Chapter 19, Problem 19.139P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ Of $\text{HF}$ solution has to be calculated before $\text{NaOH}$ is added.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution . It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Answer to Problem 19.139P

$\text{pH}$ of solution before addition of titrant $\text{NaOH}$ is 1.88.

Explanation of Solution

HF being a weak acid dissociates slowly in the solution producing less number of ${\text{H}}^{\text{+}}$.

  $\text{HF}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{F}}^{\text{-}}$

Let the dissociation constant be x.

Given that initially the concentration of HF is $\text{0}\text{.25 M}$.

  $\begin{array}{c}\text{HF }\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}\text{+}{\text{F}}^{\text{-}}\\ \\ \text{initially}\text{0}\text{.25}\text{0}\text{0}\\ \\ \text{change}\text{-x}\text{+x}\text{+x}\\ \\ \text{equilibrium}\left(\text{0}\text{.25-x}\right)\text{x}\text{ x}\end{array}$

${\text{Acid dissociation constant (K}}_{\text{a}}\text{)=}\frac{\left[{\text{H}}^{\text{+}}\right]\text{×}\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}$

Now, substituting the values,

  $\text{6}{\text{.8×10}}^{\text{-4}}\text{=}\frac{\text{x×x}}{\left(\text{0}\text{.25-x}\right)}$

  (${\text{K}}_{\text{a}}\text{=6}\text{.8}\times {\text{10}}^{-4}$, taken from text book)

$x<<1$ Hence $\left(\text{0}\text{.25}\text{-}\text{x}\right)\approx \text{0}\text{.25}$

  $\therefore {\text{x}}^{\text{2}}\text{=}\text{6}{\text{.8×10}}^{\text{-4}}\text{×0}\text{.25}$

  $\therefore$$\text{x}\text{=}\text{0}\text{.0130}\text{M}$

Hence, concentration of ${\text{H}}^{\text{+}}$,

  $\left[{\text{H}}^{\text{+}}\right]\text{=}\text{0}\text{.0130}\text{M}$

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

  $\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$

  $\begin{array}{l}\therefore \text{pH}=-\log 0.0130\\ \\ =1.88\end{array}$

Hence $\text{pH}$ of solution before addition of titrant $\text{NaOH}$ is 1.88.

(b)

Interpretation Introduction

Interpretation:

Amount of titrant ($\text{NaOH}$) is required to achieve the equivalence point has to be calculated in milliliters.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution . It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Answer to Problem 19.139P

The amount of titrant required to achieve equivalence point is $\text{57}\text{.115}\text{mL}.$

Explanation of Solution

Volume of $\text{HF}$ taken is $3\text{5}\text{mL}$.

Strength of $\text{HF}$ solution is $\text{0}\text{.25}\text{M}$.

Strength of $\text{NaOH}$ solution take is $\text{0}\text{.1532}\text{M}$.

According to volumetric titration,

  $\left[\begin{array}{l}\left(\text{Volume}\text{of}\text{acid}\text{taken}\right)\text{×}\\ \left(\text{Strength}\text{of}\text{acid}\text{taken}\right)\end{array}\right]\text{ = }\left[\begin{array}{l}\left(\text{Volume}\text{of}\text{base}\text{taken}\right)\text{×}\\ \left(\text{Strength}\text{of}\text{base}\text{taken}\right)\end{array}\right]$

Therefore,

  $\begin{array}{l}\text{Volume of NaOH taken = }\frac{\left(\text{35}\text{mL}\text{×}\text{0}\text{.25}\text{M}\right)}{\left(\text{0}\text{.1532M}\right)}\\ \\ \text{ =}\text{ 57}\text{.115}\text{mL}\end{array}$

Hence, the amount of titrant required to achieve equivalence point is $\text{57}\text{.115}\text{mL}.$

(c)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ at $0.\text{50}\text{mL}$ before equivalence point has to be calculated.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution . It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Answer to Problem 19.139P

$\text{pH}$ at $0.\text{50}\text{mL}$ before equivalence point is $5.203$.

Explanation of Solution

$\text{HF}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{F}}^{\text{-}}$

Given that $\text{35}\text{mL}\text{of}\text{0}\text{.25}\text{M}$ solution of HF is taken.

$\begin{array}{l}\text{Total no}\text{. of moles of HF taken}\text{=}\frac{\left(\text{35mL×0}\text{.25mol}\right)}{\text{1000mL}}\\ \\ \text{=}\text{0}\text{.00875}\text{mole}\end{array}$

Now according to question $\text{pH}$ at $\text{0}\text{.50}\text{ml}$ before equivalence point has to be calculated.

