Chapter 15, Problem 23PE

(a)

Interpretation Introduction

Interpretation:

Molarity of each ion found after $\text{55}\text{.5 mL}$ of $\text{0}\text{.50 M HCl}$ is mixed with $\text{75}\text{.0 mL}$ of $\text{1}\text{.25 M HCl}$ 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 molarity is given as follows:

  $M=\frac{n}{V}$

Here,

$V$ represents volume.

$M$ represents molarity.

$n$ represents number of moles.

Explanation of Solution

Since in $\text{1000 mL}$ the moles of $\text{HCl}$ present is $\text{0}\text{.50 mol}$ so, moles of $\text{HCl}$ in $\text{55}\text{.5 mL}$

is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{55}\text{.5 mL}\right)\left(\frac{\text{0}\text{.50 mol}}{\text{1000 mL}}\right)\\ =0.02775\text{ mol}\end{array}$

Similarly, moles of $\text{HCl}$ in $\text{75}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{75}\text{.0 mL}\right)\left(\frac{\text{1}\text{.25 mol}}{\text{1000 mL}}\right)\\ =0.09375\text{ mol}\end{array}$

Thus total moles of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}\text{Total moles }=0.02775\text{ mol}+0.09375\text{ mol}\\ =\text{0}\text{.1215 mol}\end{array}$

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

  $\begin{array}{c}\text{Total volume }=55.5\text{ mL}+75.0\text{ mL}\\ \text{=130}\text{.5 mL}\end{array}$

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$        (1)

Substitute $\text{0}\text{.1215 mol}$ for $n$ and $\text{130}\text{.5 mL}$ for $V$ in equation (1)

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.1215 mol}}{\text{130}\text{.5 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.93103\text{ M}\end{array}$

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

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

Since ${\text{1 M H}}^{+}$ is furnished by $\text{1 M HCl}$ so molarity of ${\text{H}}^{+}$ ion due to $0.93103\text{ M HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of H}}^{+}=\left(0.93103\text{ M HCl}\right)\left(\frac{{\text{1 M H}}^{+}}{\text{1 M HCl}}\right)\\ =0.93103{\text{ M H}}^{+}\end{array}$

Since ${\text{1 M Cl}}^{-}$ is furnished by $\text{1 M HCl}$ so molarity of ${\text{Cl}}^{-}$ ions due to $0.93103\text{ M HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(0.93103\text{ M HCl}\right)\left(\frac{{\text{1 M Cl}}^{-}}{\text{1 M HCl}}\right)\\ =0.93103{\text{ M Cl}}^{-}\end{array}$

Hence, concentration of ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ is $0.93103\text{ M}$ and $0.93103\text{ M}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Molarity of each ion found after $\text{125 mL}$ of $\text{0}{\text{.75 M CaCl}}_2$ is mixed with $\text{125 mL}$ of $\text{0}{\text{.25 M CaCl}}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ the moles of ${\text{CaCl}}_2$ present is $\text{0}\text{.75 mol}$ so, moles of ${\text{CaCl}}_2$ in $\text{125 mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of CaCl}}_2=\left(\text{125 mL}\right)\left(\frac{\text{0}\text{.75 mol}}{\text{1000 mL}}\right)\\ =0.09375\text{ mol}\end{array}$

Similarly, for $\text{0}{\text{.25 M CaCl}}_2$ solution moles of ${\text{CaCl}}_2$ in  $\text{125 mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of CaCl}}_2=\left(\text{125 mL}\right)\left(\frac{\text{0}\text{.25 mol}}{\text{1000 mL}}\right)\\ =0.03215\text{ mol}\end{array}$

Thus total moles of ${\text{CaCl}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Total moles }=0.09375\text{ mol}+0.03215\text{ mol}\\ =\text{0}\text{.125 mol}\end{array}$

Total volume of solution upon when two ${\text{CaCl}}_2$ solutions are mixed is calculated as follows:

  $\begin{array}{c}\text{Total volume}=\text{125 mL}+\text{125 mL}\\ =\text{250 mL}\end{array}$

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$        (1)

Substitute $\text{0}\text{.125 mol}$ for $n$ and $\text{200 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{\text{0}\text{.125 mol}}{\text{250 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.5\text{ M}\end{array}$

${\text{CaCl}}_2$ can be broken into ${\text{1 mol Ca}}^{2+}$ and ${\text{2 mol Cl}}^{-}$ as follows:

  ${\text{CaCl}}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{Cl}}^{-}$

Since $1{\text{ M Ca}}^{2+}$ is furnished by ${\text{1 M CaCl}}_2$ so molarity of ${\text{Ca}}^{2+}$ ion due to $0.5{\text{ M CaCl}}_2$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Ca}}^{2+}=\left(0.5{\text{ M CaCl}}_2\right)\left(\frac{1{\text{ M Ca}}^{2+}}{{\text{1 M CaCl}}_2}\right)\\ =0.5{\text{ M Ca}}^{2+}\end{array}$

