Chapter 16, Problem 16.20QA

Interpretation Introduction

To decide:

Which buffer experiences a smaller change in the $pH$ when

a) The same small quantity of strong base is added to both

b) The same small quantity of strong acid is added to both

Answer to Problem 16.20QA

Solution:

a) When the same small quantity of strong base $\left(NaOH\right)$ is added to each buffer, the buffer A experiences a smaller change in pH.

b) When the same small quantity of strong acid $\left(HCl\right)$ is added to each buffer, the buffer A experiences a smaller change in pH.

Explanation of Solution

Let’s consider acetic acid and sodium acetate buffer systems with different concentrations of conjugate acid base pairs. Let’s assume the volume for two buffers is equal as $1L$, so molarity of conjugate acid base pairs can be taken as their moles. $p{K}_a$ of acetic acid is $4.75$.

In buffer A, concentrations of conjugate acid base pair are nearly equal.

Let’s assume the moles of conjugate acid and base to be $0.1 mol$ each.

In buffer A,

$\left[C{H}_{3}CO{O}^{-}\right]=\left[C{H}_{3}COOH\right]=0.1 mol each$

Total concentration of conjugate acid base pair is

$\left[C{H}_{3}CO{O}^{-}\right]+\left[C{H}_{3}COOH\right]=0.1+0.1=0.2 mol$

So, for buffer A, the initial pH will be equal to $p{K}_a=4.75$.

In buffer B, total composition of conjugate acid base pair remains the same, but there is twice as much as weak acid as its conjugate base.

Let’s assume the moles of conjugate acid and base to be $x mol$ each.

In buffer B,

$\left[C{H}_{3}CO{O}^{-}\right]+\left[C{H}_{3}COOH\right]=x+x=0.2 mol$

$\left[C{H}_{3}CO{O}^{-}\right]\times 2=\left[C{H}_{3}COOH\right]$

Solving the above equations, we get the concentrations of $\left[C{H}_{3}CO{O}^{-}\right]and\left[C{H}_{3}COOH\right]$ as

$\left[C{H}_{3}CO{O}^{-}\right]=0.0667 mol$

$\left[C{H}_{3}CO{OH}^{ }\right]=0.133 mol$

For buffer B, the initial $pH$ can be calculated using Henderson-Hasselbalch equation.

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=4.75+\log \frac{0.0667}{0.133}=4.45$

Part a)

Consider that 0.01 moles of $NaOH$ are added to each buffer.

For buffer A:

RICE table for reaction:

Reaction	$C{H}_{3}COOH\left(aq\right)+ NaOH\left(aq\right) \to C{H}_{3}COONa\left(aq\right)+ H_{2}O\left(l\right)$
	$C{H}_{3}COOH \left(mol\right)$	$NaOH \left(mol\right)$	$C{H}_{3}COONa \left(mol\right)$
Initial	$0.1$	$0.01$	$0.1$
Change	$-0.01$	$-0.01$	$+0.01$
Final	$0.09$	$0$	$0.11$

The mole ratio of $C{H}_{3}CO{O}^{-} to C{H}_{3}COOH$ in the buffer can be substituted for the $\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]$ ratio in the Henderson-Hasselbalch equation because the components are dissolved in the same total volume of buffer. Solving for the pH, we get

$p{K}_a$ of $C{H}_{3}COOH$ as $4.75$ (from Appendix 5, table 5.1).