  $\begin{array}{l}\begin{array}{l}\text{Hence the volume of NaOH to be added }\text{=}\left(\text{57}\text{.115 – 0}\text{.50 }\right)\text{mL}\text{.}\\ \end{array}\hfill \\ \text{=}\text{56}\text{.615 mL}\hfill \end{array}$

  $\begin{array}{l}\text{Hence no}\text{. of moles of NaOH added}\text{= }\frac{\left(\text{56}\text{.615ml×0}\text{.1532mol}\right)}{\text{1000ml}}\\ \\ \text{=}\text{0}\text{.00867}\text{mole}\end{array}$

Number of moles of ${\text{F}}^{\text{-}}$ formed are equals to the no. of moles of $\text{NaOH}$ added.

Hence number of moles of ${\text{F}}^{\text{-}}$ formed $\text{=}\text{0}\text{.00867}\text{moles}$.

  $\begin{array}{l}\begin{array}{l}\text{Hence excess amount of HF formed=}\left(\text{0}\text{.00875 - 0}\text{.00867}\right)\text{moles}\\ \end{array}\hfill \\ \text{=}\text{0}\text{.00008}\text{moles}\text{.}\hfill \end{array}$

  $\begin{array}{l}\text{Volume of total solution}\text{=}\left(\text{35+57}\text{.115-0}\text{.50}\right)\text{mL}\hfill \\ \begin{array}{c}\\ \text{ =}\text{91}\text{.615}\text{mL}\\ \\ =\text{ 91}\text{.615}\times {\text{10}}^{-3}\text{ L}\end{array}\hfill \end{array}$

Thus the concentration of $\text{HF}$ is calculated as,

  $\begin{array}{l}\left[\text{HF}\right]\text{=}\frac{\left(\text{0}\text{.00008mol}\right)}{\text{91}\text{.615}\times {\text{10}}^{-3}\text{ L}}\\ \\ \text{=}\text{0}\text{.00087}\text{M}\end{array}$ (Excess $\text{HF}$)

Thus the concentration of ${\text{F}}^{\text{-}}$ is calculated as,

  $\begin{array}{l}\left[{\text{F}}^{\text{-}}\right]\text{=}\frac{\left(\text{0}\text{.00867mol}\right)}{\text{91}\text{.615}\times {\text{10}}^{-3}\text{ L}}\\ \\ \text{=}\text{0}\text{.0946}\text{M}\end{array}$

${\text{K}}_{\text{a}}\text{=6}\text{.8}\times {\text{10}}^{-4}$

Hence, ${\text{pK}}_{\text{a}}$ can be calculated as,

  $\begin{array}{l}{\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}\\ \\ =-\log (6.8\times {10}^{-4})\\ \\ =3.167\end{array}$

  ${\text{pK}}_{\text{a}}=3.167$

From Henderson-Hasselbalch equation,

  $\text{pH=}{\text{pK}}_{\text{a}}\text{+ log}\frac{\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}$

Hence, $\text{pH}$ can be calculated as,

  $\text{pH=}3.167+\text{ log}\frac{0.0946}{0.00087}$

  $\text{pH=}\text{5}\text{.203}$

Hence, $\text{pH}$ at $0.\text{50}\text{mL}$ before equivalence point is $5.203$.

(d)

Interpretation Introduction

Interpretation:

$\text{pH}$ At equivalence point has to be calculated.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution . It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Answer to Problem 19.139P

$\text{pH}$ at equivalence point is $8.07$.

Explanation of Solution

At equivalence point we are considering the equation:

  ${\text{F}}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }\text{HF}\text{+}{\text{OH}}^{\text{-}}$

At equivalence point the number of moles of  ${\text{F}}^{\text{-}}$ formed is $\text{0}\text{.00875}\text{M}$.because at equivalence point the $\text{pH}$ will be only due to the excess ${\text{OH}}^{\text{-}}$ in the medium.

