Chapter 4, Problem 72QP

Interpretation Introduction

Interpretation:

The concentration of sodium ions is to be calculated in the given solutions and the concentration of the lithium carbonate 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 combination other than $1:1$, 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 72QP

Solution:

(a) $6.50\text{ M}$, $3.56\text{ M}$, $0.585\text{ M}$.

(b) $0.298\text{ M}$

(a)

Explanation of Solution

Given information:

$\text{3}{\text{.25 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, $1.7{\text{8 M Na}}_2{\text{CO}}_3$, $0.58{\text{5 M NaHCO}}_3$

Calculate the concentration of ${\text{Na}}^{+}$ ion in $\text{3}{\text{.25 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution. ${\text{Na}}_2{\text{SO}}_4$ dissociates as follows:

${\text{Na}}_2{\text{SO}}_4\left(s\right)\to 2{\text{Na}}^{\text{+}}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

So, there are two moles of ${\text{Na}}^{+}$ for everymole of ${\text{Na}}_2{\text{SO}}_4$.

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

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=\left[{\text{Na}}_2{\text{SO}}_4\right]\times \frac{{\text{2 mol Na}}^{+}}{{\text{1 mol Na}}_2{\text{SO}}_4}\\ =3.25\text{ M}\times \frac{{\text{2 mol Na}}^{+}}{{\text{1 mol Na}}_2{\text{SO}}_4}\\ =6.50\text{ M}\end{array}$

Hence,

$\left[{\text{Na}}^{+}\right]=6.50\text{ M}$

Calculate the concentration of sodium ions in $1.7{\text{8 M Na}}_2{\text{CO}}_3$ solution.

${\text{Na}}_2{\text{CO}}_3$ dissociates as follows:

${\text{Na}}_2{\text{CO}}_3\left(s\right)\to 2{\text{Na}}^{\text{+}}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

So, there are two moles of ${\text{Na}}^{\text{+}}$ for everymole of ${\text{Na}}_2{\text{CO}}_3$.

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

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=\left[{\text{Na}}_2{\text{CO}}_3\right]\times \frac{{\text{2 mol Na}}^{+}}{{\text{1 mol Na}}_2{\text{CO}}_3}\\ =1.7\text{8 M}\times \frac{{\text{2 mol Na}}^{+}}{{\text{1 mol Na}}_2{\text{CO}}_3}\\ =3.56\text{ M}\end{array}$

Hence,

$\left[{\text{Na}}^{+}\right]=3.56\text{ M}$

Calculate the concentration of sodium ions in $0.585{\text{M NaHCO}}_3$ solution.

${\text{NaHCO}}_3$ dissociates as follows:

${\text{NaHCO}}_3\left(s\right)\to {\text{Na}}^{\text{+}}\left(aq\right)+{\text{HCO}}_3^{-}\left(aq\right)$

So, there are three moles of ${\text{Na}}^{\text{+}}$ for everymole of ${\text{NaHCO}}_3$.

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

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=\left[{\text{NaHCO}}_3\right]\times \frac{{\text{1 mol Na}}^{+}}{{\text{1 mol NaHCO}}_3}\\ =0.58\text{5 M}\times \frac{{\text{1 mol Na}}^{+}}{{\text{1 mol NaHCO}}_3}\\ =0.585\mathrm{M}\end{array}$

Hence,

$\left[{\text{Na}}^{+}\right]=0.585\text{ M}$

(b)

Given information:

$\left[{\text{Li}}^{+}\right]=0.595M$

Calculate the concentration of lithium carbonate solution, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ dissociates as follows:

${\text{Li}}_2{\text{CO}}_3\left(s\right)\to 2{\text{Li}}^{\text{+}}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

Thus, two moles of ${\text{Li}}^{\text{+}}$ are formed for everymole of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

So, use the given concentration and the stoichiometry of the reaction to calculate the concentration of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution.

$\begin{array}{c}\left[{\text{Li}}_2{\text{CO}}_3\right]=\left[{\text{Li}}^{+}\right]\times \frac{{\text{1 mol Li}}_2{\text{CO}}_3}{{\text{2 mol Li}}^{+}}\\ =0.59\text{5 M}\times \frac{{\text{1 mol Li}}_2{\text{CO}}_3}{{\text{2 mol Li}}^{+}}\\ =0.29\text{8 M}\end{array}$

Hence, the concentration of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $0.29\text{8 M}$.