Q19.145 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Because of the toxicity of mercury compounds, mercury $\mathbf{\text{(I)}}$chloride is used in antibacterial salves. The mercury $\mathbf{\text{(I)}}$ ion  ${\mathbf{(}\mathbf{Hg}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}\mathbf{)}$ consists of two bounds  ions. 

(a) What is the empirical formula of mercury $\mathbf{\text{(I)}}$chloride?

(b) Calculate ${\mathbf{(}\mathbf{Hg}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}\mathbf{)}$ in a saturated solution of mercury $\mathbf{(}\mathbf{I}\mathbf{)}$chloride $\mathbf{(}{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{)}$. 

(c) A seawater sample contains  $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{\text{lb}}$of $\mathbf{\text{NaCL}}$ per gallon. Find ${\mathbf{[}\mathbf{Hg}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}\mathbf{]}$ if the seawater is saturated with mercury $\mathbf{\text{(I)}}$chloride. 

(d) How many grams of mercury $\mathbf{\text{(I)}}$chloride is needed to saturate $\mathbf{4900}\mathbf{ }{\mathbf{km}}^{\mathbf{3}}$of pure water (the volume of Lake Michigan)? 

(e) How many grams of mercury $\mathbf{\text{(I)}}$chloride is needed to saturate $\mathbf{4900}\mathbf{ }{\mathbf{km}}^{\mathbf{3}}$ of seawater?

Step-by-Step Solution

Answer

(a) The empirical formula of mercury $\text{(I)}$chloride is $\text{HgCl}$.

(b) The equilibrium concentration of ${\mathrm{Hg}}_2^{2+}$ in saturated mercury$\text{(I)}$ chloride is $7.2112\times {10}^{-7}\mathrm{M}$.

(c) The equilibrium concentration of ${\mathrm{Hg}}_2^{2+}$in seawater is $8.9233\times {10}^{-18}\mathrm{M}$.

(d) The mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$dissolved in pure water is $1.6678\times {10}^{12} \mathrm{g}$.

(e) The mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$dissolved in sea water is $20.6378 \mathrm{g}$.

 

Step 1: Concept Introduction.

The empirical formula of a chemical compound is the simplest whole number ratio of atoms present in the substance, as defined in chemistry.

Step 2: Empirical Formula for Mercury.

(a) 

Mercury$\text{(I)}$ chloride is made up of mercury $\text{(I)}$ cations $(\mathrm{H}{\mathrm{g}}^{+})$ as well as the chloride anion $(\mathrm{C}{\mathrm{l}}^{-})$. Mercury $\left(\mathrm{I}\right)$ chloride has the chemical formula ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$. The empirical formula is the same as the molecular formula, except that subscript are the lowest whole integers possible. 

 

Therefore, $\mathrm{HgCl}$ is the empirical formula for mercury$\left(\mathrm{I}\right)$ chloride.

Step 3: Concentration of Hg 2 2 + .

(b)

Let $s$ be the solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ in pure water, then identify the equilibrium concentration of ions –

$\begin{array}{c}{[\mathrm{Hg}}_2^{2+}] = \mathrm{s}\\ [\mathrm{C}{\mathrm{l}}^{-}] = 2\mathrm{s}\end{array}$

Find the expression for the solubility product constant $\left({\mathrm{K}}_{\mathrm{sp}}\right)$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}= {[\mathrm{Hg}}_2^{2+}\left]\right[\mathrm{C}{\mathrm{l}}^{-}{]}^2\\ {\mathrm{K}}_{\mathrm{sp}}= \mathrm{s}(2\mathrm{s}{)}^2\end{array}$

Substitute the value of ${\mathrm{K}}_{\mathrm{sp}}$ and equilibrium concentration and solve for $s$.

$\begin{array}{c}1.5\times {10}^{-18}= 4{\mathrm{s}}^3\\ \mathrm{s} = \sqrt[3]{\frac{1.5\times {10}^{-18}}{4}}\\ \mathrm{s} = 7.2112\times {10}^{-7}\mathrm{M}\end{array}$

 

Therefore, the concentration value is obtained as $7.2112\times {10}^{-7}\mathrm{M}$.

 

Step 4: Concentration of Hg 2 2 + .

(c)

Convert the molar concentration of sodium chloride $(0.20\mathrm{lb}/\mathrm{gal})$in pounds per gallon to mole per litre.

