Chapter 9, Problem 9.38P

a)

Interpretation Introduction

Interpretation:

The quotient $\frac{\left[{\text{HNO}}_{\text{2}}\right]}{\left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]}$ in solution of Sodium nitrate at given $\text{pH}$ has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For acid:

Consider an acidic reaction:

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

The Henderson- Hasselbalch equation for an acidic reaction can be given as,

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

To calculate the quotient $\frac{\left[{\text{HNO}}_{\text{2}}\right]}{\left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]}$ in solution of Sodium nitrate at 2.00 $\text{pH}$

b)

Interpretation Introduction

Interpretation:

The quotient $\frac{\left[{\text{HNO}}_{\text{2}}\right]}{\left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]}$ in solution of Sodium nitrate at given $\text{pH}$ has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For acid:

Consider an acidic reaction:

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

The Henderson- Hasselbalch equation for an acidic reaction can be given as,

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

To calculate the quotient $\frac{\left[{\text{HNO}}_{\text{2}}\right]}{\left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]}$ in solution of Sodium nitrate at 10.00 $\text{pH}$