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=4.75+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$

$pH=4.75+\log \left[\frac{0.11}{0.09}\right]$

$pH=4.75+0.087=4.84$

Change in pH in buffer A $=4.84-4.75=0.09$

For buffer B:

RICE table for reaction:

Reaction	$C{H}_{3}COOH\left(aq\right)+ NaOH\left(aq\right) \to C{H}_{3}COONa\left(aq\right)+ H_{2}O\left(l\right)$
	$C{H}_{3}COOH \left(mol\right)$	$NaOH \left(mol\right)$	$C{H}_{3}COONa \left(mol\right)$
Initial	$0.133$	$0.01$	$0.0667$
Change	$-0.01$	$-0.01$	$+0.01$
Final	$0.0663$	$0$	$0.0767$

The mole ratio of $C{H}_{3}CO{O}^{-} to C{H}_{3}COOH$ in the buffer can be substituted for the $\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]$ ratio in the Henderson-Hasselbalch equation because the components are dissolved in the same total volume of buffer. Solving for the pH, we get

$p{K}_a$ of $C{H}_{3}COOH$ as $4.75$ (from Appendix 5, table 5.1).

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=4.75+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$

$pH=4.75+\log \left[\frac{0.0767}{0.0663}\right]$

$pH=4.75+0.0632=4.81$

Change in pH in buffer B $=4.81-4.45=0.36$

When the same small quantity of strong base $\left(NaOH\right)$ is added to each buffer, the buffer A experiences a smaller change in pH.

Part b)

Consider that 0.01 moles of $HCl$ are added to each buffer.

Buffer A:

RICE table for reaction:

Reaction	$C{H}_{3}CO{O}^{-} \left(aq\right)+ HCl \left(aq\right) \to C{H}_{3}COOH\left(aq\right)+ C{l}^{-} \left(aq\right)$
	$C{H}_{3}CO{O}^{-} \left(mol\right)$	$HCl \left(mol\right)$	$C{H}_{3}COOH \left(mol\right)$
Initial	$0.1$	$0.01$	$0.1$
Change	$-0.01$	$-0.01$	$+0.01$
Final	$0.09$	$0$	$0.11$

The mole ratio of $C{H}_{3}CO{O}^{-} to C{H}_{3}COOH$ in the buffer can be substituted for the $\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]$ ratio in the Henderson-Hasselbalch equation because the components are dissolved in the same total volume of buffer. Solving for the pH, we get

$p{K}_a$ of $C{H}_{3}COOH$ as $4.75$ (from Appendix 5, table 5.1).

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=4.75+\log \frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

$pH=4.75+\log \left[\frac{0.09}{0.11}\right]$

$pH=4.75+(-0.08715)=4.66$

Change in pH in buffer A $=4.75-4.66=0.09$

For buffer B:

RICE table for reaction:

Reaction	$C{H}_{3}CO{O}^{-} \left(aq\right)+ HCl \left(aq\right) \to C{H}_{3}COOH\left(aq\right)+ C{l}^{-} \left(aq\right)$
	$C{H}_{3}CO{O}^{-} \left(mol\right)$	$HCl \left(mol\right)$	$C{H}_{3}COOH \left(mol\right)$
Initial	$0.0667$	$0.01$	$0.133$
Change	$-0.01$	$-0.01$	$+0.01$
Final	$0.0567$	$0$	$0.143$

The mole ratio of $C{H}_{3}CO{O}^{-} to C{H}_{3}COOH$ in the buffer can be substituted for the $\left[C{H}_{3}CO{O}^{-}\right]/\left[C{H}_{3}COOH\right]$ ratio in the Henderson-Hasselbalch equation because the components are dissolved in the same total volume of buffer. Solving for the pH, we get

$p{K}_a$ of $C{H}_{3}COOH$ as $4.75$ (from Appendix 5, table 5.1).

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=4.75+\log \frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

$pH=4.75+\log \left[\frac{0.0567}{0.143}\right]$

$pH=4.75+(-0.4017)=4.35$

Change in pH in buffer A $=4.45-4.35=0.10$

When the same small quantity of strong acid $\left(HCl\right)$ is added to each buffer, the buffer A experiences a smaller change in pH.

Conclusion:

For buffers, the greater the concentrations of the conjugate pair components, the greater is its buffer capacity, that is, the greater is its ability to withstand additions of acid or base without a significant change in $pH$.