Chapter 16, Problem 16.58QE

Interpretation Introduction

Interpretation:

The titration curve has to be drawn and four regions of importance from the given information has to be labeled.

Explanation of Solution

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

  $\begin{array}{l}{N}_1{V}_1=N_2{V}_2\\ \\ V_1=\frac{N_2{V}_2}{N_1}=\frac{0.230M\times 100mL}{0.500M}=46mL\end{array}$

The addition of $0\%,50\%,95\%,100\%,and105\%$ of the base needed to reach the equivalent point means $0\%,50\%,95\%,100\%,and105\%$ of $\text{46}\text{mL}\text{.}$

  $\begin{array}{l}0\%\times 46mL=0mL\\ \\ 50\%\times 46mL=23mL\\ \\ 95\%\times 46mL=43.7mL\\ \\ 100\%\times 46mL=46mL\\ \\ 105\%\times 46mL=48.3mL\end{array}$

Addition of $0mLof0.500MNaOH$to $100.00mLof0.230MHF$:

$HF$ 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.

$\begin{array}{cccccccc}& HF& +& H_{2}O& \rightleftharpoons & H_3{O}^{+}& +& F^{-}\\ i\left(M\right)& 0.230& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.230-y& & & & y& & y\end{array}$

The $K_a$ value for $HF$ is $6.3\times {10}^{-4}.$ The expression for $K_a$ can be written as given below,

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[F^{-}\right]}{\left[HF\right]}=6.3\times {10}^{-4}\\ \\ \frac{y^2}{\left(0.230-y\right)}=6.3\times {10}^{-4}\end{array}$

This can be solved by approximation.  If $y\ll 0.230$, then

  $\begin{array}{l}{y}^2\approx 6.3\times {10}^{-4}\times 0.230=1.449\times {10}^{-4}\\ \\ y=\sqrt{1.449\times {10}^{-4}}=1.20\times {10}^{-2}=\left[H_3{O}^{+}\right]\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(1.20\times {10}^{-2}\right)=1.92$

Addition of $23mLof0.500MNaOH$to $100.00mLof0.230MHF$:

The neutralization reaction can be written as given,

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

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHF=100.00\cancel{mL}\times \left(\frac{0.230\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=23mi\lim ol\\ \\ AmountofO{H}^{-}=23.00\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=11.5mi\lim ol\end{array}$

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

$\begin{array}{cccccccc}& HF& +& O{H}^{-}& \to & F^{-}& +& H_{2}O\\ s\left(mmol\right)& 23& & 11.5& & 0& & Excess\\ R\left(mmol\right)& -11.5& & -11.5& & +11.5& & +11.5\\ f\left(mmol\right)& 11.5& & 0& & 11.5& & Excess\end{array}$

The entire strong base is consumed.  There are weak acid $\left(HF\right)$ and its conjugate base $\left(F^{-}\right)$ in the solution.  So this solution is a buffer.  By using Henderson-Hasselbalch equation, its $pH$ can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{n_b}{n_a}\\ \\ =p{K}_a+\log \frac{11.5}{11.5}=p{K}_a+0\\ \\ =p{K}_a=-\log \left(K_a\right)\\ \\ =-\log \left(6.3\times {10}^{-4}\right)\\ \\ =3.2\end{array}$

Addition of $43.7mLof0.500MNaOH$to $100.00mLof0.230MHF$:

Calculation of milimoles of acid and base:

  $\begin{array}{l}AmountofHF=100.00\cancel{mL}\times \left(\frac{0.230\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=23mi\lim ol\\ \\ AmountofO{H}^{-}=43.7\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=21.85mi\lim ol\end{array}$

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.