Since ${\text{2 M Cl}}^{-}$ is furnished by ${\text{1 M CaCl}}_2$ so molarity of ${\text{Cl}}^{-}$ ions due to $0.5{\text{ M CaCl}}_2$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(0.5{\text{ M CaCl}}_2\right)\left(\frac{{\text{2 M Cl}}^{-}}{{\text{1 M CaCl}}_2}\right)\\ =1{\text{ M Cl}}^{-}\end{array}$

Hence, concentration of ${\text{Ca}}^{2+}$ and ${\text{Cl}}^{-}$ is $\text{0}\text{.5 M}$ and $\text{1 M}$ respectively.

(c)

Interpretation Introduction

Interpretation:

Molarity of each ion found after $\text{35}\text{.0 mL}$ of $\text{0}\text{.333 M NaOH}$ is mixed with $\text{22}\text{.5 mL}$ of $\text{0}\text{.250 M HCl}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ the moles of $\text{NaOH}$ present is $\text{0}\text{.333 mol}$ so, moles of $\text{NaOH}$ in $\text{35}\text{.0 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{35}\text{.0 mL}\right)\left(\frac{\text{0}\text{.333 mol}}{\text{1000 mL}}\right)\\ =0.011655\text{ mol}\end{array}$

Similarly, for $\text{0}\text{.250 M HCl}$ solution moles of $\text{HCl}$ in $\text{22}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{22}\text{.5 mL}\right)\left(\frac{\text{0}\text{.25 mol}}{\text{1000 mL}}\right)\\ =0.005625\text{ mol}\end{array}$

Total volume of solution upon when two solutions are mixed is calculated as follows:

  $\begin{array}{c}\text{Total volume }=\text{35}\text{.0 mL}+\text{22}\text{.5 mL}\\ =57.5\text{ mL}\end{array}$

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$        (1)

Substitute $0.011655\text{ mol}$ for $n$ and $57.5\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.011655\text{ mol}}{57.5\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.20269\text{ M}\end{array}$

Substitute $0.005625\text{ mol}$ for $n$ and $57.5\text{ mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.005625\text{ mol}}{57.5\text{ mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.09782\text{ M}\end{array}$

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

  $\text{NaOH}\rightleftharpoons {\text{Na}}^{+}+{\text{OH}}^{-}$

Since $1{\text{ M Na}}^{+}$ is furnished by $\text{1 M NaOH}$ so molarity of ${\text{Na}}^{+}$ ion due to $0.20269\text{ M}\text{ NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(0.20269\text{ M}\text{ NaOH}\right)\left(\frac{1{\text{ M Na}}^{+}}{\text{1 M}\text{ NaOH}}\right)\\ =0.20269{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 mol OH}}^{-}$ is furnished by $\text{1 M NaOH}$ so molarity of ${\text{OH}}^{-}$ ions due to $0.20269\text{ M}\text{ NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of OH}}^{-}=\left(0.20269\text{ M}\text{ NaOH}\right)\left(\frac{1{\text{ M OH}}^{-}}{\text{1 M}\text{ NaOH}}\right)\\ =0.20269{\text{ M OH}}^{-}\end{array}$

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

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

Since $1{\text{ M H}}^{+}$ is furnished by $\text{1 M HCl}$ so molarity of ${\text{H}}^{+}$ ion due to $0.005625\text{ M HCl}$   is calculated as follows:

  $\begin{array}{c}{\text{Molarity of H}}^{+}=\left(0.005625\text{ M HCl}\right)\left(\frac{1{\text{ M H}}^{+}}{\text{1 M HCl}}\right)\\ =0.005625{\text{ M H}}^{+}\end{array}$

Since ${\text{1 M Cl}}^{-}$ is furnished by $\text{1 M NaCl}$ so molarity of ${\text{Cl}}^{-}$ ions due to $0.005625\text{ M HCl}$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(0.005625\text{ M HCl}\right)\left(\frac{1{\text{ M Cl}}^{-}}{\text{1 M HCl}}\right)\\ =0.005625{\text{ M Cl}}^{-}\end{array}$

Hence, concentration of ${\text{Na}}^{+}$, ${\text{OH}}^{-}$, ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ are $0.20269\text{ M}$, $0.20269\text{ M}$, $0.005625\text{ M}$ and $0.005625\text{ M}$ respectively.

