Chapter 12, Problem 12.28P

Interpretation Introduction

Interpretation:

Mass of silver iodide formed along with the concentration of various ions after the completion of reaction has to be calculated.

Concept Introduction:

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Answer to Problem 12.28P

Mass of silver iodide formed is $34.922\text{ g}$. Concentration of ${\text{Ca}}^{2+}$, ${\text{NO}}_3{}^{-}$ and ${\text{I}}^{-}$$0.31875\text{ mol/L}$, $0.49583\text{ mol/L}$ and $0.14166\text{ mol/L}$ respectively.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{175 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{175 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =\text{0}\text{.175 L}\end{array}$

Hence convert $\text{125 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{125 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =\text{0}\text{.125 L}\end{array}$

The formula to evaluate moles from molarity is given as follows:

  $n=MV$        (1)

Substitute $\text{0}\text{.850 mol/L}$ for $M$ and $0.175\text{ L}$ for $V$ in equation (1) to calculate moles of ${\text{AgNO}}_3$.

  $\begin{array}{c}n=\left(\text{0}\text{.850 mol/L}\right)\left(0.175\text{ L}\right)\\ =0.14875\text{ mol}\end{array}$

Substitute $\text{0}\text{.765 mol/L}$ for $M$ and $0.125\text{ L}$ for $V$ in equation (1)to calculate moles of ${\text{CaI}}_2$.

  $\begin{array}{c}n=\left(\text{0}\text{.765 mol/L}\right)\left(0.125\text{ L}\right)\\ =0.095625\text{ mol}\end{array}$

The balanced chemical reaction is given as follows:

  ${\text{2AgNO}}_3\left(aq\right)+{\text{CaI}}_2\left(aq\right)\to 2\text{AgI}\left(s\right)+\text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)$        (2)

According to the stoichiometry of the balanced equation (2), two moles of ${\text{AgNO}}_3$ can give $\text{2 mol AgI}$. Thus moles of $\text{AgI}$ formed from $0.14875{\text{ mol AgNO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Moles of AgI}=\left(0.14875{\text{ mol AgNO}}_3\right)\left(\frac{\text{2 mol AgI}}{{\text{2 mol AgNO}}_3}\right)\\ =0.14875\text{ mol AgI}\end{array}$

Similarly, one mole ${\text{CaI}}_2$ can give $\text{2 mol AgI}$. Thus moles of $\text{AgI}$ formed from $0.095625{\text{ mol CaI}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Moles of AgI}=\left(0.095625{\text{ mol CaI}}_2\right)\left(\frac{\text{2 mol AgI}}{{\text{1 mol CaI}}_2}\right)\\ =0.19125\text{ mol AgI}\end{array}$

Since moles of ${\text{AgNO}}_3$ available are less thus it is limiting reagent. Thus moles of $\text{AgI}$ formed is $0.14875\text{ mol}$.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (3)

Substitute $0.14875\text{ mol}$ for moles and $\text{234}\text{.77 g/mol}$ for molar mass of $\text{AgI}$ in equation (3).

  $\begin{array}{c}\text{Mass}=\left(0.14875\text{ mol}\right)\left(\text{234}\text{.77 g/mol}\right)\\ =34.922\text{ g}\end{array}$

Since all the ${\text{Ag}}^{+}$ present in ${\text{AgNO}}_3$ is consumed and is converted to precipitate $\text{AgI}$, Ionic species present in solution after reaction include ${\text{Ca}}^{2+}$, ${\text{NO}}_3{}^{-}$ and ${\text{I}}^{-}$ ions.

The initial concentration of ${\text{Ca}}^{2+}$ present in $\text{0}{\text{.765 M CaI}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Concentration of Ca}}^{2+}=\left(0.125\text{ L}\right)\left(\frac{\text{0}{\text{.765 mol CaI}}_2}{1\text{ L}}\right)\left(\frac{{\text{1 mol Ca}}^{2+}}{{\text{1 mol CaI}}_2}\right)\\ =0.095625\text{ mol}\end{array}$

Similarly, the initial concentration of ${\text{NO}}_3{}^{-}$  is calculated as follows:

  $\begin{array}{c}{\text{Concentration of NO}}_3{}^{-}=\left(0.175\text{ L}\right)\left(\frac{0.850{\text{ mol AgNO}}_3}{1\text{ L}}\right)\left(\frac{{\text{1 mol NO}}_3{}^{-}}{{\text{1 mol AgNO}}_3}\right)\\ =0.14875\text{ mol}\end{array}$

When solutions are mixed total volume of solution is calculated as follows:

  $\begin{array}{c}\text{Total volume}=0.175\text{ L}+0.125\text{ L}\\ =\text{0}\text{.3 L}\end{array}$

Thus the concentration of ${\text{Ca}}^{2+}$ after the reaction is calculated as follows:

  $\begin{array}{c}{\text{Concentration of Ca}}^{2+}=\left(\frac{0.095625\text{ mol}}{0.3\text{ L}}\right)\\ =0.31875\text{ mol/L}\end{array}$

The concentration of ${\text{NO}}_3{}^{-}$ the present after the reaction is calculated as follows:

  $\begin{array}{c}{\text{Concentration of NO}}_3{}^{-}=\frac{\left(0.14875\text{ mol }\right)}{0.3\text{ L}}\\ =0.49583\text{ mol/L}\end{array}$

Thus ${\text{I}}^{-}$ present in excess is calculated as follows:

  $\begin{array}{c}{\text{Concentration of Excess I}}^{-}=\left(0.095625{\text{ mol CaI}}_2\right)\left(\frac{{\text{2 mol I}}^{-}}{{\text{1 mol CaI}}_2}\right)\\ =0.19125\text{ mol}\end{array}$

Thus moles after the reaction is the difference of excess ${\text{I}}^{-}$ and ${\text{I}}^{-}$ precipitated as $\text{AgI}$ calculated as follows:

  $\begin{array}{c}{\text{Moles of I}}^{-}={\text{Excess I}}^{-}-{\text{I}}^{-}\text{precipitated}\\ =\left(0.19125\text{ mol}-0.14875\text{ mol}\right)\\ =0.0425\text{ mol}\end{array}$

The concentration of ${\text{I}}^{-}$ present after the reaction is calculated as follows:

  $\begin{array}{c}{\text{Concentration of I}}^{-}=\frac{\left(0.0425\text{ mol }\right)}{0.3\text{ L}}\\ =0.14166\text{ mol/L}\end{array}$