Chapter 16, Problem 16.92QP

Sodium benzoate, NaC₇H₅O₂, is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 M NaC₇H₅O₂ being titrated by 0.200 M HBr. Calculate the pH of the solution: a when no HBr has been added; b after the addition of 50.0 mL of the HBr solution; c at the equivalence point; d after the addition of 75.00 mL of the HBr solution. The K_b value for the benzoate ion is 1.6 × 10⁻¹⁰.

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ has to be calculated.

When no $\text{HBr}$ has been added

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

  $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

  $\text{pH + pOH = 14}$

Answer to Problem 16.92QP

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ are calculated as follows,

The pH when no $\text{HBr}$ has been added is 8.80

Explanation of Solution

To Calculate: The pH when no $\text{HBr}$ has been added

Given data:

Sodium benzoate $({\text{NaC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\text{)}$ is used as preservatives in foods.

The volume of sodium benzoate = 50.0 mL

The concentration of sodium benzoate = 0.250 M

The concentration of $\text{HBr}$= 0.200 M

The $K_b$ value for the benzoate ion is $1.6\times {10}^{-10}$

pH when no$\text{HBr}$ has been added

Construct an equilibrium table for the hydrolysis of benzoate ion as follows,

	${\text{ C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{ + OH}}^{-}$
Initial $(M)$	0.250 $-x$ 0.250-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

The $K_b$ value for the benzoate ion is $1.6\times {10}^{-10}$

Now substitute equilibrium concentrations into the equilibrium-constant expression.

  $\begin{array}{l}{\text{ K}}_b=\frac{[{\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{-}\text{]}}\\ =\frac{{\text{(}x\text{)}}^2}{(0.250-x)}\\ \text{Assume }x\text{ is very small than 0}\text{.250 and neglect it in the denominator}\\ 1.6\times {10}^{-10}\approx \frac{{\text{(}x\text{)}}^2}{(0.250)}\\ x=6.324\times {10}^{-6}\text{ M}\end{array}$

Here, x gives the concentration of hydroxide ion $6.324\times {10}^{-6}\text{ M}$

Finally calculate pOH and then the pH as follows,

  $\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{\text{-}}\text{]}\\ \text{=-log(}6.324\times {10}^{-6})\\ =5.198\end{array}$

The pH is calculated as follows,

  $\begin{array}{l}\text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 5}\text{.198}\\ \text{=8}\text{.80}\end{array}$

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ has to be calculated.

After the addition of 50.0 mL of $\text{HBr}$ solution

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

  $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

  $\text{pH + pOH = 14}$

Answer to Problem 16.92QP

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ are calculated as follows,

The pH after the addition of 50.0 mL of $\text{HBr}$ solution is 3.60

Explanation of Solution

To Calculate: The pH after the addition of 50.0 mL of $\text{HBr}$ solution

After the addition of 50.0 mL of $\text{HBr}$, the reaction will be as follows

  ${\text{ C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\stackrel{}{\to }{\text{ HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{ + H}}_{\text{2}}\text{O}$

At this point, the total volume is, $50.0\text{ mL + 50}\text{.0 mL = 100}\text{.0 mL}$

Now, the moles of ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}$ present at the initial and the moles of $\text{HBr}$ added are:

  $\begin{array}{l}{\text{mol C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}=\text{ M}\times \text{V}\\ \text{=0}\text{.250 M}\times \text{50}\times {\text{10}}^{-3}\text{ L}\\ \text{=0}\text{.01250 mol}\\ \text{mol HBr }=\text{ M}\times \text{V}\\ \text{=0}\text{.200 M}\times 50.0\times {\text{10}}^{-3}\text{ L}\\ \text{=0}\text{.01000 mol}\end{array}$

After the reaction, the moles of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ will be equal to the moles of $\text{HBr}$ added.

The moles of ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}$ present are:

  $\begin{array}{l}{\text{mol C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{= }0.01250\text{ mol - }0.01000\\ =2.50\times {10}^{-3}\text{ mol}\end{array}$

The concentrations are:

  $\begin{array}{l}{\text{[C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}\text{=}\frac{2.50\times {10}^{-3}{\text{ mol C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}}{0.1000\text{ L}}\\ =0.0250\text{ M}\\ \text{[HBr]}\text{=}\frac{0.01000\text{ mol HBr}}{0.1000\text{ L}}\\ =0.1000\text{ M}\end{array}$

Solve for ${\text{[OH}}^{\text{-}}\text{]}$ by substituting the concentrations into the ${\text{K}}_{\text{b}}$ expression

