Chapter 16, Problem 16.135QP

A 30.0-mL sample of 0.05 M HClO is titrated by a 0.0250 M KOH solution K_a for HClO is 3.5 × 10⁻⁸. Calculate a the pH when no base has been added; b the pH when 30.00 mL of the base has been added; c the pH at the equivalence point; d the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point.

Interpretation Introduction

Interpretation:

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

when no base 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 16.135QP

The pH of the given points of the titration of $\text{HClO}$ with $\text{KOH}$ are as follows,

The pH when no base has been added is 4.4

Explanation of Solution

To Calculate: The pH when no base has been added

Given data:

The volume of $\text{HClO}$ = 30.0 mL

The concentration of $\text{HClO}$ = 0.05 M

The concentration of $\text{KOH}$  = 0.0250 M

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

pH prior to the start of the titration

Construct an equilibrium table for the hydrolysis of $\text{HClO}$ as follows,

	${\text{ HClO + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_{\text{3}}{\text{O}}^{+}{\text{ + ClO}}^{-}$
Initial $(M)$	0.05 $-x$ 0.05-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

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

Now substitute equilibrium concentrations into the equilibrium-constant expression.

$\begin{array}{l}{\text{ K}}_a=\frac{{\text{(}x\text{)}}^2}{(0.05-x)}\\ \text{Assume 0}\text{.05 is very small than }x\text{ and neglect it in the denominator}\\ \text{ 3}\text{.5}\times {10}^{-8}\approx \frac{{\text{(}x\text{)}}^2}{(0.05)}\\ x=4.18\times {10}^{-5}\text{ M}\end{array}$

Here, x gives the concentration of hydronium ion $4.18\times {10}^{-5}\text{ M}$

Finally calculate pH as follows,

$\begin{array}{l}\text{pH}{\text{=-log[H}}_3{\text{O}}^{+}\text{]}\\ \text{=-log(}4.18\times {10}^{-5})\\ =4.4\end{array}$

Conclusion

The pH when no base has been added was calculated as 4.4

Interpretation Introduction

Interpretation:

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

when 30.0 mL of the base 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 16.135QP

The pH of the given points of the titration of $\text{HClO}$ with $\text{KOH}$ are as follows,

The pH when 30.0 mL of the base has been added is 7.5

Explanation of Solution

To Calculate: The pH when 30.0 mL of the base has been added

After the addition of 30.0 mL of 0.0250 M $\text{KOH}$, the reaction will be as follows

${\text{HClO + OH}}^{-}\stackrel{}{\to }{\text{ ClO}}^{-}{\text{ + H}}_{\text{2}}\text{O}$

At this point, the total volume is, $30.0\text{ mL + 30}\text{.0 mL = 60}\text{.0 mL}$

Now, the moles of $\text{HClO}$ present at the initial and the moles of ${\text{OH}}^{-}$ added are:

$\begin{array}{l}\text{mol HClO }=\text{ M}\times \text{V}\\ \text{=0}\text{.05 M}\times 3\text{0}\times {\text{10}}^{-3}\text{ L}\\ \text{=1}\text{.50}\times {\text{10}}^{-3}\text{ mol}\\ {\text{mol OH}}^{-}=\text{ M}\times \text{V}\\ \text{=0}\text{.0250 M}\times 30\times {\text{10}}^{-3}\text{ L}\\ \text{=0}\text{.7500}\times {\text{10}}^{-3}\text{ mol}\end{array}$

After the reaction, the moles of ${\text{ClO}}^{-}$ will be equal to the moles of ${\text{OH}}^{-}$ added.

The moles of $\text{HClO}$ present are:

$\begin{array}{l}\text{mol HClO }\text{= 1}\text{.50}\times {\text{10}}^{-3}\text{ mol - 0}\text{.7500}\times {\text{10}}^{-3}\\ =0.7500\times {10}^{-3}\text{ mol}\end{array}$

The concentrations are:

$\begin{array}{l}\text{[HClO]}\text{=}\frac{0.7500\times {10}^{-3}\text{ mol HClO}}{60.0\times {10}^{-3}\text{ L}}\\ =0.0125\text{ M}\\ {\text{[ClO}}^{-}\text{]}\text{=}\frac{0.7500\times {10}^{-3}{\text{ mol ClO}}^{-}}{60.0\times {10}^{-3}\text{ L}}\\ =0.0125\text{ M}\end{array}$

Solve for ${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}$ by Substituting the concentrations into the ${\text{K}}_a$ expression

