Chapter 15, Problem 52A

Interpretation Introduction

Interpretation: The concentration of each of the ion present after mixing has to be calculated.

Concept Introduction:

Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 52A

The concentration of ${\text{Co}}^{2+}$ ion is $\text{0}\text{.167}\text{M}\text{.}$

The concentration of ${\text{Ni}}^{2+}$ ion is $\text{0}\text{.117}\text{M}\text{.}$

The concentration of ${\text{Cl}}^{-}$ ion is $\text{0}\text{.567}\text{M}\text{.}$

Explanation of Solution

Given Information:

The molarity of ${\text{CoCl}}_{\text{2}}$ solution is $\text{0}\text{.250}\text{M}\text{.}$

The molarity of ${\text{NiCl}}_{\text{2}}$ solution is $\text{0}\text{.350}\text{M}\text{.}$

The volume of ${\text{CoCl}}_{\text{2}}$ solution is $\text{50}\text{.0}\text{mL}\text{.}$

The volume of ${\text{NiCl}}_{\text{2}}$ solution is $25.0\text{mL}\text{.}$

Calculation:Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

The dissolution equation is given as,

$\begin{array}{l}{\text{CoCl}}_{\text{2}}\to {\text{Co}}^{\text{2+}}{\text{+2Cl}}^{\text{-}}\\ {\text{NiCl}}_{\text{2}}\to {\text{Ni}}^{\text{2+}}{\text{+2Cl}}^{\text{-}}\end{array}$

The initial molarity of the ions is,

$\begin{array}{l}{\text{CoCl}}_{\text{2}}{\text{:Co}}^{\text{2+}}\text{=0}\text{.250}{\text{M,Cl}}^{\text{-}}\text{=2×0}\text{.250}\text{M=0}\text{.500}\text{M}\\ {\text{NiCl}}_{\text{2}}{\text{:Ni}}^{\text{2+}}\text{=0}\text{.350}{\text{M,Cl}}^{\text{-}}\text{=2×0}\text{.350}\text{M=0}\text{.700}\text{M}\end{array}$

The number of moles is calculated by multiplying the molarity with the volume.

The moles of ${\text{Co}}^{2+}$ is calculated as,

Moles= $\text{Molarity×Volume}$

Moles= $\text{0}\text{.250}\text{M×0}\text{.05}\text{L}$

Moles= $\text{0}\text{.0125}\text{mol}$

The moles of ${\text{Co}}^{2+}$ is $\text{0}\text{.0125}\text{mol}\text{.}$

The moles of ${\text{Ni}}^{2+}$ is calculated as,

Moles= $\text{Molarity×Volume}$

Moles= $\text{0}\text{.350}\text{M×0}\text{.025}\text{L}$

Moles= $\text{0}\text{.00875}\text{mol}$

The moles of ${\text{Ni}}^{2+}$ is $\text{0}\text{.00875}\text{mol}\text{.}$

The moles of ${\text{Cl}}^{-}$ is calculated as,

Moles= $\text{Molarity×Volume}$

Moles= $\left(\text{0}\text{.500}\text{M×0}\text{.05}\text{L}\right)\text{+}\left(\text{0}\text{.700}\text{M×0}\text{.025}\text{L}\right)$

Moles= $\text{0}\text{.0425}\text{mol}$

The moles of ${\text{Cl}}^{-}$ is $\text{0}\text{.0425}\text{mol}\text{.}$

The volume would be $\text{75}\text{mL=0}\text{.075}\text{L,}$ the concentration is obtained by dividing the number of moles by volume.

Concentration of ${\text{Co}}^{2+}$ is calculated as,

$\begin{array}{l}\text{Concentration=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{Concentration=}\frac{\text{0}\text{.0125}\text{mol}}{\text{0}\text{.075}\text{L}}\\ \text{Concentration=0}\text{.167}\text{M}\end{array}$

The concentration of ${\text{Co}}^{2+}$ ion is $\text{0}\text{.167}\text{M}\text{.}$

Concentration of ${\text{Ni}}^{2+}$ is calculated as,

$\begin{array}{l}\text{Concentration=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{Concentration=}\frac{\text{0}\text{.00875}\text{mol}}{\text{0}\text{.075}\text{L}}\\ \text{Concentration=0}\text{.117}\text{M}\end{array}$

The concentration of ${\text{Ni}}^{2+}$ ion is $\text{0}\text{.117}\text{M}\text{.}$

Concentration of ${\text{Cl}}^{-}$ is calculated as,

$\begin{array}{l}\text{Concentration=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{Concentration=}\frac{\text{0}\text{.0425}\text{mol}}{\text{0}\text{.075}\text{L}}\\ \text{Concentration=0}\text{.567}\text{M}\end{array}$

The concentration of ${\text{Cl}}^{-}$ ion is $\text{0}\text{.567}\text{M}\text{.}$

Conclusion

Thus,

The concentration of ${\text{Co}}^{2+}$ ion is $\text{0}\text{.167}\text{M}\text{.}$

The concentration of ${\text{Ni}}^{2+}$ ion is $\text{0}\text{.117}\text{M}\text{.}$

The concentration of ${\text{Cl}}^{-}$ ion is $\text{0}\text{.567}\text{M}\text{.}$