  $\begin{array}{cccccccc}& HF& +& O{H}^{-}& \to & F^{-}& +& H_{2}O\\ s\left(mmol\right)& 23& & 21.85& & 0& & Excess\\ R\left(mmol\right)& -21.85& & -21.85& & +21.85& & +21.85\\ f\left(mmol\right)& 1.5& & 0& & 21.85& & Excess\end{array}$

The entire strong base is consumed.  There are weak acid $\left(HF\right)$ and its conjugate base $\left(F^{-}\right)$ in the solution.  So this solution is a buffer.  By using Henderson-Hasselbalch equation, its $pH$ can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{n_b}{n_a}\\ \\ =p{K}_a+\log \frac{21.85}{1.5}=p{K}_a+1.16\\ \\ =-\log \left(K_a\right)+1.16\\ \\ =-\log \left(6.3\times {10}^{-4}\right)+1.16\\ \\ =\text{3}\text{.2}+1.16=4.36\end{array}$

Addition of $46mLof0.500MNaOH$to $100.00mLof0.230MHF$:

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHF=100.00\cancel{mL}\times \left(\frac{0.230\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=23mi\lim ol\\ \\ AmountofO{H}^{-}=46.00\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=23mi\lim ol\\ \\ Totalvolume=100+46=146mL\end{array}$

Making of sRfc table:

The table can be formed as shown below.

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

$\begin{array}{cccccccc}& HF& +& O{H}^{-}& \to & F^{-}& +& H_{2}O\\ s\left(mmol\right)& 23& & 23& & 0& & Excess\\ R\left(mmol\right)& -23& & -23& & +23& & +23\\ f\left(mmol\right)& 0& & 0& & 23& & Excess\\ c\left(M\right)& 0& & 0& & 0.1575& & \end{array}$

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.

$\begin{array}{cccccccc}& F^{-}& +& H_{2}O& \rightleftharpoons & HF& +& O{H}^{-}\\ i\left(M\right)& 0.1575& & & & 0& & 0\\ C\left(M\right)& -y& & & & +y& & +y\\ e\left(M\right)& 0.1575-y& & & & y& & y\end{array}$

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

  $K_b=\frac{\left[HF\right]\left[O{H}^{-}\right]}{\left[F^{-}\right]}=\frac{y^2}{\left(0.1575-y\right)}$

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

  $K_b=\frac{K_w}{K_a}=\frac{1.0\times {10}^{-14}}{6.3\times {10}^{-4}}=1.6\times {10}^{-11}$

This can be solved by approximation.  If $y\ll 0.1575$, then

  $\begin{array}{l}{y}^2\approx 1.6\times {10}^{-11}\times 0.1575=0.252\times {10}^{-11}\\ \\ y=\sqrt{0.252\times {10}^{-11}}=1.6\times {10}^{-6}=\left[O{H}^{-}\right]\end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(1.6\times {10}^{-6}\right)=5.8\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-5.8=8.2\end{array}$

Addition of $48.3mLof0.500MNaOH$ to $100.00mLof0.230MHF$:

Calculation of milimoles of acid and base and total volume:

  $\begin{array}{l}AmountofHF=100.00\cancel{mL}\times \left(\frac{0.230\text{milimol}{H}_3{O}^{+}}{\cancel{mL}}\right)=23mi\lim ol\\ \\ AmountofO{H}^{-}=48.3\cancel{mL}\times \left(\frac{0.500\text{milimol}O{H}^{-}}{\cancel{mL}}\right)=24.15mi\lim ol\\ \\ Totalvolume=100.00mL+48.3mL=148.3mL\end{array}$

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.

$\begin{array}{cccccccc}& HF& +& O{H}^{-}& \to & F^{-}& +& H_{2}O\\ s\left(mmol\right)& 23& & 24.15& & 0& & Excess\\ R\left(mmol\right)& -23& & -23& & +23& & +23\\ f\left(mmol\right)& 0& & 1.15& & 23& & Excess\\ c\left(M\right)& 0& & 7.75\times {10}^{-3}& & 0.1551& & \end{array}$

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

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(7.75\times {10}^{-3}\right)=2.11\\ \\ pH+pOH=14\\ \\ pH=14-pOH=14-2.11=11.89\end{array}$

Titration curve:

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

  $\begin{array}{cc}VolumeofNaOH\left(mL\right)& pH\\ 0& 1.92\\ 23& 3.2\\ 43.7& 4.36\\ 46& 8.2\\ 48.3& 11.89\end{array}$

The titration curve with four important regions is given below.

Figure