Chapter 19.2, Problem 19.4BFP

(a)

Interpretation Introduction

Interpretation:

For the titration of $30.00mL$ and 0.1000 M benzoic acid, ($K_a\text{=6}\text{.3}\times {10}^{-5}$ ) with $\text{0}\text{.1500 M NaOH}$ solution before the addition of $\text{pH}$ has to be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

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

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

(b)

Interpretation Introduction

Interpretation:

For the titration of $30.00mL$ and 0.1000 M benzoic acid, ($K_a\text{=6}\text{.3}\times {10}^{-5}$ ) with $\text{0}\text{.1500 M NaOH}$ solution after the addition $12.00mL$ of base and $\text{pH}$ has to be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Henderson-Hasselbalch equation

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

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

(c)

Interpretation Introduction

Interpretation:

For the titration of $30.00mL$ and 0.1000 M benzoic acid, ($K_a\text{=6}\text{.3}\times {10}^{-5}$ ) with $\text{0}\text{.1500 M NaOH}$ solution the equivalent point should be identified.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

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

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $30.00mL$ and 0.1000 M benzoic acid, ($K_a\text{=6}\text{.3}\times {10}^{-5}$ ) with $\text{0}\text{.1500 M NaOH}$ solution after the addition of $\text{22}\text{.00 mL}$ base concentration has to be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

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

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.