Chapter 6, Problem 62E

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Interpretation Introduction

Interpretation:

The mass of Silver chloride and concentration of each ion after the precipitation reaction has to be determined.

Concept Introduction:

The mass of a compound can be calculated using the molar mass of the compound to the given moles.  It can be given by the equation,

$\text{Mass}\text{of}\text{compound(in}\text{g)=}\text{Moles(in}\text{g)×Molar}\text{mass(in}\text{g)}$

Concentration of a solution can be defined in terms of molarity as the ratio of moles of solute to the volume of the solution.

$\text{Concentration(in}\text{M)=}\frac{\text{moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 62E

The mass of Silver chloride produced is $\text{2}\text{.9}\text{g}$.

The concentration of individual ions were found to be

${\text{M}}_{{\text{Cl}}^{\text{-}}}\text{=0}\text{.050}\text{M}{\text{Cl}}^{\text{-}}$

${\text{M}}_{{\text{NO}}_{\text{3}}{}^{\text{-}}}\text{=0}\text{.10}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}$

${\text{M}}_{{\text{Ca}}^{\text{2+}}}\text{=0}\text{.075}\text{M}{\text{Ca}}^{\text{2+}}$

Explanation of Solution

Volume of Silver nitrate         =$\text{100}\text{.0}\text{mL}$

Molarity of Silver nitrate        =$\text{0}\text{.20}\text{M}$

Volume of Calcium chloride   =$\text{100}\text{.0}\text{mL}$

Molarity of Calcium chloride =$\text{0}\text{.15}\text{M}$

The volume and molarity of Silver nitrate are recorded along with volume and molarity of Calcium chloride as shown above.

Calculate for the mass of Silver chloride is explained as follows,

Molar mass of Silver chloride= $\text{143}\text{.4}\text{g}$

The chemical equation for this precipitation reaction can be given as,

${\text{2AgNO}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+CaCl}}_{\text{2(aq)}}\to {\text{2AgCl}}_{\left(\text{s}\right)}{\text{+Ca(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}$

$\begin{array}{l}\text{Mass}\text{of}\text{AgCl=0}\text{.1000}\text{L×}\frac{\text{0}\text{.20}\text{mol}{\text{AgNO}}_{\text{3}}}{\text{L}}\text{×}\frac{\text{2}\text{mol}\text{AgCl}}{\text{2}\text{mol}{\text{AgNO}}_{\text{3}}}\text{×}\frac{\text{143}\text{.4}\text{g}\text{AgCl}}{\text{mol}\text{AgCl}}\\ \text{=2}\text{.9}\text{g}\text{AgCl}\\ \text{Mass}\text{of}\text{AgCl=0}\text{.1000}\text{L×}\frac{\text{0}\text{.15}\text{mol}{\text{CaCl}}_{\text{2}}}{\text{L}}\text{×}\frac{\text{2}\text{mol}\text{AgCl}}{\text{mol}{\text{CaCl}}_{\text{2}}}\text{×}\frac{\text{143}\text{.4}\text{g}\text{AgCl}}{\text{mol}\text{AgCl}}\\ \text{=4}\text{.3}\text{g}\text{AgCl}\end{array}$

${\text{AgNO}}_{\text{3}}$ is the limiting because it produces a smaller amount of precipitate. Therefore, the mass of $\text{AgCl}$ is $\text{2}\text{.9}\text{g}$.

Concentration of each ion in remaining solution:

Volume of Silver nitrate         =$\text{100}\text{.0}\text{mL}$

Molarity of Silver nitrate        =$\text{0}\text{.20}\text{M}$

Volume of Calcium chloride   =$\text{100}\text{.0}\text{mL}$

Molarity of Calcium chloride =$\text{0}\text{.15}\text{M}$

The volume and molarity of Silver nitrate are recorded along with volume and molarity of Calcium chloride as shown above.

Calculation for the concentration of ${\text{Cl}}^{\text{-}}$ is explained as follows,

The concentration of each ion can easily be determined from the moles of each reactant.

