Chapter 9.5, Problem 9TY

Interpretation Introduction

Interpretation:

The $\text{pH}$ if ${\text{pK}}_{\text{a}}\text{=3}\text{.00}$ has to be calculated.

Concept Introduction:

Consider a weak-acid equilibrium reaction,

$\text{HA}\stackrel{{\text{K}}_{\text{a}}}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

The acid dissociation constant ${\text{K}}_{\text{a}}$ can be given as,

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

A weak acid is one that doesn’t undergo completion.

The negative logarithm of an equilibrium constant is called as $\text{pK}$ .

The ${\text{pK}}_{\text{a}}$ of weak acid is the negative logarithm of ${\text{K}}_{\text{a}}$ and can be given as,

${\text{pK}}_{\text{a}}\text{=-log}{\text{K}}_{\text{a}}$

The value of acid dissociation constant ${\text{K}}_{\text{a}}$ decreases with increase in ${\text{pK}}_{\text{a}}$.

The value of ${\text{K}}_{\text{w}}$ is calculated by the formula,

${\text{K}}_{\text{w}}{\text{=K}}_{\text{a}}{\text{×K}}_{\text{b}}$

The $\text{pH}$ of solution can be calculated as,

$\begin{array}{l}{\text{pH=-logA}}_{{\text{H}}^{\text{+}}}\\ \\ \text{pH=-log}{\left[{\text{H}}^{\text{+}}\right]}_{{\text{γH}}^{\text{+}}}\end{array}$