Chapter 16, Problem 16.54QE

Interpretation Introduction

Interpretation:

The $pH$ during the titration of $10.00mLof0.400M$ hypochlorous acid with $0.500MKOH$ after the addition of $0\%,50\%,95\%,100\%,and105\%$ of the base needed to reach the equivalence point has to be calculated.  The titration curve has to be drawn and four regions of importance have to be labeled.

Explanation of Solution

The volume of base required to reach the equivalence point can be calculated as given below.

  $\begin{array}{l}{N}_1{V}_1=N_2{V}_2\\ \\ V_1=\frac{N_2{V}_2}{N_1}=\frac{0.400M\times 10mL}{0.500M}=8mL\end{array}$

The addition of $0\%,50\%,95\%,100\%,and105\%$ of the base needed to reach the equivalent point means $0\%,50\%,95\%,100\%,and105\%$ of $\text{8}\text{mL}\text{.}$

  $\begin{array}{l}0\%\times 8mL=0mL\\ \\ 50\%\times 8mL=4mL\\ \\ 95\%\times 8mL=7.6mL\\ \\ 100\%\times 8mL=8mL\\ \\ 105\%\times 8mL=8.4mL\end{array}$

Addition of $0mLof0.500MNaOH$ to $10.00mLof0.400MHOCl:$

$HOCl$ is a 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.

  $\begin{array}{cccccccc}& HOCl& +& H_{2}O& \rightleftharpoons & H_3{O}^{+}& +& OC{l}^{-}\\ i\left(M\right)& 0.400& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.400-y& & & & y& & y\end{array}$

The $K_a$ value for $HOCl$ is $\text{4}\text{.0}\times {10}^{-8}.$  The expression for $K_a$ can be written as given below,

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[OC{l}^{-}\right]}{\left[HOCl\right]}=4.0\times {10}^{-8}\\ \\ \frac{y^2}{\left(0.400-y\right)}=4.0\times {10}^{-8}\end{array}$

This can be solved by approximation.  If $y\ll 0.400$, then

  $\begin{array}{l}{y}^2\approx 4.0\times {10}^{-8}\times 0.400=1.6\times {10}^{-8}\\ \\ y=\sqrt{1.6\times {10}^{-8}}=1.26\times {10}^{-4}=\left[H_3{O}^{+}\right]\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(1.26\times {10}^{-4}\right)=3.89.$

Addition of $4mLof0.500MNaOH$ to $10.00mLof0.400MHOCl:$

The neutralization reaction can be written as given,

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

Calculation of milimole of acid and base:

  $\begin{array}{l}AmountofHOCl=10.00\cancel{mL}\times \left(\frac{0.400\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=4mi\lim ol\\ \\ AmountofO{H}^{-}=4.00\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=2mi\lim ol\end{array}$

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.

  $\begin{array}{cccccccc}& HOCl& +& O{H}^{-}& \to & OC{l}^{-}& +& H_{2}O\\ s\left(mmol\right)& 4& & 2& & 0& & Excess\\ R\left(mmol\right)& -2& & -2& & +2& & +2\\ f\left(mmol\right)& 2& & 0& & 2& & Excess\end{array}$

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

  $\begin{array}{c}pH=p{K}_a+\log \frac{n_b}{n_a}\\ \\ =p{K}_a+\log \frac{2}{2}=p{K}_a+0\\ \\ =p{K}_a=-\log \left(K_a\right)\\ \\ =-\log \left(4.0\times {10}^{-8}\right)\\ \\ =7.34\end{array}$

Addition of $7.6mLof0.500MNaOH$ to $10.00mLof0.400MHOCl:$

Calculation of milimole of acid and base:

  $\begin{array}{l}AmountofHOCl=10.00\cancel{mL}\times \left(\frac{0.400\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=4mi\lim ol\\ \\ AmountofO{H}^{-}=7.6\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=3.8mi\lim ol\end{array}$

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.

  $\begin{array}{cccccccc}& HOCl& +& O{H}^{-}& \to & OC{l}^{-}& +& H_{2}O\\ s\left(mmol\right)& 4& & 3.8& & 0& & Excess\\ R\left(mmol\right)& -3.8& & -3.8& & +3.8& & +3.8\\ f\left(mmol\right)& 0.2& & 0& & 3.8& & Excess\end{array}$

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

  $\begin{array}{c}pH=p{K}_a+\log \frac{n_b}{n_a}\\ \\ =p{K}_a+\log \frac{3.8}{0.2}=p{K}_a+1.278\\ \\ =-\log \left(K_a\right)+1.278\\ \\ =-\log \left(4.0\times {10}^{-8}\right)+1.278\\ \\ =7.34+1.278=8.62.\end{array}$

Addition of $8mLof0.500MNaOH$ to $10.00mLof0.400MHOCl:$

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHOCl=10.00\cancel{mL}\times \left(\frac{0.400\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=4mi\lim ol\\ \\ AmountofO{H}^{-}=8\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=4mi\lim ol\\ \\ Totalvolume=18mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

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

  $\begin{array}{cccccccc}& HOCl& +& O{H}^{-}& \to & OC{l}^{-}& +& H_{2}O\\ s\left(mmol\right)& 4& & 4& & 0& & Excess\\ R\left(mmol\right)& -4& & -4& & +4& & +4\\ f\left(mmol\right)& 0& & 0& & 4& & Excess\\ c\left(M\right)& 0& & 0& & 0.2222& & Excess\end{array}$

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.

$\begin{array}{cccccccc}& OC{l}^{-}& +& H_{2}O& \rightleftharpoons & HOCl& +& O{H}^{-}\\ i\left(M\right)& 0.2222& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.2222-y& & & & y& & y\end{array}$

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

  $K_b=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[OC{l}^{-}\right]}=\frac{y^2}{\left(0.2222-y\right)}$

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

  $K_b=\frac{K_w}{K_a}=\frac{1.0\times {10}^{-14}}{4.0\times {10}^{-8}}=0.25\times {10}^{-6}$

This can be solved by approximation.  If $y\ll 0.2222$, then

  $\begin{array}{l}{y}^2\approx 0.25\times {10}^{-6}\times 0.2222=0.056\times {10}^{-6}\\ \\ y=\sqrt{0.056\times {10}^{-6}}=2.4\times {10}^{-4}=\left[O{H}^{-}\right]\end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.4\times {10}^{-4}\right)=3.62\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-3.62=10.38\end{array}$

Addition of $8.4mLof0.500MNaOH$ to $10.00mLof0.400MHOCl:$

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHOCl=10.00\cancel{mL}\times \left(\frac{0.400\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=4mi\lim ol\\ \\ AmountofO{H}^{-}=8.4\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=4.2mi\lim ol\\ \\ Totalvolume=18.4mL\end{array}$

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.

$\begin{array}{cccccccc}& HOCl& +& O{H}^{-}& \to & OC{l}^{-}& +& H_{2}O\\ s\left(mmol\right)& 4& & 4.2& & 0& & Excess\\ R\left(mmol\right)& -4& & -4& & +4& & +4\\ f\left(mmol\right)& 0& & 0.2& & 4& & Excess\\ c\left(M\right)& 0& & 0.011& & 0.217& & Excess\end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.011\right)=1.96\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-1.96=12.04\end{array}$

Titration curve:

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

  $\begin{array}{cc}VolumeofNaOH\left(mL\right)& pH\\ 0& 3.89\\ 4& 7.34\\ 7.6& 8.62\\ 8& 10.38\\ 8.4& 12.04\end{array}$

The titration curve with four important regions is given below.

Figure