Chapter 15, Problem 24PE

(a)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after $\text{45}\text{.5 mL}$ of $\text{0}\text{.10 M NaCl}$ is mixed with $\text{60}\text{.5 mL}$ of $\text{0}\text{.35 M NaCl}$ has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example $\text{0}{\text{.070 M AlCl}}_3$ indicates that in $1\text{ L}$ the moles of ${\text{AlCl}}_3$ is $\text{0}\text{.070 mol}$. The formula to evaluate volume from molarity is given as follows:

  $M=\frac{n}{V}$

Where,

$V$ represents volume.

$M$ represents molarity.

$n$ represents number of moles.

Explanation of Solution

Since in $\text{1000 mL}$ the moles of $\text{NaCl}$ present is $\text{0}\text{.10 mol}$ so, moles of $\text{NaCl}$ in $\text{45}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaCl}=\left(\text{45}\text{.5 mL}\right)\left(\frac{\text{0}\text{.10 mol}}{\text{1000 mL}}\right)\\ =0.0045\text{ mol}\end{array}$

Similarly for other solution, in $\text{1000 mL}$ moles of $\text{NaCl}$ present is $\text{0}\text{.35 mol}$ so, moles of $\text{NaCl}$ in $\text{60}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaCl}=\left(\text{60}\text{.5 mL}\right)\left(\frac{\text{0}\text{.35 mol}}{\text{1000 mL}}\right)\\ =0.02117\text{ mol}\end{array}$

Total moles of solution when two $\text{NaCl}$ solutions is  mixed are calculated as follows:

  $\begin{array}{c}\text{Total moles}=0.0045\text{ mol}+0.02117\text{ mol}\\ =0.02567\text{mol}\end{array}$

Total volume of solution when two $\text{NaCl}$ solutions are mixed is calculated as follows:

  $\begin{array}{c}\text{Total volume }=45.5\text{ mL}+60.5\text{ mL}\\ \text{=106 mL}\end{array}$

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

  $M=\frac{n}{V}$        (1)

Substitute $0.02567\text{mol}$ for $n$ and $\text{106 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.02567\text{mol}}{\text{106 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.2421\text{ M}\end{array}$

$\text{NaCl}$ can be broken into ${\text{1 mol Na}}^{+}$ and ${\text{1 mol Cl}}^{-}$ as follows:

  $\text{NaCl}\rightleftharpoons {\text{Na}}^{+}+{\text{Cl}}^{-}$

Since ${\text{1 M Na}}^{+}$ is furnished by $\text{1 M NaCl}$ so molarity of ${\text{Na}}^{+}$ ion due to $0.2421\text{ M NaCl}$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(0.2421\text{ M NaCl}\right)\left(\frac{{\text{1 M Na}}^{+}}{\text{1 M NaCl}}\right)\\ =0.2421{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 M Cl}}^{-}$ is furnished by $\text{1 M NaCl}$ so molarity of ${\text{Cl}}^{-}$ ions due to $0.2421\text{ M}\text{NaCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(0.2421\text{ M NaCl}\right)\left(\frac{{\text{1 M Cl}}^{-}}{\text{1 M NaCl}}\right)\\ =0.2421{\text{ M Cl}}^{-}\end{array}$

Hence molar concentration of both ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ is $0.2421\text{ M}$ and $0.2421\text{ M}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after $95.5\text{ mL}$ of $\text{1}\text{.25 M HCl}$ is mixed with $125.5\text{ mL}$ of $\text{2}\text{.50 M HCl}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{1}\text{.25 M HCl}$ indicates that in $1\text{ L}$ the moles of $\text{HCl}$ is $\text{1}\text{.25 mol}$, Similarly $\text{2}\text{.50 M HCl}$ indicates that in $1\text{ L}$ the moles of $\text{HCl}$ is $\text{2}\text{.50 mol}$.

Since in $\text{1000 mL}$ the moles of $\text{HCl}$ present is $\text{1}\text{.25 mol}$ so, moles of $\text{HCl}$ in $95.5\text{ mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(95.5\text{ mL}\right)\left(\frac{\text{1}\text{.25 mol}}{\text{1000 mL}}\right)\\ =0.119375\text{ mol}\end{array}$

Similarly for other solution, in $\text{1000 mL}$ moles of $\text{HCl}$ present is $\text{2}\text{.50 mol}$ so, moles of $\text{HCl}$ in $125.5\text{ mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(125.5\text{ mL}\right)\left(\frac{\text{2}\text{.50 mol}}{\text{1000 mL}}\right)\\ =0.31375\text{ mol}\end{array}$

Total volume of solution upon when two $\text{HCl}$ solutions are mixed is calculated as follows:

  $\begin{array}{c}\text{Total volume }=95.5\text{ mL}+125.5\text{ mL}\\ =\text{221 mL}\end{array}$