$\begin{array}{c}\mathrm{M} =\left(\frac{0.20\mathrm{lbNaCl}}{\mathrm{gal}}\right)\times \left(\frac{453.59\mathrm{gal}}{\mathrm{lb}}\right)\times \left(\frac{1\mathrm{molNaCl}}{58.44\mathrm{gNaCl}}\right)\times \left(\frac{1\mathrm{gal}}{3.785\mathrm{L}}\right)\\ \mathrm{M} = 0.4101\frac{\mathrm{molNaCl}}{\mathrm{L}}\end{array}$

The initial concentration of chloride ion is $0.4101 \mathrm{M}$ when mercury$\text{(I)}$ chloride dissolves in $0.4101 \mathrm{M}$sodium chloride.

Let $s$ be the solubility of mercury$\text{(I)}$ chloride in $0.4101 \mathrm{M}$sodium chloride, then identify the equilibrium concentrations of ions.

$\begin{array}{c}{[\mathrm{Hg}}_2^{2+}]=\mathrm{s}\\ [\mathrm{C}{\mathrm{l}}^{-}]=0.41+2\mathrm{s}\end{array}$

Substitute the value of ${\mathrm{K}}_{\mathrm{sp}}$and equilibrium concentration and solve for $s$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}= {[\mathrm{Hg}}_2^{2+}\left]\right[\mathrm{C}{\mathrm{l}}^{-}{]}^2\\ 1.5\times {10}^{-18} = \mathrm{s}(0.41+2\mathrm{s}{)}^2\\ \mathrm{s} = 8.9233\times {10}^{-18}\mathrm{M}\end{array}$

 

Therefore, the concentration value is obtained as $8.9233\times {10}^{-18} \mathrm{M}$.

Step 5: Mass of Mercury Chloride.

(d)

The solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ in pure water is $7.2112\times {10}^{-7} \mathrm{M}$. Identify from the volume of pure water and solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ the value of mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$dissolved.

Convert the volume of water $(4900 \mathrm{k}{\mathrm{m}}^3)$in ${\mathrm{km}}^3$ to $\text{L}$.

$\begin{array}{c}\mathrm{V} = \left(4900{\mathrm{km}}^3\right)\times \left(\frac{{10}^{15}{\mathrm{cm}}^3}{1{\mathrm{km}}^3}\right)\times \left(\frac{1\mathrm{mL}}{1{\mathrm{cm}}^3}\right)\times \left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right)\\ \mathrm{V} = 4.9\times {10}^{15}\mathrm{L}\end{array}$

The mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ dissolved is calculated by multiplying the solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$in pure water by the volume of water and the molar mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$.

$\begin{array}{c}\mathrm{mass} = \left(4.9\times {10}^{15}\mathrm{L}\right)\times \left(\frac{7.2112\times {10}^{-7}\mathrm{mol}}{\mathrm{L}}\right)\times \left(\frac{472{\mathrm{gHg}}_2{\mathrm{Cl}}_2}{{\mathrm{molHg}}_2{\mathrm{Cl}}_2}\right)\\ \mathrm{mass} = 1.6678\times {10}^{12}\mathrm{g}\end{array}$

 

Therefore, the value for mass is obtained as $1.6678\times {10}^{12} \mathrm{g}$.

Step 6: Mass of Mercury Chloride.

(e)

The solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$in sea water is $8.9233\times {10}^{-18}\mathrm{M}$. Identify from the volume of sea water and solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ the value of mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ dissolved.

Convert the volume of water $(4900 \mathrm{k}{\mathrm{m}}^3)$in $\mathrm{k}{\mathrm{m}}^3$to $\text{L}$.

$\begin{array}{c}\mathrm{V}= \left(4900 {\mathrm{km}}^3\right)\times \left(\frac{{10}^{15}{\mathrm{cm}}^3}{1 {\mathrm{km}}^3}\right)\times \left(\frac{1 \mathrm{mL}}{1 {\mathrm{cm}}^3}\right)\times \left(\frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right)\\ \mathrm{V} = 4.9\times {10}^{15} \mathrm{L}\end{array}$

The mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ dissolved is calculated by multiplying the solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ in sea water by the volume of sea water and the molar mass of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$.

$\begin{array}{c}\mathrm{mass} = \left(4.9\times {10}^{15} \mathrm{L}\right)\times \left(\frac{8.9233\times {10}^{-18} \mathrm{mol}}{\mathrm{L}}\right)\times \left(\frac{472 \mathrm{g} {\mathrm{Hg}}_2{\mathrm{Cl}}_2}{\mathrm{mol} {\mathrm{Hg}}_2{\mathrm{Cl}}_2}\right)\\ \mathrm{mass} = 20.6378 \mathrm{g}\end{array}$

 

Therefore, the value for mass is obtained as $20.6378 \mathrm{g}$.