Chapter 2, Problem 5Q

A.

Summary Introduction

To explain: Whether there is any ${\text{H}}_3{\text{O}}^{+}$ ions present in water at neutral pH.

Introduction:

Water is an inorganic compound used as a universal solvent. ${\text{H}}_{\text{2}}\text{O}$ is the chemical formula of water. It is a tasteless and odorless liquid at room temperature. It is polar in nature and each molecule is composed of one proton $\left({\text{H}}_{\text{2}}\text{O}\right)$ and one molecule of hydroxyl ion $\left({\text{OH}}^{\text{-}}\right)$. Upon dissociation; the proton molecule binds with water molecule giving rise to a hydronium ion $\left({\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right)$

Answer to Problem 5Q

Yes, at neutral pH; $\left(\text{pH=7}\right)$ there will be hydronium ion in ${\text{H}}_3{\text{O}}^{+}$ water. But, its presence has no significant effect. This is due to the fact that at the neutral pH, the concentration of ${\text{H}}_3{\text{O}}^{+}$ and ${\text{OH}}^{\text{-}}$ are equal.

Explanation of Solution

Water is a polar inorganic solvent; which is amphoteric in nature. It means that it can act as a base and an acid at the same time. By self-ionization, it produces ${\text{H}}^{\text{+}}$ (acid) and ${\text{OH}}^{{}^{\text{-}}}$ (base) ions .The proton ${\text{H}}^{\text{+}}$ thus formed will bind with water molecule ${\text{H}}_{\text{2}}\text{O}$ results in the production of hydronium ion $\left({\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right)$. Increased ${\text{H}}_3{\text{O}}^{+}$ concentration always results in acidity and in ${\text{OH}}^{{}^{\text{-}}}$ basicity respectively.

But at neutral pH, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ will be equal and it would not show any acidic character. The dissociation and association of water molecule are mentioned below:

$\begin{array}{l}{\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}\\ {\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}\end{array}$

B.

Summary Introduction

To explain: The ratio of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions to ${\text{H}}_{\text{2}}\text{O}$ molecules in water at neutral pH.

Introduction:

Water is an inorganic compound used as a universal solvent. ${\text{H}}_{\text{2}}\text{O}$ is the chemical formula of water. It is a tasteless and odorless liquid at room temperature. It is polar in nature and each molecule is composed of one proton $\left({\text{H}}_{\text{2}}\text{O}\right)$ and one molecule of hydroxyl ion $\left({\text{OH}}^{\text{-}}\right)$. Upon dissociation; the proton molecule binds with water molecule giving rise to a hydronium ion $\left({\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right)$

Answer to Problem 5Q

The ratio of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions to ${\text{H}}_{\text{2}}\text{O}$ molecules is $\frac{{10}^{-7}}{55.6}$

Explanation of Solution

The ratio of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions to ${\text{H}}_{\text{2}}\text{O}$ molecules at neutral pH will be a very negligible value and it has no relevance

To find the ratio of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions with ${\text{H}}_{\text{2}}\text{O}$ molecules; it is needed to divide the molecular weight of water with the weight of 1Kg of water. The calculation is shown below:

The molecular weight of water $=$ $\frac{\text{18gm}}{\text{1mol}}$.

1 kg of water = 1 liter of water

1 kg = 1000g

Therefore,

$\begin{array}{c}\text{The concentration of water in 1 liter=}\frac{\text{Weight of 1 liter of water}}{\text{molecular weight of water}}\\ \text{=}\frac{\frac{\text{1000}}{\text{1gm}}}{\frac{\text{18gm}}{\text{1mol}}}\\ \text{=}\text{55}\text{.56}\text{M}\end{array}$

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in water at neutral pH is ${\text{10}}^{\text{-7}}$.

Therefore, the ratio of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions to ${\text{H}}_{\text{2}}\text{O}$ molecules will be $=\frac{{\text{10}}^{\text{-7}}}{\text{55}\text{.6}}\text{or}\text{1}\text{.8}\times {10}^{-9}$

So, only a fewer water molecules got dissociated at neutral pH