Chapter 16, Problem 20A

Interpretation Introduction

Interpretation: Ion product constant of water and its significance has to be explained with the chemical reaction that can be used to derive the ionization constant.

Concept introduction: A very simple definition of an acid is that any substance which furnishes ${\text{H}}^{\text{+}}$ ions when added to ${\text{H}}_2\text{O}$ can be put into a class of acid. On the counterpart, if any substance accepts ${\text{H}}^{\text{+}}$ ions when added to ${\text{H}}_2\text{O}$ , can be classified as a base.

Conjugate acid-base pair contains 2 components, one that accepts ${\text{H}}^{\text{+}}$ and other which donates shown as follows: $\text{HA}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

Where,

• $\text{HA}$ and ${\text{A}}^{-}$ is acid-conjugate base pair.
• ${\text{H}}_2\text{O}$ and ${\text{H}}_3{\text{O}}^{+}$ is base- conjugate pair.

Answer to Problem 20A

  $K_w$ is ion product of water. At ${25}^{°}\text{C}$ $K_w$ that is a product of $K_{\text{eq}}$ and $\left[{\text{H}}_2\text{O}\right]$ is ${10}^{-14}$ . To estimate the balance of acid and base component in any solution, the ion product of water is significant. The equation used to derive the ion product of water:

  ${\text{H}}_2\text{O}+{\text{H}}_2\text{O}\leftrightarrow {\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$ .

Explanation of Solution

To estimate the balance of acid and base component in any solution, the ion product of water is significant. Since water is amphoteric, one water molecule donates ${\text{H}}^{+}$

to another water molecule and vice versa. This leads to the following reaction:

  ${\text{H}}_2\text{O}+{\text{H}}_2\text{O}\leftrightarrow {\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$

The arrow indicates that ions again react together to form water molecules.

So equilibrium reaction for the reaction is:

  $K_{\text{eq}}\text{=}\frac{\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2\text{O}\right]}$

The extent to which water ionizes is very low, so the concentration of water remains the same. So the above equation reduces to $K_w\text{=}\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$ where $K_w$ is a product of $K_{\text{eq}}$ and $\left[{\text{H}}_2\text{O}\right]$ and known as ionic product. At ${25}^{°}\text{C}$ $K_w$ is ${10}^{-14}$ .

Conclusion

  $K_w$ is an ion product of water. At ${25}^{°}\text{C}$ $K_w$ that is a product of $K_{\text{eq}}$ and $\left[{\text{H}}_2\text{O}\right]$ is ${10}^{-14}$ . To estimate the balance of acid and base component in any solution, the ion product of water is significant. The equation used to derive the ion product of water:

  ${\text{H}}_2\text{O}+{\text{H}}_2\text{O}\leftrightarrow {\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$ .