Chapter 4, Problem 4.18P

(a)

Interpretation Introduction

Interpretation:The molarity of each ion in the given solution needs to be determined.

Concept introduction:Molarity is a way of expressing the concentration in terms of number of moles present in volume of solution (in litres). The formula for molarity is:

  $\text{Molarity = }\frac{\text{moles of solute}}{\text{Volume of solution (L)}}$

Explanation of Solution

Given:

Mass of:

  $\begin{array}{l}\text{NaCl = 26}\text{.5 g}\\ {\text{MgCl}}_{\text{2 }}=\text{ 2}\text{.40 g}\\ {\text{MgSO}}_{\text{4}}\text{ = 3}\text{.35 g}\\ {\text{CaCl}}_{\text{2}}\text{ = 1}\text{.20 g}\\ \text{KCl = 1}\text{.05 g}\\ {\text{NaHCO}}_3\text{ = 0}\text{.315 g}\\ \text{NaBr = 0}\text{.098 g}\end{array}$

Density of the solution = $1.025{\text{ g/cm}}^{\text{3}}$

Mass of sample = 1.00 kg

The volume of the solution is calculated as:

  $\text{Density = }\frac{\text{mass}}{\text{Volume}}$

Substituting the values:

  $\begin{array}{l}\text{1}{\text{.025 g/cm}}^{\text{3}}\text{ = }\frac{\text{1 kg}}{\text{Volume}}\\ \text{Volume = }\frac{\text{1 kg}}{\text{1}{\text{.025 g/cm}}^{\text{3}}}\\ \text{Volume = 1}\times {\text{10}}^{\text{3}}\text{ g}\times \frac{{\text{cm}}^{\text{3}}}{\text{1}\text{.025 g}}\times \frac{1\text{ L}}{1000{\text{ cm}}^{\text{3}}}\\ \text{Volume = 0}\text{.976 L}\end{array}$

Since, all the compounds are present in the same solution so, the volume of for all is 0.976 L.

Now, calculating the number of moles of each compound using formula:

  $\text{Mole = }\frac{\text{given mass}}{\text{Molar mass}}$

Substituting the values in above formula and calculating the moles for:

$\text{NaCl}$:

  $\text{moles of NaCl = }\frac{\text{26}\text{.5 g}}{\text{58}\text{.45 g/mol}}\text{ = 0}\text{.4533 mol}$

${\text{MgCl}}_2$:

  ${\text{moles of MgCl}}_{\text{2}}\text{ = }\frac{\text{2}\text{.40 g}}{\text{95}\text{.3 g/mol}}\text{ = 0}\text{.02518 mol}$

${\text{MgSO}}_4$:

  ${\text{moles of MgSO}}_4\text{ = }\frac{3.35\text{ g}}{120.0\text{ g/mol}}\text{ = 0}\text{.0279 mol}$

${\text{CaCl}}_2$:

  ${\text{moles of CaCl}}_2\text{ = }\frac{1.20\text{ g}}{111.0\text{ g/mol}}\text{ = 0}\text{.0108 mol}$

$\text{KCl}$:

  $\text{moles of KCl = }\frac{1.05\text{ g}}{74.55\text{ g/mol}}\text{ = 0}\text{.01408 mol}$

${\text{NaHCO}}_3$:

  ${\text{moles of NaHCO}}_3\text{ = }\frac{0.315\text{ g}}{84.0\text{ g/mol}}\text{ = 0}\text{.00375 mol}$

$\text{NaBr}$:

  $\text{moles of NaBr = }\frac{0.098\text{ g}}{102.9\text{ g/mol}}\text{ = 9}\text{.52 }\times {\text{ 10}}^{-4}\text{ mol}$

Now, calculating the molarity of each compound as:

$\text{NaCl}$:

