Chapter 16, Problem 16.57QE

Interpretation Introduction

Interpretation:

The curve for the titration of $50mLofa0.10M$ weak base with a $0.20M$ strong acid has to be sketched.  On the same axes, the titration curve for the same volume and concentration of $NaOH$ has to be sketched.

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where $HCl$ has been taken as strong acid.

Addition of $0mLof0.20MHCl$to$50.00mLof0.10M$weak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the $pH$ of the solution.

  $\begin{array}{cccccccc}& B& +& H_{2}O& \rightleftharpoons & O{H}^{-}& +& HB\\ i\left(M\right)& 0.1& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.1-y& & & & y& & y\end{array}$

Where, $B$ represents weak base and $HB$ represents the conjugate acid of weak base.

The $K_b$ value for weak base is $1.0\times {10}^{-5}.$  The expression for $K_b$ can be written as given below,

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[HB\right]}{\left[B\right]}=1.0\times {10}^{-5}\\ \\ \frac{y^2}{\left(0.1-y\right)}=1.0\times {10}^{-5}\end{array}$

This can be solved by approximation.  If $y\ll 0.10$, then

  $\begin{array}{l}{y}^2\approx 1.0\times {10}^{-5}\times 0.10=1.0\times {10}^{-6}\\ \\ y=\sqrt{1.0\times {10}^{-6}}=1.0\times {10}^{-3}=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as given below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(1.0\times {10}^{-3}\right)=\text{3}\text{.00}\\ \\ pH=14-pOH=14-3.00=11.00.\end{array}$

Addition of $10mLof0.20MHCl$to$50.00mLof0.10M$weak base:

The neutralization reaction can be written as given,

  $B\left(aq\right)+H_3{O}^{+}\left(aq\right)\to HB\left(aq\right)+H_{2}O\left(l\right)$

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHCl=10.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=2mi\lim ol\\ \\ AmountofB=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}}{\cancel{mL}}\right)=5mi\lim ol\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$ So the limiting reactant is $H_3{O}^{+}$ ion.

  $\begin{array}{cccccccc}& B& +& H_3{O}^{+}& \to & HB& +& H_{2}O\\ s\left(mmol\right)& 5& & 2& & 0& & Excess\\ R\left(mmol\right)& -2& & -2& & +2& & +2\\ f\left(mmol\right)& 3& & 0& & 2& & Excess\end{array}$

All the strong acid is consumed.  There are weak base $\left(B\right)$ and its conjugate acid $\left(HB\right)$ in the solution.  So this solution is a basic buffer.  By using Henderson-Hasselbalch equation, it’s $pH$ can be calculated.

  $\begin{array}{c}pOH=p{K}_b+\log \frac{n_a}{n_b}\\ \\ =p{K}_b+\log \frac{2.0}{3.0}\\ \\ =-\log \left(K_b\right)-0.176\\ \\ =-\log \left(1.0\times {10}^{-5}\right)-0.176\\ \\ =5.00-0.176\\ \\ =4.824\\ \\ pH=14-pOH=14-4.824=9.176.\end{array}$

Addition of $25mLof0.20MHCl$to$50.00mLof0.10M$weak base:

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHCl=25.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ AmountofB=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ Totalvolume=25.00+50.00=75.00mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$

  $\begin{array}{cccccccc}& B& +& H_3{O}^{+}& \to & HB& +& H_{2}O\\ s\left(mmol\right)& 5& & 5& & 0& & Excess\\ R\left(mmol\right)& -5& & -5& & +5& & +5\\ f\left(mmol\right)& 0& & 0& & 5& & Excess\\ c\left(M\right)& 0& & 0& & 0.067& & \end{array}$

Both $H_3{O}^{+}$ and weak base have been completely consumed, the titration is at the equivalence point.  There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the $pH$ of the solution.

  $\begin{array}{cccccccc}& HB& +& H_{2}O& \rightleftharpoons & B& +& H_3{O}^{+}\\ i\left(M\right)& 0.067& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.067-y& & & & y& & y\end{array}$

The expression for $K_a$ can be written as given below,

  $K_a=\frac{\left[B\right]\left[H_3{O}^{+}\right]}{\left[HB\right]}=\frac{y^2}{\left(0.067-y\right)}$

The value of $K_a$ can be calculated as given below.

