Q4.39P - Chemistry: Molecular Nature Of Matter And Change

Question

The mass percent of ${\mathbf{\text{Cl}}}^{\mathbf{\text{-}}}$  in a seawater sample is determined by titrating 25.00 mL of seawater with  ${\mathbf{\text{AgNO}}}_{\mathbf{\text{3}}}$  solution, causing a precipitation reaction. An indicator is used to detect the endpoint, which occurs when a free ${\mathbf{\text{Ag}}}^{\mathbf{\text{+}}}$   ion is present in the solution after all the ${\mathbf{\text{Cl}}}^{\mathbf{\text{-}}}$  has reacted. If 53.63 mL of 0.2970 M ${\mathbf{\text{AgNO}}}_{\mathbf{\text{3}}}$   is required to reach the endpoint, what is the mass percent of  ${\mathbf{\text{Cl}}}^{\mathbf{\text{-}}}$ in the seawater (d of seawater = 1.024 g/mL)?

Step-by-Step Solution

Answer

The mass percent of ${\mathrm{Cl}}^{-}$  in the seawater is 2.20 %.

Step 1: Calculate moles of the AgNO 3 in 25ml of seawater

The balanced chemical equation is:

 ${\mathrm{AgNO}}_3\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$

The number of moles of ${\mathrm{AgNO}}_3$  is:

$\begin{array}{c}\mathrm{moles} \mathrm{of} {\mathrm{AgNO}}_3=0.2970\mathrm{moles}/\mathrm{l}\times 53.63\mathrm{ml}\times \frac{1\mathrm{lit}}{1000\mathrm{ml}}\\ =0.0159\mathrm{mol}{\mathrm{AgNO}}_3\end{array}$ 

Step 2: Calculate the mass of Cl - ion in seawater

The sea water is titrated with the silver nitrate solution.

Therefore, 

  $\begin{array}{l}\text{V}\left({\text{Cl}}^{\text{-}}\right)\text{×S}\left({\text{Cl}}^{\text{-}}\right)\text{=V}\left({\text{AgNO}}_{\text{3}}\right)\text{×S}\left({\text{AgNO}}_{\text{3}}\right)\\ \Rightarrow \text{S}\left({\text{Cl}}^{\text{-}}\right)=\frac{\text{V}\left({\text{AgNO}}_{\text{3}}\right)\text{×S}\left({\text{AgNO}}_{\text{3}}\right)}{\text{V}\left({\text{Cl}}^{\text{-}}\right)}\\ \Rightarrow \text{S}\left({\text{Cl}}^{\text{-}}\right)=\frac{\text{0.2970\hspace{0.17em}moles/l\hspace{0.17em}\hspace{0.17em}× 53.63ml}}{\text{25ml}}\\ \Rightarrow \text{S}\left({\text{Cl}}^{\text{-}}\right)=\text{0.637moles/l}\end{array}$

Here, 

  $\begin{array}{l}\text{V=volume of\hspace{0.17em}the\hspace{0.17em}\hspace{0.17em}solution}\\ \text{S=strength\hspace{0.17em}of the\hspace{0.17em}\hspace{0.17em}solution}\end{array}$

Therefore, number of moles of ${\text{Cl}}^{\text{-}}$ present in 25 ml sea water is calculated.  

$\begin{array}{l}{\text{number of moles of Cl}}^{-}\text{=0.637moles/l\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}× 25ml ×}\frac{\text{1l}}{\text{1000ml}}\\ =\text{0.0159\hspace{0.17em}\hspace{0.17em}moles}\end{array}$

The mass of ${\mathrm{Cl}}^{-}$  in seawater is calculated as:

  $\begin{array}{c}\text{Mass\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}}1{\text{\hspace{0.17em}\hspace{0.17em} mole of Cl}}^{-}=35.45\text{\hspace{0.17em}g }\\ {\text{Mass of \hspace{0.17em}0.0159 mole\hspace{0.17em}of Cl}}^{-}=35.45\text{\hspace{0.17em}g}\times 0.0159\\ =0.563\text{\hspace{0.17em}g}\end{array}$

Step 3: Calculate the mass% of Cl - in seawater

The mass % of  ${\mathrm{Cl}}^{-}$  is calculated as:

 

 $\begin{array}{c}\mathrm{mass}\mathrm{\%}=\frac{\mathrm{mass} \mathrm{of} {\mathrm{Cl}}^{-}}{\mathrm{mass} \mathrm{of} \mathrm{sea} \mathrm{water}}\mathrm{x}100\mathrm{\%}\\ \mathrm{mass}\mathrm{\%}=\frac{0.5637\mathrm{gm}{\mathrm{Cl}}^{-}}{(\mathrm{density} \mathrm{of} \mathrm{sea} \mathrm{water})\times (\mathrm{volume} \mathrm{of} \mathrm{sea} \mathrm{water})}\mathrm{x}100\mathrm{\%}\\ =\frac{0.5637\mathrm{gm}}{1.024\mathrm{gm}/\mathrm{ml}\times 25\mathrm{ml}}\mathrm{x}100\mathrm{\%}\\ =2.20\mathrm{\%}{\mathrm{Cl}}^{-}\end{array}$