Question

Based on a concentration of $\mathbf{0}\mathbf{.}\mathbf{065}\mathbf{}{\mathbf{gBr}}^{\mathbf{-}}\mathbf{/}\mathbf{L}$the ocean is estimated to have localid="1663415120962" $\mathbf{\text{\$ 391,000,000}}$of localid="1663415161974" ${\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{/}{\mathbf{\text{mi}}}^{\mathbf{3}}$at recent prices.

a) What is the price per gram of ${\mathbf{\text{Br}}}_{\mathbf{\text{2}}}\mathbf{\text{?}}$

b) Find the mass (in g) of ${\mathbf{\text{AgNO}}}_{\mathbf{\text{3}}}$needed to precipitate $\mathbf{\text{90.0\%}}$of the in of ocean water.

c) Use Appendix C to find how low $\left[{\text{Cl}}^{\text{-}}\right]$must be to prevent the unwanted precipitation of

d) Given a $\left[{\text{Cl}}^{\text{-}}\right]$of about 0.3M, is precipitation with silver ion a practical way to obtain bromine from the ocean? Explain.

Short answer

a) The price per gram of ${\text{Br}}_{\text{2}}$is $\$1.44\times {10}^{-3}$.

b) The mass (in g) of ${\text{AgNO}}_{\text{3}}$needed to precipitate 90% of the ${\text{Br}}^{\text{-}}$in 1 L of ocean water is role="math" $\text{m}\left({\text{AgNO}}_{\text{3}}\right)\text{= 0.124g}$.

c)$2.46\times {10}^{-7}\text{mol}$above $\left[{\text{Cl}}^{\text{-}}\right]$must be used to prevent the unwanted precipitation of $\text{AgCl}$.

d) No, both compounds would precipitate if silver ions were added to ocean water since they are nearly insoluble in water.

Step-by-step solution

 Concept Introduction

Bromine is the second component in the halogen group, and it is found in minute levels in crustal rock as bromide salts, from which it has been leached and then collected in the seas.

 Compute the price per gram of  Br2

(a)

We can compute the price of one gram of${\mathrm{Br}}_2$ if one cubic mile of ocean water contains $\$391,000,000$worth of ${\mathrm{Br}}_2$and we assume that the mass concentration of ${\mathrm{Br}}^{-}$is $0.065\text{g}/L$:

$$\begin{gathered} 1\text{mile}=16093.44\text{decimeters} \\ 1m{i}^3=4.17\times {10}^{12}d{m}^3 \\ 1d{m}^3=1L \\ \frac{\$391000000}{1m{i}^3}=\frac{\$391000000}{4.17\times {10}^{12}{L}^3} \\ =\frac{9.38\times {10}^{-5}}{1L} \end{gathered}$$

The amount of ${\mathrm{Br}}_2$in one litre of ocean water is $\$9.38\times {10}^{-5}$.

We may determine the price per gram of ${\mathrm{Br}}_2$using mass concentration:

$$\begin{gathered} 2Br\left(aq\right)\to B{r}_2\left(aq\right)+2c \\ m\left(Br\right)=0.065g \\ n\left(B{r}_2\right)=\frac{1}{2}n\left(Br\right) \\ M\left(B{r}_2\right)=2A\left(Br\right) \\ m\left(B{r}_2\right)=m\left(B{r}^{-}\right) \\ =0.065g \end{gathered}$$

We can deduct from the chemical equation that the chemical amount of${\mathrm{Br}}_2$ is twice as tiny as the chemical amount of $B{r}^{-}$. However, because the molar mass of ${\mathrm{Br}}_2$ is two times greater than the atomic mass of ${\mathrm{Br}}^{-}$, the mass of ${\mathrm{Br}}_2$ is the same as the mass of ${\mathrm{Br}}^{-}$. So, there is of ${\mathrm{Br}}_2$ in of ocean water, which is worth (per gram):

$\frac{\$9.38\times {10}^{-5}}{1L}=\frac{\$9.38\times {10}^{-5}}{0.065\text{g}}=\frac{\$1.44\times {10}^{-3}}{1g}$

Therefore, the price per gram of ${\mathrm{Br}}_2$is $\$1.44\times {10}^{-3}$.