  $K_a=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-5}}=1.0\times {10}^{-9}$

This can be solved by approximation.  If $y\ll 0.067$, then

  $\begin{array}{l}{y}^2\approx 1.0\times {10}^{-9}\times 0.067=0.067\times {10}^{-9}\\ \\ y=\sqrt{0.067\times {10}^{-9}}=8.2\times {10}^{-6}=\left[H_3{O}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as given below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(8.2\times {10}^{-6}\right)=5.1.$

Addition of $50mLof0.20MHCl$to$50.00mLof0.10M$weak base:

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHCl=50.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofB=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ Totalvolume=50.00+50.00=100.00mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$  So the limiting reactant is the base.

$\begin{array}{cccccccc}& B& +& H_3{O}^{+}& \to & HB& +& H_{2}O\\ s\left(mmol\right)& 5& & 10& & 0& & Excess\\ R\left(mmol\right)& -5& & -5& & +5& & +5\\ f\left(mmol\right)& 0& & 5& & 5& & Excess\\ c\left(M\right)& 0& & 0.05& & 0.05& & \end{array}$

Now, the $pH$ of the system can be calculated as given below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.05\right)=1.30.$

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where $HCl$ has been taken as strong acid.

Addition of $0mLof0.20MHCl$ to$50.00mLof0.10MNaOH:$

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As $NaOH$ is a strong base, the hydroxide ion concentration is $0.10M.$

The $pH$ of the solution can be calculated as given below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.1\right)=\text{1}\\ \\ pH=14-pOH=14-1=13.\end{array}$

Addition of $10mLof0.20MHCl$ to$50.00mLof0.10MNaOH:$

The neutralization reaction can be written as given,

  $H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)$

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHCl=10.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=2mi\lim ol\\ \\ AmountofO{H}^{-}=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ Totalvolume=60mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$ So the limiting reactant is $H_3{O}^{+}$ ion.

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 2& & 5& & Excess\\ R\left(mmol\right)& -2& & -2& & +2\\ f\left(mmol\right)& 0& & 3& & Excess\\ C\left(M\right)& 0& & 0.05& & \end{array}$

The $pH$ of the solution can be calculated as given below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.05\right)=\text{1}\text{.30}\\ \\ pH=14-pOH=14-1.30=12.7.\end{array}$

Addition of $25mLof0.20MHCl$ to$50.00mLof0.10MNaOH:$

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHCl=25.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ AmountofO{H}^{-}=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ Totalvolume=75mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 5& & 5& & Excess\\ R\left(mmol\right)& -5& & -5& & +5\\ f\left(mmol\right)& 0& & 0& & Excess\\ C\left(M\right)& 0& & 0& & Excess\end{array}$

Both $H_3{O}^{+}$ and $O{H}^{-}$ ions have been completely consumed, the titration is at the equivalence point.  Therefore, the $pH$ of the system is $7.$

Addition of $50mLof0.20MHCl$ to$50.00mLof0.10MNaOH:$

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHCl=50.00\cancel{mL}\times \left(\frac{0.20\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=50.00\cancel{mL}\times \left(\frac{0.10\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=5mi\lim ol\\ \\ Totalvolume=100mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is $1:1.$ So the limiting reactant is $O{H}^{-}$ ion.

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 10& & 5& & Excess\\ R\left(mmol\right)& -5& & -5& & +5\\ f\left(mmol\right)& 5& & 0& & Excess\\ C\left(M\right)& 0.05& & 0& & Excess\end{array}$

Now, the $pH$ of the system can be calculated as given below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.05\right)=1.30.$

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding $pH$ values.

For weak base:

  $\begin{array}{cc}VolumeofHCl\left(mL\right)& pH\\ 0& 11.00\\ 10.00& 9.176\\ 25.00& 5.10\\ 50.00& 1.30\end{array}$

For strong base:

  $\begin{array}{cc}VolumeofNaOH\left(mL\right)& pH\\ 0& 13.00\\ 10.00& 12.70\\ 25.00& 7.00\\ 50.00& 1.30\end{array}$

The titration curve with four important regions is given below.

Figure