Chapter 16, Problem 16.119QP

a Draw a pH titration curve that represents the titration of 50.0 mL of 0.10 M NH₃ by the addition of 0.10 M HCl from a buret. Label the axes and put a scale on each axis. Show where the equivalence point and the buffer region are on the titration curve. You should do calculations for the 0%, 30%, 50%, and 100% titration points. b Is the solution neutral, acidic, or basic at the equivalence point? Why?

Interpretation Introduction

Interpretation:

For titration of 50.0 mL of 0.10 M ${\text{NH}}_{\text{3}}$ by the addition of 0.10 M $\text{HCl}$,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 30 %, 50% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.119QP

A pH titration curve showing the equivalence point and buffer region is given in Figure 1 as follows,

Figure 1

(a)

The pH at the 0% titration point is 11.13

The pH at the 30% titration point is 9.62

The pH at the 50% titration point is 9.26

The pH at the 100% titration point is 5.28

(b)

The solution at the equivalence point is acidic

Explanation of Solution

To Calculate: The pH of the titration points for the 0%, 30 %, 50% and 100%

Given data:

Titration of 0.10 M ${\text{NH}}_3$ with 0.10 M $\text{HCl}$

pH at the 0% titration point:

Construct an equilibrium table with x as unknown concentration

	${\text{NH}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ NH}}_4{}^{\text{+}}{\text{ + OH}}^{-}$
Initial $(M)$	0.10 $-x$ 0.10-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Substitute equilibrium concentrations into the equilibrium-constant equation.

The ${\text{K}}_b$ for ammonia is $1.8\times {10}^{-5}$

Assume x is negligible compared to 0.10 M

$\begin{array}{l}{\text{ K}}_b=\frac{{\text{[NH}}_4{}^{+}{\text{][OH}}^{-}]}{{\text{[NH}}_3\text{]}}\\ =\frac{{\text{(}x\text{)}}^{\text{2}}}{(0.10-x)}\\ \text{1}\text{.8}\times {10}^{-5}\approx \frac{{\text{(}x\text{)}}^2}{0.10}\\ \text{Rearrange and solve for }x\\ x=\sqrt{(0.10)(\text{1}\text{.8}\times {10}^{-5})}\\ =1.34\times {10}^{-3}\text{ M}\end{array}$

Therefore, the concentration of hydroxide ion ${\text{[OH}}^{-}]$ is $1.34\times {10}^{-3}\text{ M}$

In the end, pH is calculated as follows,

$\begin{array}{l}\text{ pOH}\text{=}-\log {\text{[OH}}^{-}]\\ =-\log (1.34\times {10}^{-3})\\ =2.873\\ \text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 2}\text{.873}\\ \text{=11}\text{.13}\end{array}$

Therefore, the pH at the 0% titration point is 11.13

pH at the 30% titration point:

For convenience, express the concentrations as percents.

$\begin{array}{l}{\text{[NH}}_3]\approx 70\%\\ {\text{[NH}}_4{}^{+}]\approx 30\%\end{array}$

Substitute the concentrations into the equilibrium expression.

$\begin{array}{l}{\text{ K}}_b=\frac{{\text{[NH}}_4{}^{+}{\text{][OH}}^{-}]}{{\text{[NH}}_3\text{]}}\\ =\frac{\text{(30\%)(}x\text{)}}{(70\%)}\\ \text{1}\text{.8}\times {10}^{-5}=\frac{\text{(30\%)(}x\text{)}}{(70\%)}\\ \text{Rearrange and solve for }x\\ x=\text{1}\text{.8}\times {10}^{-5}\times \frac{70\%}{30\%}\\ =4.20\times {10}^{-5}\text{ M}\end{array}$

Therefore, the concentration of hydroxide ion ${\text{[OH}}^{-}]$ is $4.20\times {10}^{-5}\text{ M}$

In the end, pH is calculated as follows,

$\begin{array}{l}\text{ pOH}\text{=}-\log {\text{[OH}}^{-}]\\ =-\log (4.20\times {10}^{-5})\\ =4.377\\ \text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 4}\text{.377}\\ \text{=9}\text{.62}\end{array}$