$\text{0}\text{.1000}\text{L(0}\text{.20}\text{mol}{\text{AgNO}}_{\text{3}}{\text{L}}^{\text{-1}}\text{)=0}\text{.020}\text{mol}{\text{AgNO}}_{\text{3}}$ dissolves $\text{0}\text{.020}\text{mol}{\text{Ag}}^{\text{+}}\text{and}\text{0}\text{.020}\text{mol}{\text{NO}}_{\text{3}}{}^{\text{-}}$

Similarly,

$\text{0}\text{.1000}\text{L(0}\text{.15}\text{mol}{\text{CaCl}}_{\text{2}}{\text{L}}^{\text{-1}}\text{)=0}\text{.015}\text{mol}{\text{CaCl}}_{\text{2}}$ dissolves $\text{0}\text{.015}\text{mol}{\text{Ca}}^{\text{2+}}\text{and}\text{2(0}\text{.015)=0}\text{.030}\text{mol}{\text{Cl}}^{\text{-}}$

To get $\text{2}\text{.9}\text{g}$ of $\text{AgCl}$ precipitate $\text{0}\text{.020}\text{mol}{\text{Ag}}^{\text{+}}$ will react with $\text{0}\text{.020}\text{mol}$${\text{Cl}}^{\text{-}}$ to form $\text{0}\text{.020}\text{mol}\text{AgCl}$

$\begin{array}{l}\text{Mol}\text{unreacted}{\text{Cl}}^{\text{-}}\text{=0}\text{.030}\text{mol}{\text{Cl}}^{\text{-}}\text{initially-0}\text{.020}\text{mol}{\text{Cl}}^{\text{-}}\text{reacted}\\ \text{=0}\text{.010}\text{mol}{\text{Cl}}^{\text{-}}\end{array}$

  $\begin{array}{l}{\text{M}}_{{\text{Cl}}^{\text{-}}}\text{=}\frac{\text{0}\text{.010}\text{mol}{\text{Cl}}^{\text{-}}}{\text{total}\text{volume}}\\ \text{=}\frac{\text{0}\text{.010}\text{mol}{\text{Cl}}^{\text{-}}}{\text{0}\text{.1000}\text{L+0}\text{.1000}\text{L}}\\ \text{=0}\text{.050}\text{M}{\text{Cl}}^{\text{-}}\end{array}$

Concentration of ${\text{Cl}}^{\text{-}}$= $\text{0}\text{.050}\text{M}$

The concentration of ${\text{Cl}}^{\text{-}}$ is calculated by taking in the moles of ${\text{Cl}}^{\text{-}}$ reacted initially to moles of ${\text{Cl}}^{\text{-}}$ reacted finally. The concentration of ${\text{Cl}}^{\text{-}}$ is calculated by plugging in the difference in the moles of ${\text{Cl}}^{\text{-}}$ to the total volume.  The concentration of ${\text{Cl}}^{\text{-}}$ is found to be $\text{0}\text{.050}\text{M}$

Calculation for the concentration of spectator ions is as follows,

  $\begin{array}{l}\text{Concentration}\text{of}{\text{NO}}_{\text{3}}{}^{\text{-}}\text{=}\frac{\text{0}\text{.0200}\text{mol}{\text{NO}}_{\text{3}}{}^{\text{-}}}{\text{0}\text{.2000}\text{L}}\\ \text{=0}\text{.10}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}\\ \text{Concentration}\text{of}{\text{Ca}}^{\text{2+}}\text{=}\frac{\text{0}\text{.015}\text{mol}{\text{Ca}}^{\text{2+}}}{\text{0}\text{.2000}\text{L}}\\ \text{=0}\text{.075}\text{M}{\text{Ca}}^{\text{2+}}\\ \end{array}$

Concentration of ${\text{NO}}_{\text{3}}{}^{\text{-}}\text{=0}\text{.10}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}$

Concentration of ${\text{Ca}}^{\text{2+}}\text{=0}\text{.075}\text{M}{\text{Ca}}^{\text{2+}}$

The concentration of the spectator ions is calculated by plugging in the values of molarities and volumes of Calcium chloride and Silver nitrate.  The concentration of ${\text{NO}}_{\text{3}}{}^{\text{-}}$ and ${\text{Ca}}^{\text{2+}}$ ions were found to be $\text{0}\text{.10}\text{M}$ and $\text{0}\text{.075}$ M respectively.