Q19.144 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A $\mathbf{35}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{mL}$ solution of  $\mathbf{0}\mathbf{.}\mathbf{2500}\mathbf{\text{M }}\mathbf{HF}$is titrated with a standardized  $\mathbf{0}\mathbf{.}\mathbf{1532}\mathbf{ }\mathbf{M}$ solution of   $\mathbf{\text{NaoH}}$at ${\mathbf{25}}^{\mathbf{\circ }}\mathit{C}$. 

(a) What is the $\mathbf{pH}$ of the $\mathbf{HF}$ solution before titrant is added? 

(b) How many millilitres of titrant are required to reach the equivalence point? 

(c) What is the  $\mathbf{pH}$ at $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{mL}$  before the equivalence point? 

(d) What is the $\mathbf{pH}$  at the equivalence point? 

(e) What is the  $\mathbf{pH}$ at $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{mL}$  after the equivalence point?

Step-by-Step Solution

Answer

(a) The $\text{pH}$ of the $\text{HF}$solution before titrant is added is  $\text{pH=1.88}$.

(b) The volume of titrant required to reach the equivalence point is $57.115 \text{mL}$ .

(c) The  $\text{pH}$ at $0.50 \text{mL}$ before the equivalence point is  $\text{pH}=5.2$.

(d) The  $\text{pH}$ at the equivalence point is $\text{pH=8.07}$ .

(e) The  $\text{pH}$ at   after the equivalence point is  $\text{pH=5.27}$.

Step 1: Concept Introduction.

Titration, also known as titrimetry, is a chemical qualitative analytical technique for determining the concentration of a specific analyte in a mixture. Titration is a crucial technique in analytical chemistry, and it's also known as volumetric analysis.

The amount of titrant applied is just enough to totally neutralize the analyte solution at the equivalence point in the titration.

Step 2: pH of HF solution.

(a)

 $\text{HF}$being a weak acid dissociates slowly in the solution producing a smaller number of ${\mathrm{H}}^{+}$. The reaction is expressed as –

$\mathrm{HF}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{F}}^{-}$

Given that initially the concentration of  $\text{HF}$is $0.25 \text{M}$.

	$\text{HF}$	${\text{H}}^{+}$	${\text{F}}^{-}$
Initial	$0.25 \text{M}$	$0$	$0$
Change	$-\text{X}$	$+\mathrm{x}$	$+\mathrm{x}$
Equilibrium	$0.25 \text{M-x}$	$+\mathrm{x}$	$+\mathrm{x}$

Using the information above, the  ${\text{K}}_a$ can be calculated by –

 $\begin{array}{c}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}\\ 6.8\times {10}^4=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{(0.25-\mathrm{x})}\\ \mathrm{x}=0.0130\mathrm{M}\end{array}$

The  $\text{pH}$ value can be calculated using the formula below –

$\begin{array}{c}\mathrm{pH} = -\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pH} = -\log (0.0130)\\ \mathrm{pH} = 1.88\end{array}$

Therefore, the value for $\text{pH}$ is obtained as $\text{pH=1.88}$ .

Step 3: Volume of titrant required.

(b)

Volume of $\text{HF}$  taken is $35 \text{mL}$.

Strength of $\text{HF}$  solution is $0.25 \text{M}$.

Strength of   $\text{NaOH}$ solution taken is $0.1532 \text{M}$.

The Volume of $\text{NaOH}$ that was taken can be calculated using the formula below –

 $\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{NaOH} = \frac{\left(35\mathrm{mL}\right)(0.25\mathrm{M})}{0.1532\mathrm{mL}}\\ = 57.115\mathrm{mL}\end{array}$

 

Therefore, the value for volume is obtained as  $57.115 \text{mL}$.

Step 4: before the equivalence point.

$35 \text{mL}$of$0.25 \text{M}$solution of$\text{HF}$is taken

The Number of Moles of$\text{HF}$taken can be calculate by –

$\begin{array}{c}\mathrm{Number} \mathrm{of} \mathrm{Moles} = \frac{\left(35\mathrm{mL}\right)(0.25\mathrm{M})}{1000\mathrm{mL}}\\ = 0.00875\mathrm{mol}\end{array}$

The$\text{pH}$at $0.50 \text{mL}$before equivalence point has to be calculated by –

$\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{NaOH} \mathrm{to} \mathrm{be} \mathrm{added} = (57.115-0.50) = 56.615 \mathrm{mL}\\ \mathrm{Number} \mathrm{of} \mathrm{NaOH} \mathrm{moles} \mathrm{to} \mathrm{be} \mathrm{added} = \frac{(56.615\mathrm{mL}\left)\right(0.1532\mathrm{mol})}{1000\mathrm{ml}}\\ = 0.00867 \mathrm{mol}\\ \mathrm{Excess} \mathrm{amount} \mathrm{of} \mathrm{HF} \mathrm{formed} = (0.00875-0.00867) \mathrm{moles}\\ = 0.00008 \mathrm{moles}\\ \mathrm{Volume} \mathrm{of} \mathrm{Total} \mathrm{Solution} = (35+57.115-0.50)\\ = 91.615\times {10}^{-3} \mathrm{L}\end{array}$

Thus, the concentration of$\text{HF}$and$\text{F}$is calculated by –

$\begin{array}{c}\left[\mathrm{HF}\right] = \frac{0.00008 \mathrm{moles}}{91.615\times {10}^{-3} \mathrm{L}} = 0.00087 \mathrm{M}\\ \left[{\mathrm{F}}^{-}\right] =\frac{0.0867 \mathrm{mol}}{91.615\times {10}^{-3} \mathrm{L}} = 0.0946 \mathrm{M}\end{array}$