Substitute $0.119375\text{ mol}$ for $n$ and $\text{221 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.119375\text{ mol}}{\text{221 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.54015\text{ M}\end{array}$

Substitute $0.31375\text{ mol}$ for $n$ and $\text{221 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.31375\text{ mol}}{\text{221 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =1.41968\text{ M}\end{array}$

Thus total molar concentration of $\text{HCl}$ is sum of concentrations calculated as follows:

  $\begin{array}{c}\text{Total concentration of HCl}=0.54015\text{ M}+1.41968\text{ M}\\ =\text{1}\text{.9598 M}\end{array}$

$\text{HCl}$ can be broken into ${\text{1 mol H}}^{+}$ and ${\text{1 mol Cl}}^{-}$ as follows:

  $\text{HCl}\rightleftharpoons {\text{H}}^{+}+{\text{Cl}}^{-}$

Since ${\text{1 M H}}^{+}$ is furnished by $\text{1 M HCl}$ so molarity of ${\text{H}}^{+}$ ion due to $\text{1}\text{.9598 M HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of H}}^{+}=\left(\text{1}\text{.9598 M HCl}\right)\left(\frac{{\text{1 M H}}^{+}}{\text{1 M HCl}}\right)\\ =\text{1}{\text{.9598 M H}}^{+}\end{array}$

Since ${\text{1 M Cl}}^{-}$ is furnished by $\text{1 M HCl}$ so molarity of ${\text{Cl}}^{-}$ ions due to $\text{1}\text{.9598 M HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(\text{1}\text{.9598 M HCl}\right)\left(\frac{{\text{1 M Cl}}^{-}}{\text{1 M HCl}}\right)\\ =\text{1}{\text{.9598 M Cl}}^{-}\end{array}$

Hence molar concentration of both ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ is $\text{1}\text{.9598 M}$.

(c)

Interpretation Introduction

Interpretation:

Molar concentration each ion found after $\text{15}\text{.5 mL}$ of $\text{0}\text{.10 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ is mixed with $\text{10}\text{.5 mL}$ of $\text{0}{\text{.20 M AgNO}}_3$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{0}\text{.10 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ indicates that in $1\text{ L}$ the moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\text{0}\text{.10 mol}$. Similarly $\text{0}{\text{.20 M AgNO}}_3$ indicates that in the moles of ${\text{AgNO}}_3$ is $0.\text{20 mol}$.

Since in $\text{1000 mL}$ moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ present is $\text{0}\text{.10 mol}$ so, moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ in $\text{15}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Ba}{\left({\text{NO}}_{\text{3}}\right)}_2=\left(\text{15}\text{.5 mL}\right)\left(\frac{0.10\text{ mol}}{\text{1000 mL}}\right)\\ =0.00155\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of ${\text{AgNO}}_3$ present is $\text{0}\text{.20 mol}$ so, moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ in $\text{10}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of AgNO}}_3=\left(\text{10}\text{.5 mL}\right)\left(\frac{0.20\text{ mol}}{\text{1000 mL}}\right)\\ =0.0021\text{ mol}\end{array}$

$\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ can be broken into ${\text{1 mol Ba}}^{2+}$ and $2{\text{ mol NO}}_3^{-}$ as follows:

  $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_2\rightleftharpoons {\text{Ba}}^{2+}+2{\text{NO}}_3^{-}$

Since ${\text{1 M Ba}}^{2+}$ is furnished by $\text{1 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ so molarity of ${\text{Ba}}^{2+}$ ion due to $0.00155\text{ M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Ba}}^{2+}=\left(0.00155\text{ M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2\right)\left(\frac{{\text{1 M Ba}}^{2+}}{\text{1 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)\\ =0.00155{\text{ M Ba}}^{2+}\end{array}$

Since $2{\text{ M NO}}_3^{-}$ is furnished by $\text{1 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ so molarity of ${\text{NO}}_3^{-}$ ions due to $0.00155\text{ M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of NO}}_3^{-}=\left(0.00155\text{ M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2\right)\left(\frac{2{\text{ M NO}}_3^{-}}{\text{1 M Ba}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)\\ =0.0031{\text{ M NO}}_3^{-}\end{array}$

${\text{AgNO}}_3$ can be broken into ${\text{1 mol Ag}}^{2+}$ and ${\text{1 mol NO}}_3^{-}$ as follows:

  ${\text{AgNO}}_3\rightleftharpoons {\text{Ag}}^{+}+{\text{NO}}_3^{-}$

Since ${\text{1 M Ag}}^{+}$ is furnished by ${\text{1 M AgNO}}_3$ so molarity of ${\text{Ag}}^{+}$ ion due to $0.0021{\text{ M AgNO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Ag}}^{+}=\left(0.0021{\text{ M AgNO}}_3\right)\left(\frac{1{\text{ M Ag}}^{+}}{{\text{1 M AgNO}}_3}\right)\\ =0.0021{\text{ M Ag}}^{+}\end{array}$

