Chapter 4, Problem 71QP

Interpretation Introduction

Interpretation:

The concentration of chloride ion is to be calculated in the given solutions and the concentration of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is to be determined.

Concept introduction:

Soluble ionic compounds dissociate completely in the solution upon dissolution, forming ions.

The compounds with $1:1$ combination of constituent ions dissociate to produce one mole of each ion.

For compounds with a combination otherthan $1:1$, subscripts in the chemical formula of the compound are used to calculate the concentration of each ion in the solution.

The molar concentrations of the species in the solution are expressed using square brackets.

Answer to Problem 71QP

Solution:

$BaC{l}_2:0.300MC{l}^{-}:NaCl:0.566MC{l}^{-}:AlC{l}_3:3.606MC{l}^{-}$

$\text{1}\text{.28 M}\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

$0.15{\text{0 M BaCl}}_2$, $0.56\text{6 M NaCl}$, $1.20{\text{2 M AlCl}}_3$

Explanation of Solution

Calculate the concentration of chloride ion in $0.150{\text{M BaCl}}_2$ solution.

${\text{BaCl}}_2$ dissociates as follows:

${\text{BaCl}}_2\left(s\right)\to {\text{Ba}}^{\text{2+}}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

So, there are two moles of ${\text{Cl}}^{-}$ for one mole of ${\text{BaCl}}_2$.

Next, use the given concentration and the stoichiometry as indicated by the molecular formula to calculate the concentration of chloride ion.

$\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\left[{\text{BaCl}}_2\right]\times \frac{{\text{2 mol Cl}}^{-}}{{\text{1 mol BaCl}}_2}\\ =0.15\text{0 M}\times \frac{{\text{2 mol Cl}}^{-}}{{\text{1 mol BaCl}}_2}\\ =0.300\text{ M}\end{array}$

Hence,

$\left[{\text{Cl}}^{-}\right]=0.300\text{ M}$

Calculate the concentration of chloride ions in $0.56\text{6 M NaCl}$ solution.

$\text{NaCl}$ dissociates as follows:

$\text{NaCl}\left(s\right)\to {\text{Na}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

So, there is one mole of ${\text{Cl}}^{-}$ for one mole of $\text{NaCl}$.

Next, use the given concentration and the stoichiometry as indicated by the molecular formula to calculate the concentration of chloride ions.

$\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\left[\text{NaCl}\right]\times \frac{{\text{1 mol Cl}}^{-}}{\text{1 mol NaCl}}\\ =0.566M\times \frac{{\text{1 mol Cl}}^{-}}{\text{1 mol NaCl}}\\ =0.566\text{ M}\end{array}$

Hence,

$\left[{\text{Cl}}^{-}\right]=0.56\text{6 M}$

Calculate the concentration of chloride ions in $1.20{\text{2 M AlCl}}_3$ solution.

${\text{AlCl}}_{\text{3}}$ dissociates as follows:

${\text{AlCl}}_3\left(s\right)\to {\text{Al}}^{\text{3+}}\left(aq\right)+3{\text{Cl}}^{-}\left(aq\right)$

So, there are three moles of ${\text{Cl}}^{-}$ for one mole of ${\text{AlCl}}_{\text{3}}$.

Then, use the given concentration and the stoichiometry as indicated by the molecular formula to calculate the concentration of chloride ions.

$\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\left[{\text{AlCl}}_3\right]\times \frac{{\text{3 mol Cl}}^{-}}{{\text{1 mol AlCl}}_3}\\ =1.20\text{2 M}\times \frac{{\text{3 mol Cl}}^{-}}{{\text{1 mol AlCl}}_3}\\ =3.60\text{6 M}\end{array}$

Hence,

$\left[{\text{Cl}}^{-}\right]=3.60\text{6 M}$

$\left[{\text{NO}}_3^{-}\right]=2.5\text{5 M}$

Calculate the concentration of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution. $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ dissociates as follows:

$\text{Sr}{\left({\text{NO}}_3\right)}_2\left(s\right)\to {\text{Sr}}^{\text{2+}}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)$

Thus, two moles of ${\text{NO}}_{\text{3}}^{\text{-}}$ are formed for every mole of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

So, use the given concentration and the stoichiometry of the reaction to calculate the concentration of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution.

$\begin{array}{c}\left[\text{Sr}{\left({\text{NO}}_3\right)}_2\right]=\left[{\text{NO}}_3^{-}\right]\times \frac{\text{1 mol Sr}{\left({\text{NO}}_3\right)}_2}{{\text{2 mol NO}}_3^{-}}\\ =2.5\text{5 M}\times \frac{\text{1 mol Sr}{\left({\text{NO}}_3\right)}_2}{{\text{2 mol NO}}_3^{-}}\\ \text{=1}\text{.28 M}\end{array}$

Hence, the concentration of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $1.2\text{8 M}$.