Therefore, the pH at the 30% titration point is 9.62

pH at the 50% titration point:

$\begin{array}{l}{\text{[NH}}_4{}^{+}\text{]}={\text{[NH}}_3\text{]}\\ \text{Therefore,}\\ {\text{K}}_b=\frac{{\text{[NH}}_4{}^{+}{\text{][OH}}^{-}]}{{\text{[NH}}_3\text{]}}\\ {\text{[OH}}^{-}]=\text{1}\text{.8}\times {10}^{-5}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{ pOH}\text{=}-\log {\text{[OH}}^{-}]\\ =-\log (1.8\times {10}^{-5})\\ =4.744\\ \text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 4}\text{.744}\\ \text{=9}\text{.26}\end{array}$

Therefore, the pH at the 50% titration point is 9.26

pH at the 100% titration point:

As a result of titration ${\text{NH}}_4\text{Cl}$ has got produced.

The ${\text{NH}}_4\text{Cl}$ that is produced has undergone a twofold dilution.

Therefore, ${\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]=0}\text{.0500 M}$

Construct an equilibrium table with x as unknown concentration

	${\text{NH}}_4{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ NH}}_{\text{3}}{\text{ + H}}_3{\text{O}}^{+}$
Initial $(M)$	0.0500 $-x$ 0.0500-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Now, calculate ${\text{K}}_{\text{a}}$ using ${\text{K}}_w$ as follows,

$\begin{array}{l}{\text{K}}_a=\frac{{\text{K}}_w}{{\text{K}}_b}\\ =\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =5.55\times {10}^{-10}\end{array}$

Substitute into the equilibrium constant expression.

$\begin{array}{l}{\text{K}}_a=\frac{{\text{[NH}}_3{\text{][H}}_{\text{3}}{\text{O}}^{+}]}{{\text{[NH}}_4{}^{+}\text{]}}\\ 5.55\times {10}^{-10}=\frac{{(x)}^2}{(0.0500-x)}\\ \text{Assume }x\text{ is very small compared to 0}\text{.0500 and neglect it}\\ \cong \frac{{(x)}^2}{(0.0500)}\\ \text{solve for }x,\\ x=\sqrt{(5.55\times {10}^{-10})(0.0500)}\\ =5.27\times {10}^{-6}\text{ M}\end{array}$

Here, x gives the concentration of hydronium ion, ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=5.27\times {10}^{-6}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (5.27\times {10}^{-6})\\ =5.28\end{array}$

Therefore, the pH at the 100% titration point is 5.28

Conclusion

The pH at the 0% titration point was calculated as 11.13

The pH at the 30% titration point was calculated as 9.62

The pH at the 50% titration point was calculated as 9.26

The pH at the 100% titration point was calculated as 5.28

Interpretation Introduction

Interpretation:

For titration of 50.0 mL of 0.10 M ${\text{NH}}_{\text{3}}$ by the addition of 0.10 M $\text{HCl}$,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 30 %, 50% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.119QP

A pH titration curve showing the equivalence point and buffer region is given in Figure 1 as follows,

Figure 1

(a)

The pH at the 0% titration point is 11.13

The pH at the 30% titration point is 9.62

The pH at the 50% titration point is 9.26

The pH at the 100% titration point is 5.28

(b)

The solution at the equivalence point is acidic

Explanation of Solution

To Explain: Whether the solution at the equivalence point is neutral, acidic or basic

As a result of titration ${\text{NH}}_4\text{Cl}$ salt is produced.

${\text{NH}}_4\text{Cl}$ has ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{Cl}}^{-}$ ions.

Ammonium chloride salt is the salt of weak base and a strong acid

${\text{NH}}_{\text{4}}{}^{\text{+}}$ ion reacts with water to produce acid whereas ${\text{Cl}}^{-}$ does not undergo hydrolysis.

Therefore, the given solution is acidic

Conclusion

The solution at the equivalence point was found as acidic