  $\begin{array}{l}\text{Molarity of NaCl = }\frac{\text{Number of moles of NaCl}}{\text{Volume of the solution in (L)}}\\ \text{Molarity of NaCl = }\frac{0.4533\text{ mol}}{0.976\text{ L}}\\ \text{Molarity of NaCl = 0}\text{.464 M}\end{array}$

Since, moles of ${\text{Na}}^{+}$ and ${\text{Cl}}^{\text{-}}$ in $\text{NaCl}$ is in 1:1 ratio so,

  ${\text{[Na}}^{+}]\text{ = 0}{\text{.464 M and [Cl}}^{-}]\text{ = 0}\text{.464 M}$

${\text{MgCl}}_{\text{2 }}$:

  $\begin{array}{l}{\text{Molarity of MgCl}}_2\text{ = }\frac{0.02518\text{ mol}}{0.976\text{ L}}\\ {\text{Molarity of MgCl}}_2\text{ = 0}\text{.0258 M}\end{array}$

Since, moles of ${\text{Mg}}^{\text{2+}}$ and ${\text{Cl}}^{\text{-}}$ in ${\text{MgCl}}_{\text{2}}$ is in 1:2 ratio so,

  ${\text{[Mg}}^{+2}]\text{ = 0}{\text{.0258 M and [Cl}}^{-}]\text{ = 2 }\times \text{ 0}\text{.0258 M = 0}\text{.0516 M}$

${\text{MgSO}}_{\text{4}}$:

  $\begin{array}{l}{\text{Molarity of MgSO}}_4\text{ = }\frac{0.0279\text{ mol}}{0.976\text{ L}}\\ {\text{Molarity of MgSO}}_4\text{ = 0}\text{.0286 M}\end{array}$

Since, moles of ${\text{Mg}}^{\text{2+}}$ and ${\text{SO}}_4^{2-}$ in ${\text{MgSO}}_4$ is in 1:1 ratio so,

  ${\text{[Mg}}^{+2}]\text{ = 0}{\text{.0286 M and [SO}}_4^{-2}]\text{ = 0}\text{.0286 M}$

${\text{CaCl}}_{\text{2}}$:

  $\begin{array}{l}{\text{Molarity of CaCl}}_2\text{ = }\frac{0.0108\text{ mol}}{0.976\text{ L}}\\ {\text{Molarity of CaCl}}_2\text{= 0}\text{.0111 M}\end{array}$

Since, moles of ${\text{Ca}}^{\text{2+}}$ and ${\text{Cl}}^{\text{-}}$ in ${\text{CaCl}}_2$ is in 1:2 ratio so,

  $[{\text{Ca}}^{\text{2+}}]=\text{0}\text{.0110 M and }[{\text{Cl}}^{-}]\text{ = 2 }\times \text{ 0}\text{.0110 M = 0}\text{.022 M}$

$\text{KCl}$:

  $\begin{array}{l}\text{Molarity of KCl = }\frac{0.01408\text{ mol}}{0.976\text{ L}}\\ \text{Molarity of KCl = 0}\text{.0144 M}\end{array}$

Since, moles of ${\text{K}}^{\text{+}}$ and ${\text{Cl}}^{\text{-}}$ in $\text{KCl}$ is in 1:1 ratio so,

  $[{\text{K}}^{+}]\text{ = 0}{\text{.0144 M and [Cl}}^{-}]\text{ = 0}\text{.0144 M}$

${\text{NaHCO}}_3$:

  $\begin{array}{l}{\text{Molarity of NaHCO}}_3\text{ = }\frac{0.00375\text{ mol}}{0.976\text{ L}}\\ {\text{Molarity of NaHCO}}_3\text{ = 0}\text{.0038 M}\end{array}$

Since, moles of ${\text{Na}}^{\text{+}}$ and ${\text{HCO}}_3^{-}$ in ${\text{NaHCO}}_3$ is in 1:1 ratio so,

  $[{\text{Na}}^{+}]\text{ = 0}{\text{.0038 M and [HCO}}_3^{-}]\text{ = 0}\text{.0038 M}$

