Chapter 16, Problem 16.30QE

(a)

Interpretation Introduction

Interpretation:

pH of 0.20 M acetic acid and 0.50 M sodium acetate solution has to be calculated.

Concept introduction:

The pH of solution is calculated by using Henderson – Hasselbalch equation,

  $\text{pH = pKa + log }\left(\frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}\right)$

  ${\text{pK}}_{\text{a}}\text{ = -log}\left[{\text{K}}_{\text{a}}\right]$

A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have $\text{pH < 7}$, basic have $\text{pH > 7}$ and netural have $\text{pH=7}$.

(b)

Interpretation Introduction

Interpretation:

pH of 0.25 M hydrofluoric acid and 0.10 M potassium fluoride solution has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Interpretation Introduction

Interpretation:

pH of 0.0250 mol sodium nitrite in 250.0 mL of 0.0410 M nitrous acid has to be calculated.

Concept introduction:

Refer to part (a)