$\begin{array}{l}{\text{ K}}_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{+}{\text{][ClO}}^{\text{-}}\text{]}}{[\text{HClO]}}\\ \text{ 3}\text{.5}\times {10}^{-8}=\frac{(0.0125+x)(x)}{(0.0125-x)}\\ \text{Assume that }x\text{ is small compared to 0}\text{.0125 and neglect it}\\ \approx \frac{(0.0125)(x)}{(0.125)}\\ x=\text{ 3}\text{.5}\times {10}^{-8}\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=-log(3}\text{.5}\times {\text{10}}^{\text{-8}})\\ =7.45\end{array}$

Conclusion

The pH when 30.0 mL of the base has been added was calculated as 7.5

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{HClO}$  with $\text{KOH}$ 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 16.135QP

The pH of the given points of the titration of $\text{HClO}$ with $\text{KOH}$ are as follows,

The pH at the equivalence point is 9.8

Explanation of Solution

To Calculate: The pH at the equivalence point

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

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

$\begin{array}{l}\text{Volume of KOH}\text{= }\frac{{\text{M}}_{acid}{\text{ V}}_{acid}}{{\text{M}}_{base}}\\ =\frac{\text{(0}\text{.05 M)(30}\text{.0 mL)}}{\text{(0}\text{.0250 M)}}\\ =\text{ 60}\text{.0 mL}\end{array}$

Hence, the total volume is as follows,

$\begin{array}{l}\text{Total volume}\text{= 30}\text{.0 mL + 60}\text{.0 mL}\\ =90.\text{0 mL}\\ \text{= 0}\text{.090 L}\end{array}$

The equilibrium reaction is,

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

$\begin{array}{l}{\text{[ClO}}^{-}\text{]}\text{=}\frac{1.50\times {10}^{-3}\text{ mol}}{90.0\times {10}^{-3}\text{ L}}\\ =0.0166\text{ M}\end{array}$

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

$\begin{array}{l}{\text{K}}_b=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\\ =\frac{\text{1}\text{.00}\times {\text{10}}^{\text{-14}}}{3.5\times {\text{10}}^{\text{-8}}}\\ =2.85\times {10}^{-7}\end{array}$

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{HClO][OH}}^{-}\text{]}}{[{\text{ClO}}^{\text{-}}\text{]}}\\ 2.85\times {10}^{-7}=\frac{x^2}{0.0166-x}\\ x=6.90\times {10}^{-5}\text{ M}\end{array}$

Here, x gives the hydroxide ion concentration.

The pH can be calculated from pOH as follows,

$\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{-}\text{]}\\ \text{=-log(6}\text{.90}\times {\text{10}}^{\text{-5}})\\ =4.161\\ \text{pH + pOH }\text{=14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14- }4.161\\ =9.8\end{array}$

Conclusion

The pH at the equivalence point was calculated as 9.8

Interpretation Introduction

Interpretation:

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

When an additional 4.00 mL of the $\text{KOH}$ solution 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 16.135QP

The pH of the given points of the titration of $\text{HClO}$ with $\text{KOH}$ are as follows,

The pH when an additional 4.00 mL of the $\text{KOH}$ solution has been added beyond the equivalence point is 11.0

Explanation of Solution

To Calculate: The pH when an additional 4.00 mL of the $\text{KOH}$ solution has been added beyond the equivalence point

Calculate the moles of base added.

$(0.0250\text{ M})\times 94.0\times {10}^{-3}\text{L = 1}\text{.60}\times {\text{10}}^{\text{-3}}\text{ mol}$

$\begin{array}{l}\text{Moles of acid remaining }\text{= 1}\text{.60}\times {\text{10}}^{\text{-3}}\text{ mol - 1}\text{.5}\times {\text{10}}^{\text{-3}}\text{ mol}\\ \text{= 0}\text{.10}\times {\text{10}}^{\text{-3}}\text{ mol}\end{array}$

The total volume after the addition of 4.00 mL of the $\text{KOH}$ solution beyond the equivalence point: $\text{90}\text{.0 mL + 4}\text{.0 mL = 94}\text{.0 mL}$

The hydroxide ion concentration and the pH are:

$\begin{array}{l}{\text{[OH}}^{-}\text{]}\text{=}\frac{0.10\times {\text{10}}^{\text{-3}}\text{ mol}}{94.0\times {\text{10}}^{\text{-3}}\text{ L}}\\ =0.00106\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{-}\text{]}\\ \text{=-log(0}\text{.00106})\\ =2.973\\ \text{pH + pOH }\text{=14}\\ \text{pH}\text{=14 - 2}\text{.973}\\ =11.0\end{array}$

Conclusion

The pH when an additional 4.00 mL of the $\text{KOH}$ solution has been added beyond the equivalence point was calculated as 11.o