$\text{NaBr}$:

  $\begin{array}{l}\text{Molarity of NaBr = }\frac{9.52\times {\text{ 10}}^{-4}\text{ mol}}{0.976\text{ L}}\\ \text{Molarity of NaBr }=\text{ 9}\text{.754 }\times {\text{ 10}}^{-4}\text{ M}\end{array}$

Since, moles of ${\text{Na}}^{\text{+}}$ and ${\text{Br}}^{\text{-}}$ in $\text{NaBr}$ is in 1:1 ratio so,

  $[{\text{Na}}^{+}]\text{ = 9}\text{.764 }\times {\text{ 10}}^{-4}{\text{ M and [Br}}^{-}]\text{ = 9}\text{.764 }\times {\text{ 10}}^{-4}\text{ M}$

(b)

Interpretation Introduction

Interpretation:The total molarity of alkali metal ions needs to be calculated.

Concept introduction: Molarity is a way of expressing the concentration in terms of number of moles present in volume of solution (in litres). The formula for molarity is:

  $\text{Molarity = }\frac{\text{moles of solute}}{\text{Volume of solution (L)}}$

Explanation of Solution

Alkali metals are group 1 metals of the periodic table. So, from the given compounds the alkali metal ions are:

  ${\text{Na}}^{+}$ form $\text{NaCl}$ , ${\text{K}}^{+}$ from $\text{KCl}$ , ${\text{Na}}^{+}$ from ${\text{NaHCO}}_3$ and ${\text{Na}}^{+}$ from $\text{NaBr}$ .

So, adding up the values of each concentration as:

  $\begin{array}{l}\text{= 0}\text{.464 M + 0}\text{.0144 M + 0}\text{.0038 M + 0}\text{.0009764 M}\\ \text{= 0}\text{.483 M}\end{array}$

(c)

Interpretation Introduction

Interpretation: The total molarity of alkaline earth metal ions needs to be calculated.

Concept introduction: Molarity is a way of expressing the concentration in terms of number of moles present in volume of solution (in litres). The formula for molarity is:

  $\text{Molarity = }\frac{\text{moles of solute}}{\text{Volume of solution (L)}}$

Explanation of Solution

Alkaline earth metals are group 2 metals of the periodic table. So, from the given compounds the alkaline metal ions are:

  ${\text{Mg}}^{+2}$ form ${\text{MgCl}}_2$ , ${\text{Mg}}^{+2}$ from ${\text{MgSO}}_4$ and ${\text{Ca}}^{+2}$ from ${\text{CaCl}}_2$ .

So, adding up the values of each concentration as:

  $\begin{array}{l}\text{= 0}\text{.0258 M + 0}\text{.0286 M + 0}\text{.0110 M}\\ \text{= 0}\text{.0654 M}\end{array}$

(d)

Interpretation Introduction

Interpretation:Total molarity of anions needs to be calculated.

Concept introduction: Molarity is a way of expressing the concentration in terms of number of moles present in volume of solution (in litres). The formula for molarity is:

  $\text{Molarity = }\frac{\text{moles of solute}}{\text{Volume of solution (L)}}$

Explanation of Solution

Anions are negatively charged species due to gain of electrons. So, from the given compounds the sum of addition of concentration of anions are:

  $\begin{array}{l}{\text{[Cl}}^{-}]{\text{ + [Cl}}^{-}]{\text{ + [SO}}_4^{-2}{\text{] + [Cl}}^{-}]{\text{ + [Cl}}^{-}]{\text{ + [HCO}}_3^{-}{\text{] + [Br}}^{-}\text{]}\\ \text{= 0}\text{.464 M + 0}\text{.0516 M + 0}\text{.0286 M + 0}\text{.022 M + 0}\text{.0144 M + 0}\text{.0038 M + 0}\text{.0009764 M}\\ \text{= 0}\text{.585 M }\end{array}$