  $\begin{array}{l}{\text{ K}}_{\text{b}}=\frac{[{\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{-}\text{]}}\\ \text{ 1}\text{.6}\times {10}^{-10}=\frac{(0.100+x)(x)}{(0.0250-x)}\\ \approx \frac{(0.1000)(x)}{(0.0250)}\\ x=\frac{(0.0250)(1.6\times {10}^{-10})}{(0.1000)}\\ =4.00\times {10}^{-11}\text{ M}\end{array}$

The pH is calculated from pOH as follows,

  $\begin{array}{l}\text{pOH}\text{=-log(}4.00\times {10}^{-11})\\ =10.398\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 10}\text{.398}\\ \text{=3}\text{.60}\end{array}$

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ has to be calculated.

At the equivalence point

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

  $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

  $\text{pH + pOH = 14}$

Answer to Problem 16.92QP

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ are calculated as follows,

The pH at the equivalence point is 2.58

Explanation of Solution

To Calculate: The pH at the equivalence point

Calculate the volume of $\text{HBr}$ added to reach the equivalence point

The volume of $\text{HBr}$ added is calculated as follows,

  $\begin{array}{l}\text{Volume of HBr}\text{= }\frac{{\text{M}}_{base}{\text{ V}}_{base}}{{\text{M}}_{acid}}\\ =\frac{(0.250\text{ M)(50}\text{.0 mL)}}{0.200\text{ M}}\\ =\text{ 62}\text{.50 mL}\end{array}$

Hence, the total volume is as follows,

  $\begin{array}{l}\text{Total volume}\text{= 50}\text{.0 mL + 62}\text{.50 mL}\\ =\text{ 112}\text{.5 mL}\\ \text{= 0}\text{.1125 L}\end{array}$

The equilibrium reaction is,

  ${\text{ HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{+}$

  $\begin{array}{l}{\text{[HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\text{]}\text{=}\frac{\text{0}\text{.01250 mol}}{\text{0}\text{.1125 L}}\\ =0.1111\text{ M}\end{array}$

The value of ${\text{K}}_{\text{a}}$ is calculated from ${\text{K}}_{\text{b}}$ as follows,

  $\begin{array}{l}{\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\\ =\frac{\text{1}\text{.00}\times {\text{10}}^{\text{-14}}}{1.6\times {\text{10}}^{\text{-10}}}\\ =6.25\times {10}^{-5}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{a}}=\frac{[{\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[{\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\text{]}}\\ 6.25\times {10}^{-5}=\frac{x^2}{0.1111-x}\\ x=2.635\times {10}^{-3}\text{ M}\end{array}$

Here, x gives the hydronium ion concentration.

  $\begin{array}{l}\text{pH}{\text{=-log[H}}^{\text{+}}\text{]}\\ \text{=-log(2}\text{.635}\times {\text{10}}^{\text{-6}})\\ =2.579\end{array}$

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ has to be calculated.

After the addition of 75.00 mL of $\text{HBr}$ solution

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

  $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

  $\text{pH + pOH = 14}$

Answer to Problem 16.92QP

The pH of the given points of the titration of sodium benzoate with $\text{HBr}$ are calculated as follows,

The pH after the addition of 75.00 mL of $\text{HBr}$ solution is 1.70

Explanation of Solution

To Calculate: The pH after the addition of 75.00 mL of $\text{HBr}$ solution

Calculate the moles of acid added.

  $\begin{array}{l}\text{mol HBr added}\text{= M}\times \text{V}\\ =(0.200\text{ M})\times (0.075\text{ L)}\\ \text{= }0.01500\text{ mol}\end{array}$

  $\begin{array}{l}\text{Moles of acid remaining }\text{= }0.01500\text{ mol - }0.01250\text{ mol}\\ \text{= 2}\text{.50}\times {\text{10}}^{\text{-3}}\text{ mol}\end{array}$

The total volume after the addition of 75.0 mL of $\text{HBr}$ is:

  $\text{50}\text{.0 mL + 75}\text{.0 mL = 125}\text{.0 mL}$

The hydronium ion concentration and the pH are:

  $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\frac{2.50\times {\text{10}}^{\text{-3}}\text{ mol}}{\text{0}\text{.1250 L}}\\ =0.0200\text{ M}\end{array}$

The pH is calculated as follows,

  $\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}^{+}]\\ =-\log (0.0200)\\ =1.699\\ =1.70\end{array}$