The  ${\text{pK}}_a$ value can be calculated as –

$\begin{array}{c}{\mathrm{pK}}_{\mathrm{a}}= -{\mathrm{logK}}_{\mathrm{a}} = -\log \left(6.8\times {10}^{-4}\right)\\ {\mathrm{pK}}_{\mathrm{a}}= 3.167\end{array}$

Using the Henderson-Hasselbalch equation, it is obtained that –

$\begin{array}{c}\mathrm{pH} = {\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{\left[{\mathrm{F}}^{-}\right.}{\left[\mathrm{HF}\right]}\right) = (3.167)+\log \left(\frac{0.0946}{0.00087}\right)\\ \mathrm{pH} = 5.2\end{array}$

Therefore, the value for  $\text{pH}$ is obtained as  $\text{pH=5.2}$.

Step 5: pH at the equivalence point.

(d)

At equivalence point the total volume can calculated as –

$\begin{array}{c}\mathrm{Total} \mathrm{Volume} = (35+51.115)= 92.115 \mathrm{mL}\\ \mathrm{Molarity} \mathrm{of} {\mathrm{F}}^{-}= \frac{(0.00087\left)\right(1000 \mathrm{mL})}{92.115 \mathrm{mL}}= 0.09499 \mathrm{M}\end{array}$

The acid dissociation of the base can be calculated as –

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}} =\frac{1\times {10}^{-14}}{6.8\times {10}^{-4}}\\ {\mathrm{K}}_{\mathrm{b}}= 1.47\times {10}^{-11}\end{array}$

Thus, calculating the concentration of ${\text{OH}}^{-}$will go as follows –

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}}=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{0.09499-\mathrm{x}}\\ \mathrm{x} = 1.182\times {10}^{-6} \mathrm{M}\\ \left[{\mathrm{OH}}^{-}\right] = 1.182\times {10}^{-6} \mathrm{M}\end{array}$

Concentration of ${\text{H}}^{+}$can be calculated by –

$\begin{array}{c}\left[{\mathrm{H}}^{+}\right]=\frac{1\times {10}^4}{1.182\times {10}^{-6}}\\ =8.46\times {10}^{-9}\end{array}$

Thus, $\text{pH}$ can be calculated as –

$\begin{array}{c}\mathrm{pH}= -\log \left[{\mathrm{H}}^{+}\right]= -\log \left(8.46\times {10}^{-9}\right)\\ \mathrm{pH}= 8.07\end{array}$

Therefore, the value for $\text{pH}$ is obtained as $\text{pH=8.07}$.

Step 6: pH after the equivalence point.

(e)$35 \text{mL}$of$0.25 \text{M}$solution of $\text{HF}$is taken

The Number of Moles of $\text{HF}$taken can be calculate by –

 $\mathrm{Number} \mathrm{of} \mathrm{Moles} =\frac{(35 \mathrm{mL})(0.25 \mathrm{M})}{1000 \mathrm{mL}} = 0.00875 \mathrm{mol}$

The$\text{pH}$at$0.5 \text{mL}$after equivalence point must be calculated by –

$\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{NaOH} \mathrm{to} \mathrm{be} \mathrm{added} = (57.115+0.50)\\ = 57.615 \mathrm{mL}\\ \mathrm{Number} \mathrm{of} \mathrm{NaOH} \mathrm{moles} \mathrm{to} \mathrm{be} \mathrm{added} = \frac{(57.615 \mathrm{mL}\left)\right(0.1532 \mathrm{mol})}{1000 \mathrm{ml}}\\ = 0.00882 \mathrm{mol}\\ \mathrm{Excess} \mathrm{amount} \mathrm{of} \mathrm{HF} \mathrm{formed} = (0.00882-0.00875) \mathrm{moles}\\ = 0.00007 \mathrm{moles}\\ \mathrm{Volume} \mathrm{of} \mathrm{Total} \mathrm{Solution} = (35+57.115+0.50)\\ = 92.615\times {10}^{-3} \mathrm{L}\end{array}$

Thus, the concentration of $\text{HF}$and $\text{F}$is calculated by –

$\begin{array}{c}\left[\mathrm{HF}\right]=\frac{0.00007 \mathrm{moles}}{92.615\times {10}^{-3} \mathrm{L}}\\ =0.00075 \mathrm{M}\\ \left[{\mathrm{F}}^{-}\right]=\frac{0.0882 \mathrm{mol}}{92.615\times {10}^{-3} \mathrm{L}}\\ =0.0952 \mathrm{M}\end{array}$

The ${\text{pK}}_a$ value can be calculated as –

 $\begin{array}{c}{\mathrm{pK}}_{\mathrm{a}}= -{\mathrm{logK}}_{\mathrm{a}}\\ {\mathrm{pK}}_{\mathrm{a}}= -\log \left(6.8\times {10}^{-4}\right)\\ {\mathrm{pK}}_{\mathrm{a}}= 3.167\end{array}$

Using the Henderson-Hasselbalch equation, it is obtained that –

 $\begin{array}{c}\mathrm{pH} = {\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{\left[{\mathrm{F}}^{-}\right.}{\left[\mathrm{HF}\right]}\right) = (3.167)+\log \left(\frac{0.0952}{0.00075}\right)\\ \mathrm{pH} = 5.27\end{array}$

 

Therefore, the value for $\text{pH}$ is obtained as $\text{pH=5.27}$.