\text{Mg}}^{\text{2+}}$ ions, $\text{100\%}$ of ${\text{Al}}^{\text{3+}}$ and ${\text{Fe}}^{\text{3+}}$ ions will precipitate.

Mass of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ (precipitate) can be calculated from concentration of ion, percentage precipitation, molar mass of precipitate and volume of sea water sample as follows:

$\begin{array}{c}\text{Mass}\text{of}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}=\frac{0.056\cancel{\text{mol}{\text{Mg}}^{2+}}}{\cancel{1\text{L}\text{sample}}}\times \cancel{1\text{L}\text{sample}}\times \frac{\text{50}\cancel{\text{mol}{\text{Mg}}^{2+}}}{\text{100}\cancel{\text{mol}{\text{Mg}}^{2+}}}\times \frac{\cancel{1\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}}{1\cancel{\text{mol}{\text{Mg}}^{2+}}}\times \frac{58.32\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}{\cancel{1\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}}\\ =1.63\text{g}\end{array}$

Hence, mass of precipitate $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is $1.63\text{g}$.

Now, mass of $\text{Fe}{\left(\text{OH}\right)}_3$ (precipitate) can be calculated form concentration of ion, percentage precipitation, molar mass of precipitate and volume of sea water sample as follows:

$\begin{array}{c}\text{Mass}\text{of}\text{Fe}{\left(\text{OH}\right)}_3=\frac{2\times {10}^{-7}\cancel{\text{mol}{\text{Fe}}^{+3}}}{\cancel{1\text{L}\text{sample}}}\times \cancel{1\text{L}\text{sample}}\times \frac{\text{100}\cancel{\text{mol}{\text{Fe}}^{+3}}}{\text{100}\cancel{\text{mol}{\text{Fe}}^{+3}}}\times \frac{\cancel{1\text{mol}\text{Fe}{\left(\text{OH}\right)}_3}}{1\cancel{\text{mol}{\text{Fe}}^{+3}}}\times \frac{106.867\text{g}\text{Fe}{\left(\text{OH}\right)}_3}{\cancel{1\text{mol}\text{Fe}{\left(\text{OH}\right)}_3}}\\ =2.137\times {10}^{-7}\text{g}\end{array}$

Hence, mass of precipitate $\text{Fe}{\left(\text{OH}\right)}_3$ is $2.137\times {10}^{-7}\text{g}$.

Similarly, mass of $\text{Al}{\left(\text{OH}\right)}_3$ (precipitate) can be calculated form concentration of ion, percentage precipitation, molar mass of precipitate and volume of sea water sample as follows:

$\begin{array}{c}\text{Mass}\text{of}\text{Al}{\left(\text{OH}\right)}_3=\frac{4\times {10}^{-7}\cancel{\text{mol}{\text{Al}}^{+3}}}{\cancel{1\text{L}\text{sample}}}\times \cancel{1\text{L}\text{sample}}\times \frac{\text{100}\cancel{\text{mol}{\text{Al}}^{+3}}}{\text{100}\cancel{\text{mol}{\text{Al}}^{+3}}}\times \frac{\cancel{1\text{mol}\text{Al}{\left(\text{OH}\right)}_3}}{1\cancel{\text{mol}{\text{Al}}^{+3}}}\times \frac{78.003\text{g}\text{Al}{\left(\text{OH}\right)}_3}{\cancel{1\text{mol}\text{Al}{\left(\text{OH}\right)}_3}}\\ =3.125\times {10}^{-5}\text{g}\end{array}$

Hence, mass of precipitate $\text{Al}{\left(\text{OH}\right)}_3$ is $3.125\times {10}^{-5}\text{g}$.

Total mass of precipitate can be calculated as follows:

$\begin{array}{c}\text{Total}\text{mass}\text{of}\text{precipitate}=1.63\text{g+}2.137\times {10}^{-7}\text{g}+3.125\times {10}^{-5}\text{g}\\ \text{=1}\text{.63}\text{g}\end{array}$

Hence, mass of precipitate is $\text{1}\text{.63}\text{g}$ in one lire seawater.