Chapter 9, Problem 9.9P

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from oxoacetic acid and potassium oxoacetate

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged form of equilibrium constant, $K_{\text{a}}$. It relates $\text{pH}$ of buffer solution and $\text{p}{K}_{\text{a}}$ value. It also helps to find equilibrium $\text{pH}$ in acid-base reactions. Consider an equation of dissociation of an acid as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{pH}$ is as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)$

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant of weak acid $\text{HA}$.

$\left[{\text{A}}^{-}\right]$ is concentration of conjugate base ${\text{A}}^{-}$

$\left[\text{HA}\right]$ is concentration of acid $\text{HA}$.

Concentration of a solution is expressed by molality and is denoted by $m$. It is defined as the ratio of number of moles of solute to mass of solvent in $1\text{ kg}$ or $1000\text{ g}$.

The formula of number of moles of solute is as follows:

  $\text{Number of moles of solute}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Molality is expressed as follows:

  $m=\left(\frac{\text{mass of solute}}{\text{molar mass of solute}}\right)\left(\frac{1000\text{ g}}{\text{mass of solvent}\left(\text{g}\right)}\right)$