Since ${\text{1 M Ag}}^{+}$ is furnished by ${\text{1 M AgNO}}_3$ so molarity of ${\text{NO}}_3^{-}$ ion due to $0.0021{\text{ M AgNO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of NO}}_3^{-}=\left(0.0021{\text{ M AgNO}}_3\right)\left(\frac{1{\text{ M NO}}_3^{-}}{{\text{1 M AgNO}}_3}\right)\\ =0.0021{\text{ M NO}}_3^{-}\end{array}$

Thus total molar concentration of ${\text{NO}}_3^{-}$ is calculated as follows:

  $\begin{array}{c}\text{Total molar concentration }=0.0031\text{ M}+0.0021\text{ M}\\ =0.0052\text{ M}\end{array}$

Hence molar concentration of ${\text{Ba}}^{2+}$, ${\text{Ag}}^{+}$ and ${\text{NO}}_3^{-}$ are $0.00155\text{ M}$, $0.0021\text{ M}$ and $0.0052\text{ M}$.

(d)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after $\text{25}\text{.5 mL}$ of $\text{0}\text{.25 M NaCl}$ is mixed with $\text{15}\text{.5 mL}$ of $\text{0}\text{.15 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{0}\text{.333 M NaCl}$ indicates that in $1000\text{ mL}$ moles of $\text{NaCl}$ is $\text{0}\text{.25 mol}$. Similarly $\text{0}\text{.15 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ indicates that in $1000\text{ mL}$ the moles of $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ is $0.\text{15 mol}$.

Since in $\text{1000 mL}$ moles of $\text{NaCl}$ present is $\text{0}\text{.25 mol}$ so, moles of $\text{NaCl}$ in $\text{25}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaCl}=\left(\text{25}\text{.5 mL}\right)\left(\frac{0.25\text{ mol}}{\text{1000 mL}}\right)\\ =0.006375\text{ mol}\end{array}$

Since in $\text{1000 mL}$ moles of $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ present is $\text{0}\text{.20 mol}$ so, moles of $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ in $\text{15}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2=\left(\text{15}\text{.5 mL}\right)\left(\frac{0.15\text{ mol}}{\text{1000 mL}}\right)\\ =0.002325\text{ mol}\end{array}$

$\text{NaCl}$ can be broken into ${\text{1 mol Na}}^{+}$ and ${\text{1 mol Cl}}^{-}$ as follows:

  $\text{NaCl}\rightleftharpoons {\text{Na}}^{+}+{\text{Cl}}^{-}$

Since ${\text{1 M Na}}^{+}$ is furnished by $\text{1 M NaCl}$ so molarity of ${\text{Na}}^{+}$ ion due to $0.006375\text{ M NaCl}$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(0.006375\text{ M NaCl}\right)\left(\frac{{\text{1 M Na}}^{+}}{\text{1 M NaCl}}\right)\\ =0.006375{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 M Cl}}^{-}$ is furnished by $\text{1 M NaCl}$ so molarity of ${\text{Cl}}^{-}$ ions due to $0.006375\text{ M NaCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(0.006375\text{ M NaCl}\right)\left(\frac{{\text{1 M Cl}}^{-}}{\text{1 M NaCl}}\right)\\ =0.006375{\text{ M Cl}}^{-}\end{array}$

$\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ can be broken into ${\text{1 mol Ca}}^{2+}$ and $2{\text{ mol CH}}_{\text{3}}{\text{COO}}^{-}$ as follows:

  $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{CH}}_{\text{3}}{\text{COO}}^{-}$

Since ${\text{1 M Ca}}^{2+}$ is furnished by $\text{1 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ so molarity of ${\text{Ca}}^{2+}$ ion due to $0.002325\text{ M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Ca}}^{2+}=\left(0.002325\text{ M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2\right)\left(\frac{{\text{1 M Ca}}^{2+}}{\text{1 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2}\right)\\ =0.002325{\text{ M Ca}}^{2+}\end{array}$

Since $2{\text{ M CH}}_{\text{3}}{\text{COO}}^{-}$ is furnished by $\text{1 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ so molarity of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ions due to $0.002325\text{ M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of CH}}_{\text{3}}{\text{COO}}^{-}=\left(0.002325\text{ M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2\right)\left(\frac{2{\text{ M CH}}_{\text{3}}{\text{COO}}^{-}}{\text{1 M Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2}\right)\\ =0.00465{\text{ M CH}}_{\text{3}}{\text{COO}}^{-}\end{array}$

Hence molar concentration of ${\text{Na}}^{+}$, ${\text{Cl}}^{-}$ is $0.006375\text{ M}$ while molar concentration of ${\text{Ca}}^{2+}$ and ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ions are $0.002325\text{ M}$ and $0.00465\text{ M}$ respectively.