(d)

Interpretation Introduction

Interpretation:

Molarity of each ion found after $\text{12}\text{.5 mL}$ of $\text{0}{\text{.500 M H}}_2{\text{SO}}_{\text{4}}$ is mixed with $\text{23}\text{.5 mL}$ of $\text{0}\text{.175 M NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since in $\text{1000 mL}$ moles of $\text{NaOH}$ present is $\text{0}\text{.175 mol}$ so, moles of $\text{NaOH}$ in $\text{23}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{23}\text{.5 mL}\right)\left(\frac{\text{0}\text{.175 mol}}{\text{1000 mL}}\right)\\ =0.004112\text{ mol}\end{array}$

Similarly, for $\text{0}{\text{.500 M H}}_2{\text{SO}}_{\text{4}}$ solution moles of ${\text{H}}_2{\text{SO}}_{\text{4}}$ in $\text{12}\text{.5 mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_{\text{4}}=\left(\text{12}\text{.5 mL}\right)\left(\frac{\text{0}\text{.500 mol}}{\text{1000 mL}}\right)\\ =0.00625\text{ mol}\end{array}$

Total volume of solution upon when two solutions are mixed is calculated as follows:

  $\begin{array}{c}\text{Total volume }=\text{23}\text{.5 mL}+\text{12}\text{.5 mL}\\ =36\text{ mL}\end{array}$

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$        (1)

Substitute $0.004112\text{ mol}$ for $n$ and $\text{36 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.004112\text{ mol}}{\text{36 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.11422\text{ M}\end{array}$

Substitute $0.00625\text{ mol}$ for $n$ and $\text{36 mL}$ for $V$ in equation (1).

  $\begin{array}{c}M=\left(\frac{0.00625\text{ mol}}{\text{36 mL}}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =0.1736\text{ M}\end{array}$

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

  $\text{NaOH}\rightleftharpoons {\text{Na}}^{+}+{\text{OH}}^{-}$

Since $1{\text{ M Na}}^{+}$ is furnished by $\text{1 M NaOH}$ so molarity of ${\text{Na}}^{+}$ ion due to $0.11422\text{ M}\text{ NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(0.11422\text{ M}\text{ NaOH}\right)\left(\frac{1{\text{ M Na}}^{+}}{\text{1 M}\text{ NaOH}}\right)\\ =0.11422{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 mol OH}}^{-}$ is furnished by $\text{1 M NaOH}$ so molarity of ${\text{OH}}^{-}$ ions due to $0.11422\text{ M}\text{ NaOH}$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of OH}}^{-}=\left(0.11422\text{ M}\text{ NaOH}\right)\left(\frac{1{\text{ M OH}}^{-}}{\text{1 M}\text{ NaOH}}\right)\\ =0.11422{\text{ M OH}}^{-}\end{array}$

${\text{H}}_2{\text{SO}}_{\text{4}}$ can be broken into ${\text{2 mol H}}^{+}$ and ${\text{1 mol SO}}_4^{2-}$ as follows:

  ${\text{H}}_2{\text{SO}}_{\text{4}}\rightleftharpoons 2{\text{H}}^{+}+{\text{SO}}_4^{2-}$

Since ${\text{2 M H}}^{+}$ is furnished by ${\text{1 M H}}_2{\text{SO}}_{\text{4}}$ so molarity of ${\text{H}}^{+}$ ion due to $0.1736{\text{ M H}}_2{\text{SO}}_{\text{4}}$   is calculated as follows:

  $\begin{array}{c}{\text{Molarity of H}}^{+}=\left(0.1736{\text{ M H}}_2{\text{SO}}_{\text{4}}\right)\left(\frac{{\text{2 M H}}^{+}}{{\text{1 M Na}}_2{\text{SO}}_{\text{4}}}\right)\\ =0.3472{\text{ M H}}^{+}\end{array}$

Since ${\text{1 M SO}}_4^{2-}$ is furnished by ${\text{1 M H}}_2{\text{SO}}_{\text{4}}$ so molarity of ${\text{SO}}_4^{2-}$ ions due to $0.1736{\text{ M H}}_2{\text{SO}}_{\text{4}}$  is calculated as follows:

  $\begin{array}{c}{\text{Molarity of SO}}_4^{2-}=\left(0.1736{\text{ M H}}_2{\text{SO}}_{\text{4}}\right)\left(\frac{{\text{1 M SO}}_4^{2-}}{{\text{1 M H}}_2{\text{SO}}_{\text{4}}}\right)\\ =0.1736{\text{ M SO}}_4^{2-}\end{array}$

Hence, concentration of ${\text{Na}}^{+}$, ${\text{OH}}^{-}$, ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ are $0.11422\text{ M}$, $0.11422\text{ M}$, $0.3472\text{ M}$ and $0.1736\text{ M}$ respectively.