Chapter 6, Problem 52Q

(a)

Interpretation Introduction

Interpretation:

Molar concentration of all the molecular and ionic species present in $\text{1}\text{.0}\text{M}$ solution of $\text{NaOH}$ has to be calculated.

Explanation of Solution

Sodium hydroxide is completely soluble in water and hence it dissociates completely into ions.  This can be represented as,

    $\text{NaOH}\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{Na}}^{\text{+}}+{\text{OH}}^{-}$

Concentration of $\text{NaOH}$ is $\text{1}\text{.0}\text{M}$ and therefore, $\text{1}\text{.0}\text{M}$ of ${\text{Na}}^{\text{+}}$ ions and $\text{1}\text{.0}\text{M}$ of ${\text{OH}}^{-}$ ions will be present after dissociation.  Equilibrium concentration of the ions in solution is given as,

    $\begin{array}{l}{\text{[Na}}^{\text{+}}\text{]}\text{=}{\text{[OH}}^{\text{-}}\text{]}\\ \text{=}\text{1}\text{.0}\text{M}\end{array}$

Relation between the concentration of hydroxide ion and hydrogen ion in aqueous solution can be used to calculated the concentration of hydrogen ions,

    $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{x}{\text{10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]}=\frac{\text{1}\text{x}{\text{10}}^{\text{-14}}}{\text{1}\text{.0}}\\ =\text{1}\text{x}{\text{10}}^{\text{-14}}\end{array}$

The concentration of all the ionic species is calculated as,

    $\begin{array}{l}{\text{[Na}}^{\text{+}}\text{]}\text{=}\text{1}\text{.0}\text{M}\\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.0}\text{M}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}\text{1}\text{x}{\text{10}}^{\text{-14}}\text{M}\end{array}$

Molecular species that is involved is water molecule.  The molar concentration of water can be calculated by assuming the density of solution to be $\text{1}\text{.0}\text{g/mL}$.

Ratio of number of moles of solute that is present in one liter of solution is known as molarity.

    $\text{Molarity}\text{(M)}\text{=}\frac{\text{moles of solute}}{\text{liter of solution}}$

The moles of water that is present in one liter of solution can be calculated using density and molar mass as given below,

    $\begin{array}{l}\text{[H2O]}\text{=}\frac{\left(\text{1}\text{.0}\text{L}\text{x}\frac{{\text{10}}^{\text{3}}\text{mL}}{\text{1}\text{L}}\right)\left(\frac{\text{1}\text{.0}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1 mL solution}}\text{x}\frac{\text{1}{\text{mol H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\right)}{\text{1}\text{.0}\text{L}}\\ \text{=}\text{0}\text{.555}\text{x}{\text{10}}^{\text{3}}\text{M}\\ \text{=}\text{56}\text{M}\end{array}$

Therefore, the concentration of molecular species is calculated as $\text{56}\text{M}$.

Conclusion

Concentration of all species in $\text{1}\text{.0}\text{M}$ solution of $\text{NaOH}$ is calculated.

(b)

Interpretation Introduction

Interpretation:

Molar concentration of all the molecular and ionic species present in $\text{1}\text{.0}\text{M}$ solution of $\text{HCl}$ has to be calculated.

Explanation of Solution

Hydrogen chloride is completely soluble in water and hence it dissociates completely into ions.  This can be represented as,

    $\text{HCl}\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{H}}^{\text{+}}+{\text{Cl}}^{-}$

Concentration of $\text{HCl}$ is $\text{1}\text{.0}\text{M}$ and therefore, $\text{1}\text{.0}\text{M}$ of ${\text{H}}^{\text{+}}$ ions and $\text{1}\text{.0}\text{M}$ of ${\text{Cl}}^{-}$ ions will be present after dissociation.  Equilibrium concentration of the ions in solution is given as,

    $\begin{array}{l}{\text{[H}}^{\text{+}}\text{]}\text{=}{\text{[Cl}}^{\text{-}}\text{]}\\ \text{=}\text{1}\text{.0}\text{M}\end{array}$

Relation between the concentration of hydroxide ion and hydrogen ion in aqueous solution can be used to calculated the concentration of hydrogen ions,

    $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{x}{\text{10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]}=\frac{\text{1}\text{x}{\text{10}}^{\text{-14}}}{\text{1}\text{.0}}\text{M}\\ =\text{1}\text{x}{\text{10}}^{\text{-14}}\text{M}\end{array}$

The concentration of all the ionic species is calculated as,

    $\begin{array}{l}{\text{[Cl}}^{\text{-}}\text{]}\text{=}\text{1}\text{.0}\text{M}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}\text{1}\text{.0}\text{M}\\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{x}{\text{10}}^{\text{-14}}\text{M}\end{array}$

Molecular species that is involved is water molecule.  The molar concentration of water can be calculated by assuming the density of solution to be $\text{1}\text{.0}\text{g/mL}$.

Ratio of number of moles of solute that is present in one liter of solution is known as molarity.

    $\text{Molarity}\text{(M)}\text{=}\frac{\text{moles of solute}}{\text{liter of solution}}$

The moles of water that is present in one liter of solution can be calculated using density and molar mass as given below,

    $\begin{array}{l}\text{[H2O]}\text{=}\frac{\left(\text{1}\text{.0}\text{L}\text{x}\frac{{\text{10}}^{\text{3}}\text{mL}}{\text{1}\text{L}}\right)\left(\frac{\text{1}\text{.0}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1 mL solution}}\text{x}\frac{\text{1}{\text{mol H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\right)}{\text{1}\text{.0}\text{L}}\\ \text{=}\text{0}\text{.555}\text{x}{\text{10}}^{\text{3}}\text{M}\\ \text{=}\text{56}\text{M}\end{array}$

Therefore, the concentration of molecular species is calculated as $\text{56}\text{M}$.

Conclusion

Concentration of all species in $\text{1}\text{.0}\text{M}$ solution of $\text{HCl}$ is calculated.