Question

Veterinarians use Donovan's solution to treat skin diseases in animals. The solution is prepared by mixing $1.00g$of${\text{AsI}}_3\left(s\right)$ , $1.00g$ of$0.209\text{atm}$ localid="1663416331008" ${\mathbf{HgI}}_{\mathbf{2}}$ , andlocalid="1663416361403" $0.900\text{g}$ oflocalid="1663416371349" ${\text{NaHCO}}_3\left(s\right)$ in enough water to make a total volume oflocalid="1663416378613" $100.0mL$ .

(a) Compute the total mass of iodine per litre of Donovan's solution, in grams per litre.

(b) You need a lot of Donovan's solution to treat an outbreak of rash in an elephant herd. You have plenty of mercury (II) iodide and sodium hydrogen carbonate, but the only arsenic (III) iodide you can find is localid="1663416386804" $1.50L$of alocalid="1663416393818" $0.100\text{M}$ aqueous solution. Explain how to preparelocalid="1663416400727" $3.50L$ . of Donovan's solution starting with these materials.

Short answer

a) The mass of iodine present in 1 litre of solution is $14.0g$

b) To prepare $3.50L$ of Donovan's solution we have to take $35.0g$ of ${\text{AsI}}_3(\text{s})$ and $35.0\text{g}$of ${\text{HgI}}_2\left(s\right)$ and $31.5\text{g}$ of ${\text{NaHCO}}_3\left(s\right)$.

Step-by-step solution

Given data

Molar mass of ${\text{AsI}}_3=455.63501{\text{gmol}}^{-1}$

Molar mass of ${\text{HgI}}_2=454.39894{\text{gmol}}^{-1}$

Molar mass of ${\text{NgICO}}_3=84.0066g{\text{mol}}^{-1}$

Concept of Donovan's solution

Donovan's solution is an inorganic compound prepared from arsenic triiodide and mercuric iodide

Calculate the number of moles of each component

The number of moles of each component is calculated as:

Moles of ${\text{HgI}}_2=\frac{1.00g}{454.39894{\text{gmol}}^{-1}}$

$=2.20\times {10}^{-3}\text{mol}$

Moles of ${\text{NaHCO}}_3=\frac{0.900\text{g}}{84.0066{\text{gmol}}^{-1}}$

$=1.07\times {10}^{-2}\text{mol}$

Calculate the total number of moles

One mole of ${\text{AsI}}_3$ is corresponding to three moles of iodine and one mole of data-custom-editor="chemistry" ${\text{HgI}}_2$ corresponds to two moles of iodine.

$$\begin{gathered} Total\text{number of moles}=2.20\times {10}^{-3}{\text{mol AsI}}_3\left(\frac{3\text{molI}}{1{\text{molAsI}}_3}\right)+2.20\times {10}^{-3}{\text{mol HgI}}_2\left(\frac{2\text{molI}}{1{\text{molHgI}}_2}\right) \\ Total\text{number of mole}=1.10\times {10}^{-2}\text{mol} \end{gathered}$$

Find the total mass of iodine per liter of Donovan's solution

$$\begin{gathered} Total\text{mass}=126.90447{\text{gmol}}^{-1}\times 1.10\times {10}^{-2}\text{mol} \\ Total\text{mass}=1.40\text{g} \end{gathered}$$

The mass per liter is,

$1.40\text{g}\times \left(\frac{1000\text{mL}}{1000\text{mL}}\right)=14.0\text{g}$

Preparation of 3.5L of Donovan's solution

Masses of components present in $3.5L$ of solution

$1.00g{\text{HgI}}_2\left(\frac{3500\text{mL}}{100.0\text{mL}}\right)=35.0\text{g}$

$0.900{\text{gNaHCO}}_3\left(\frac{3500\text{mL}}{100.0\text{mL}}\right)=31.5\text{g}$

Mass of ${\text{AsI}}_3\left(s\right)$ present in $1.50L$ of $0.100\text{M}$ solution

$1.50\text{L}\times 0.100\frac{\text{mol}}{L}\times 455.63504\frac{\text{g}}{\text{mol}}=63.8{\text{gAsI}}_3$