 Compute the mass of  AgNO3

(b)

Let's multiply 90% by the chemical amount of bromine ions in 1L:

$$\begin{gathered} V\left(\text{oceanwater}\right)=1.00L \\ m\left(B{r}^{-}\text{in}1L\text{of ocean water}\right)=0.065g \\ n\left(B{r}^{-}\right)=\frac{m\left(B{r}^{-}\right)}{A\left(B{r}^{-}\right)} \\ n\left(B{r}^{-}\right)=\frac{0.065g}{79.904gmol} \\ n\left(B{r}^{-}\right)=8.13\times {10}^{-4}\text{moln}{\left(B{r}^{-}\right)}_{90\%} \\ n\left(B{r}^{-}\right)=8.13\times {10}^{-4}\times 0.90 \\ n{\left(B{r}^{-}\right)}_{90\%}=7.32\times {10}^{-4}mol \end{gathered}$$

Let's replace${\text{Br}}^{-}$ions with${\text{AgNO}}_3$in the equation and compute the mass of${\text{AgNO}}_3$:

$$\begin{gathered} {\text{Br}}^{-}\left(aq\right)+{\text{AgNO}}_3\left(aq\right)\to \text{AgBr}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(aq\right) \\ n\left({\text{AgNO}}_3\right)=n{\left({\text{Br}}^{-}\right)}_{90\%} \\ n\left({\text{AgNO}}_3\right)=7.32\times {10}^{-4}\text{mol} \\ m\left({\text{AgNO}}_3\right)=n\left(\text{AgNO}3\right)·\text{M}\left({\text{AgNO}}_3\right) \\ m\left({\text{AgNO}}_3\right)=7.32\times {10}^{-4}\text{mol}\times 169.87g{\text{mol}}^{-1} \\ m\left({\text{AgNO}}_3\right)=0.124g \end{gathered}$$

Therefore, the mass (in g) of ${\text{AgNO}}_{\text{3}}$needed to precipitate 90% of the ${\text{Br}}^{-}$ in $1L$of ocean water is $\text{m}\left({\text{AgNO}}_{\text{3}}\right)\text{= 0.124g}$.

Compute how long  CI- must be used to prevent unwanted precipitation of  AgCI

(c)

We can find the solubility constant for AgCI by looking at Appendix C.

$K_{sp}=1.8\times {10}^{-10}$

We can utilise $K_{sp}$to figure out how low $\left[{\text{Cl}}^{-}\right]$is. To avoid undesirable precipitation of must be:

$$\begin{gathered} AgCl\left(s\right)\to A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right) \\ {K}_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right] \\ \left[C{l}^{-}\right]=\frac{K_{sp}}{\left[A{g}^{+}\right]} \\ \left[{\text{Cl}}^{-}\right]=\frac{1.8\times {10}^{-10}}{7.32\times {10}^{-4}} \\ \left[C{l}^{-}\right]=2.46\times {10}^{-7}\text{mol} \end{gathered}$$

Given that$\text{AgCl}$ is very insoluble in water.

Therefore, any chlorine ion concentration greater than$2.46\times {10}^{-7}$ mole would cause precipitation in the circumstances described below (b).

 Add silver ion with ocean water

(d)

Bromine can't be obtained from the ocean by precipitation with silver ion. If we look at Appendix C again, we can see that${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgCl}\right)=1.8\times {10}^{-10}$and ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgBr}\right)=5.0\times {10}^{-13}.$While$\mathrm{AgBr}$is less soluble in water than$\text{AgCl.}$

Therefore, both compounds would precipitate if silver ions were added to ocean water because they are nearly insoluble in water.