Chapter 16, Problem 16.144QP

(a)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

Prior to the addition of any $\text{HCl}$

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 16.144QP

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

The pH prior to the addition of any $\text{HCl}$ is 10.32

Explanation of Solution

To Calculate: The pH prior to the addition of any $\text{HCl}$

Given data:

A 0.150 M solution of $\text{NaClO}$ is prepared by dissolving $\text{NaClO}$ in water and 50.0 mL of this sample is titrated against $\text{HCl}$

The volume of $\text{NaClO}$ = 50.0 mL

The concentration of $\text{NaClO}$ = 0.150 M

The concentration of $\text{HCl}$= 0.100 M

The $K_a$ value for $\text{HClO}$ is $3.5\times {10}^{-8}$

The pH prior to the addition of any $\text{HCl}$

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

	${\text{ ClO}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ HClO + OH}}^{-}$
Initial $(M)$	0.150 $-x$ 0.150-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

The $K_a$ value for $\text{HClO}$ is $3.5\times {10}^{-8}$

The $K_b$ value is calculated using $K_a$ as follows,

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_a}\\ =\frac{1.00\times {10}^{-14}}{3.5\times {10}^{-8}}\\ =2.85\times {10}^{-7}\end{array}$

Now substitute equilibrium concentrations into the equilibrium-constant expression.

$\begin{array}{l}{\text{ K}}_b=\frac{[{\text{HClO][OH}}^{\text{-}}\text{]}}{{\text{[ClO}}^{-}\text{]}}\\ =\frac{{\text{(}x\text{)}}^2}{(0.150-x)}\\ \text{Assume 0}\text{.150 is very small than }x\text{ and neglect it in the denominator}\\ \text{ 2}\text{.85}\times {10}^{-7}\approx \frac{{\text{(}x\text{)}}^2}{(0.150)}\\ x=2.070\times {10}^{-4}\text{ M}\end{array}$

Here, x gives the concentration of hydroxide ion $2.070\times {10}^{-4}\text{ M}$

Finally, calculate pOH and then the pH as follows,

$\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{\text{-}}\text{]}\\ \text{=-log(}2.070\times {10}^{-4})\\ =3.683\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 3}\text{.683}\\ \text{=10}\text{.32}\end{array}$

Conclusion

The pH prior to the addition of any $\text{HCl}$ was calculated as 10.32

(b)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

Halfway to 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 16.144QP

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

The pH halfway to the equivalence point is 7.45

Explanation of Solution

To Calculate: The pH halfway to the equivalence point

At the half-way point, ${\text{[ClO}}^{-}]\text{ = [HClO]}$

This gives,

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{HClO][OH}}^{\text{-}}\text{]}}{{\text{[ClO}}^{-}\text{]}}\\ 2.85\times {10}^{-7}=\frac{[{\text{HClO][OH}}^{\text{-}}\text{]}}{{\text{[ClO}}^{-}\text{]}}\\ {\text{[OH}}^{\text{-}}\text{]}=2.85\times {10}^{-7}\text{}\text{M}\end{array}$

The pOH is calculated from the obtained hydroxide ion concentration as follows,

$\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{\text{-}}\text{]}\\ \text{=-log(}2.85\times {10}^{-7})\\ =6.545\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 6}\text{.545}\\ \text{=7}\text{.45}\end{array}$

Conclusion

The pH halfway to the equivalence point was calculated as 7.45

(c)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ 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 16.144QP

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

The pH at the equivalence point is 4.34

Explanation of Solution

To Calculate: The pH at the equivalence point

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

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

$\begin{array}{l}\text{Volume of HCl}\text{= }\frac{{\text{M}}_{base}{\text{ V}}_{base}}{{\text{M}}_{acid}}\\ =\frac{(0.150\text{ M)(50}\text{.0 mL)}}{0.100\text{ M}}\\ =\text{ 75}\text{.00 mL}\end{array}$

Hence, the total volume is as follows,

$\begin{array}{l}\text{Total volume}\text{= 50}\text{.0 mL + 75}\text{.00 mL}\\ =\text{ 125}\text{.00 mL}\\ \text{= 0}\text{.125 L}\end{array}$

The equilibrium reaction is,

${\text{ HClO + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ ClO}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{+}$

The moles of $\text{HClO}$ present are the same as the moles of ${\text{ClO}}^{-}$ present initially.

The concentration of $\text{HClO}$ is:

$\begin{array}{l}\text{[HClO]}\text{=}\frac{(0.150\text{ M)(50}\text{.0 mL)}}{\text{125 mL}}\\ =0.06000\text{ M}\end{array}$

Substitute into the equilibrium-constant expression and solve.

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{[{\text{ClO}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[\text{HClO]}}\\ 3.5\times {10}^{-8}=\frac{x^2}{0.06000-x}\\ \text{Assume }x\text{ is very small compared to 0}\text{.06000 and neglect it}\text{.}\\ \cong \frac{x^2}{0.06000}\\ x=4.58\times {10}^{-5}\text{ M}\end{array}$

Here, x gives the hydronium ion concentration.

$\begin{array}{l}\text{pH}{\text{=-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{=-log(}4.58\times {10}^{-5})\\ =4.34\end{array}$

Conclusion

The pH at the equivalence point was calculated as 4.34

(d)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

After 5.00 mL of $\text{HCl}$ has been added beyond 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 16.144QP

The pH of the given points of the titration of $\text{NaClO}$ with $\text{HCl}$ has to be calculated.

The pH after 5.00 mL of $\text{HCl}$ has been added beyond the equivalence point is 2.41

Explanation of Solution

To Calculate: The pH after 5.00 mL of $\text{HCl}$ has been added beyond the equivalence point

The total volume after the addition of 5.00 mL of $\text{HCl}$ is:

$\text{125}\text{.0 mL + 5}\text{.00 mL = 130}\text{.0 mL}$

Calculate the concentration of excess ${\text{H}}_{\text{3}}{\text{O}}^{+}$

The hydronium ion concentration and the pH are:

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\frac{({\text{mol HCl added) - (mol ClO}}^{\text{-}}\text{ at start)}}{\text{Total volume}}\\ =\frac{(\text{0}\text{.100 M)(80}\text{.00 mL) - (0}\text{.150 M)(50}\text{.0 mL)}}{130.00\text{ mL}}\\ =3.846\times {10}^{-3}\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (3.846\times {10}^{-3})\\ =2.41\end{array}$

Conclusion

The pH after 5.00 mL of $\text{HCl}$ has been added beyond the equivalence point was calculated as 2.41