Chapter 15, Problem 15.39QE

Interpretation Introduction

Interpretation:

The following table has to be filled. Also, whether the solution is acidic basic or neutral has to be identified.

  $\begin{array}{ccccc}\text{S}\text{.NO}& \text{pH}& \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]M& \text{pOH}& \left[{\text{OH}}^{-}\right]M\\ \left(\text{a}\right)& -1.04& & & \\ \left(\text{b}\right)& & & 0.34& \\ \left(\text{c}\right)& & 1.98\times {10}^{-7}& & \\ \left(\text{d}\right)& & & & 4.42\times {10}^{-2}\end{array}$

Concept Introduction:

Water is a compound that can act as an acid or a base. A water molecule can donate a proton as well as it accepts a proton. Water molecules can react with another molecule of water to produce hydronium ion and hydroxide ion. This reaction is called as autoionization of water. In this reaction, water can be an acid or base.

The reaction of autoionization of water is as follows:

  ${\text{H}}_{\text{2}}\text{O}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{OH}}^{-}$

The equilibrium constant for the autoionization of water is denoted by $K\text{'}$.

The expression for $K\text{'}$ is as follows:

    $K\text{'}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{{\left[{\text{H}}_{\text{2}}\text{O}\right]}^2}$        (1)

Also, water is pure liquid. Its concentration does not appear. Therefore, the equilibrium constant $K\text{'}$ will become $K_{\text{w}}$ for pure water.

The expression for $K_{\text{w}}$ is as follows:

    $K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (2)

Also at $25°\text{C}$, $K_{\text{w}}$ is equal to $1.0\times {10}^{-14}$. The equation (2) will be modified as follows:

  $1.0\times {10}^{-14}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (3)

The negative logarithm of the molar concentration of hydronium ion is called $\text{pH}$. The expression for pH is as follows:

    $\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$        (4)

Also, the relation between pH and concentration is as follows:

    $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\text{10}}^{-\text{pH}}$        (5)

The expression for $\text{pOH}$ is as follows:

  $\text{pOH}=-{\text{log}}_{\text{10}}\left[{\text{OH}}^{-}\right]$        (6)

Also, the relation between $\text{pOH}$ and concentration is as follows:

    $\left[{\text{OH}}^{-}\right]={\text{10}}^{-\text{pOH}}$        (7)

The relation between $\text{pH}$ and $\text{pOH}$ at $25°\text{C}$ is as follows:

  $\text{pH}+\text{pOH}=14$        (8)

For $\text{pH}$ equation (8) will be modified as follows:

  $\text{pH}=14-\text{pOH}$        (9)

For $\text{pOH}$ equation (8) will be modified as follows:

  $\text{pOH}=14-\text{pH}$        (10)