Chapter 16, Problem 16.53QE

Interpretation Introduction

Interpretation:

The curve for the titration of $100mLofa0.10M$ weak acid with a $0.20M$ strong base has to be sketched.  On the same axes, the titration curve for the same volume and concentration of $HCl$ has to be sketched.

Explanation of Solution

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

Addition of $0mLof0.20MNaOH$to$100.00mLof0.10M$weak acid:

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

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

  $\begin{array}{cccccccc}& HA& +& H_{2}O& \rightleftharpoons & H_3{O}^{+}& +& A^{-}\\ i\left(M\right)& 0.10& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.10-y& & & & y& & y\end{array}$

Where, $HA$ represents weak acid and $A^{-}$ represents the conjugate base of weak acid.

The $K_a$ value for weak acid is $1.0\times {10}^{-4}.$  The expression for $K_a$ can be written as given below,

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=1.0\times {10}^{-4}\\ \\ \frac{y^2}{\left(0.10-y\right)}=1.0\times {10}^{-4}\end{array}$

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

  $\begin{array}{l}{y}^2\approx 1.0\times {10}^{-4}\times 0.10=0.1\times {10}^{-4}\\ \\ y=\sqrt{0.1\times {10}^{-4}}=3.16\times {10}^{-3}=\left[H_3{O}^{+}\right]\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(3.16\times {10}^{-3}\right)=2.50.$

Addition of $10mLof0.20MNaOH$to$100.00mLof0.10M$weak acid:

The neutralization reaction can be written as given,

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

Calculation of milimole of acid and base:

  $\begin{array}{l}AmountofHA=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=10.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=2.0mi\lim ol\end{array}$

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 $O{H}^{-}$ ion.

  $\begin{array}{cccccccc}& HA& +& O{H}^{-}& \to & A^{-}& +& H_{2}O\\ s\left(mmol\right)& 10& & 2.0& & 0& & Excess\\ R\left(mmol\right)& -2.0& & -2.0& & +2.0& & +2.0\\ f\left(mmol\right)& 8.0& & 0& & 2.0& & Excess\end{array}$

All the strong base is consumed.  There are weak acid $\left(HA\right)$ and its conjugate base $\left(A^{-}\right)$ in the solution.  So this solution is an acidic buffer.  By using Henderson-Hasselbalch equation, it’s $pH$ can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{n_b}{n_a}\\ \\ =p{K}_a+\log \frac{2.0}{8}\\ \\ =-\log \left(K_a\right)-0.60\\ \\ =-\log \left(1.0\times {10}^{-4}\right)-0.60\\ \\ =4.00-0.60\\ \\ =3.4.\end{array}$

Addition of $50mLof0.20MNaOH$to$100.00mLof0.10M$weak acid:

Calculation of milimole of acid and base:

  $\begin{array}{l}AmountofHA=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=50.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ Totalvolume=100.00+50.00=150.00mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

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

  $\begin{array}{cccccccc}& HA& +& O{H}^{-}& \to & A^{-}& +& H_{2}O\\ s\left(mmol\right)& 10& & 10& & 0& & Excess\\ R\left(mmol\right)& -10& & -10& & +10& & +10\\ f\left(mmol\right)& 0& & 0& & 10& & Excess\\ c\left(M\right)& 0& & 0& & 0.067& & \end{array}$

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

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

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

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

  $K_b=\frac{\left[HA\right]\left[O{H}^{-}\right]}{\left[A^{-}\right]}=\frac{y^2}{\left(0.067-y\right)}$

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

  $K_b=\frac{K_w}{K_a}=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-4}}=1.0\times {10}^{-10}$

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

  $\begin{array}{l}{y}^2\approx 1.0\times {10}^{-10}\times 0.067=0.067\times {10}^{-10}\\ \\ y=\sqrt{0.067\times {10}^{-10}}=2.6\times {10}^{-6}=\left[O{H}^{-}\right]\end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.6\times {10}^{-6}\right)=5.6\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-5.6=8.4.\end{array}$

Addition of $70mLof0.20MNaOH$to$100.00mLof0.10M$weak acid:

Calculation of milimole of acid and base and total volume:

  $\begin{array}{l}AmountofHA=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=70.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=14mi\lim ol\\ \\ Totalvolume=100.00+70.00=170.00mL\end{array}$

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 $H_3{O}^{+}$ ion.

$\begin{array}{cccccccc}& HA& +& O{H}^{-}& \to & A^{-}& +& H_{2}O\\ s\left(mmol\right)& 10& & 14& & 0& & Excess\\ R\left(mmol\right)& -10& & -10& & +10& & +10\\ f\left(mmol\right)& 0& & 4& & 10& & Excess\\ c\left(M\right)& 0& & 0.0235& & 0.06& & \end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.0235\right)=1.63\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-1.63=12.37.\end{array}$

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

Addition of $0mLof0.20MNaOH$ to$100.00mLof0.10MHCl:$

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

As $HCl$ is a strong acid, the hydronium ion concentration is $0.10M.$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.1\right)=1.$

Addition of $10mLof0.20MNaOH$ to$100.00mLof0.10MHCl:$

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 milimole of acid and base:

  $\begin{array}{l}AmountofHCl=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=10.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=2.0mi\lim ol\\ \\ Totalvolume=110mL\end{array}$

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.

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 10& & 2.0& & Excess\\ R\left(mmol\right)& -2.0& & -2.0& & +2.0\\ f\left(mmol\right)& 8.0& & 0& & Excess\\ C\left(M\right)& 0.073& & & & \end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.073\right)=1.14.$

Addition of $50mLof0.20MNaOH$ to$100.00mLof0.10MHCl:$

Calculation of milimole of acid and base:

  $\begin{array}{l}AmountofHA=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=50.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ Totalvolume=100.00+50.00=150.00mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

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

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 10& & 10& & Excess\\ R\left(mmol\right)& -10& & -10& & +2.0\\ f\left(mmol\right)& 0& & 0& & Excess\\ C\left(M\right)& 0& & 0& & Excess\end{array}$

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 $70mLof0.20MNaOH$ to$100.00mLof0.10MHCl:$

Calculation of milimole of acid and base and total volume:

  $\begin{array}{l}AmountofHA=100.00\cancel{mL}\times \left(\frac{0.10\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=10mi\lim ol\\ \\ AmountofO{H}^{-}=70.00\cancel{mL}\times \left(\frac{0.20\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=14mi\lim ol\\ \\ Totalvolume=100.00+70.00=170.00mL\end{array}$

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 $H_3{O}^{+}$ ion.

  $\begin{array}{cccccc}& H_3{O}^{+}& +& O{H}^{-}& \to & 2H_{2}O\\ s\left(mmol\right)& 10& & 14& & Excess\\ R\left(mmol\right)& -10& & -10& & +2.0\\ f\left(mmol\right)& 0& & 4& & Excess\\ C\left(M\right)& 0& & 0.0235& & Excess\end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.0235\right)=1.63\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-1.63=12.37.\end{array}$

Titration curve:

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

For weak acid:

  $\begin{array}{cc}VolumeofNaOH\left(mL\right)& pH\\ 0& 2.50\\ 10.00& 3.4\\ 50.00& 8.4\\ 70.00& 12.37\end{array}$

For strong acid:

  $\begin{array}{cc}VolumeofNaOH\left(mL\right)& pH\\ 0& 1.00\\ 10.00& 1.14\\ 50.00& 7.00\\ 70.00& 12.37\end{array}$

The titration curve with four important regions is given below.

Figure