  $\begin{array}{l}\text{At equivalence point the total volume is}\text{=}\left(\text{35 + 57}\text{.115}\right)\text{ml }\\ \text{=}\text{92}\text{.115}\text{ml}\end{array}$

  $\begin{array}{c}\text{Molarity of}{\text{F}}^{\text{-}}\text{= }\frac{\left(\text{0}\text{.00875mol}\text{×}\text{1000mL}\right)}{\text{92}\text{.115mL}}\\ \\ \text{=}\text{0}\text{.09499}\text{M}\end{array}$

Now, ${\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{ = K}}_{\text{w}}$

Thus the acid dissociation constant of the base can be calculated as,

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{6}{\text{.8×10}}^{\text{-4}}}\\ \left({\text{at 25}}^{\text{o}}\text{C}{\text{K}}_{\text{w}}\text{=}{\text{1×10}}^{\text{-14}}\right)\\ \text{=}\text{1}{\text{.47×10}}^{\text{-11}}\end{array}$

  ${\text{K}}_{\text{b}}=\frac{\left[\text{HF}\right]\times \left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{F}}^{\text{-}}\right]}$

Say concentration of ${\text{OH}}^{\text{-}}$ at equilibrium is x.

  $\begin{array}{l}{\text{F}}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }\text{HF}\text{+}{\text{OH}}^{\text{-}}\\ \\ \text{initially}\text{0}\text{.09499}\text{0}\text{0}\\ \\ \text{change}\text{-x}\text{+x}\text{+x}\\ \\ \text{equilibrium}\left(\text{0}\text{.09499-x}\right)\text{x}\text{ x}\end{array}$

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x}\text{×}\text{x}}{\left(\text{0}\text{.09499-x}\right)}$

  $\therefore \text{x}\text{=}1.182\times {10}^{-6}\text{M}$

Thus the concentration of ${\text{OH}}^{\text{-}}$ is,

  $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.182×10}}^{\text{-6}}\text{M}$

At ${\text{25}}^{\text{ο}}\text{C}$,

  $\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]={\text{K}}_{\text{w}}$

Hence, concentration of ${\text{H}}^{+}$ is,

  $\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}$

  $\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{1}{\text{.182×10}}^{\text{-6}}}\\ \\ \text{=}\text{8}{\text{.46×10}}^{\text{-9}}\text{M}\end{array}$

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

  $\text{pH}=\text{-log}\left[{\text{H}}^{\text{+}}\right]$

  $\begin{array}{l}\therefore \text{pH}=-\log \left(8.46\times {10}^{-9}\right)\\ \\ =8.07\end{array}$

Hence $\text{pH}$ at equivalence point is $8.07$.

(e)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ at $0.\text{50}\text{mL}$ has to be calculated after equivalence point.

Concept introduction:

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Equivalence point:

Equivalence point in the titration reaction is the point where the amount of titrant added is absolutely enough to neutralize completely the analyte. The moles of titrant and the moles of analyte are same at this point.

pH:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution . It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Answer to Problem 19.139P

The value of $\text{pH}$ at $0.\text{50}\text{mL}$ after equivalence point is $10.92$.

Explanation of Solution

After equivalence point there will be excess $\text{NaOH}$ and so the $\text{pH}$ will be due to the excess ${\text{OH}}^{\text{-}}$ in the solution.

  $\begin{array}{c}\text{After}\text{0}\text{.50}\text{ml of equivalence point the total volume =}\left(\text{35 + 57}\text{.115 + 0}\text{.50}\right)\text{mL }\\ \\ \text{=}\text{92}\text{.615}\text{mL}\end{array}$

  $\begin{array}{l}\text{Moles of excess NaOH}\text{=}\frac{\left(\text{0}\text{.50ml}\text{×}\text{0}\text{.1532mol}\right)}{\text{1000mL}}\\ \\ \text{=}\text{7}{\text{.66×10}}^{\text{-5}}\text{moles}\end{array}$

Thus concentration of ${\text{OH}}^{\text{-}}$ is,

  $\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\left(\text{7}{\text{.66×10}}^{\text{-5}}\text{mol}\right)}{\text{92}\text{.615}\times {\text{10}}^{-3}\text{L}}\\ \\ \text{=}\text{8}{\text{.271×10}}^{\text{-4}}\text{M}\end{array}$

At ${\text{25}}^{\text{ο}}\text{C}$,

  $\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]={\text{K}}_{\text{w}}$

Hence, concentration of ${\text{H}}^{+}$ can be calculated as,

  $\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}$

  $\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{8}{\text{.271×10}}^{\text{-4}}}\text{M}\\ \\ \text{=}\text{1}{\text{.209×10}}^{\text{-11}}\text{M}\end{array}$

Thus $\text{pH}$ becomes,

  $\text{pH}=\text{-log}\left[{\text{H}}^{\text{+}}\right]$

$\begin{array}{l} \text{pH}=-\log (1.209\times {10}^{-11})\\ \\ =10.92\end{array}$

The value of $\text{pH}$ at $0.\text{50}\text{mL}$ after equivalence point is $10.92$.