Q19.141 CP - Chemistry: Molecular Nature Of Matter And Change

Question

An eco-botanist separates the components of a tropical bark extract by chromatography. She discovers a large proportion of quinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogen’s $(K_{b1}=4.0\times {10}^{-6}$and ${\mathbf{K}}_{\mathbf{b}\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$). To measure the concentration, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogen’s with excess  and titrates the acidified solution with standardized base. A $\mathbf{33}\mathbf{.}\mathbf{85}$-mg sample of quinidine $\mathbf{(}\mathbf{M}\mathbf{=}\mathbf{324}\mathbf{.}\mathbf{41}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)}$is acidified with $\mathbf{6}\mathbf{.}\mathbf{55}\mathbf{mL}$ of $\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{ }\mathbf{M}\mathbf{ }\mathbf{HCl}$.

(a) How many milliliters of $\mathbf{0}\mathbf{.}\mathbf{0133}\mathbf{ }\mathbf{M}\mathbf{ }\mathbf{NaOH}$are needed to titrate the excess $\mathbf{HCl}$?

(b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride?

(c) What is the  at the first equivalence point?

Step-by-Step Solution

Answer

1. The amount of   needed for the neutralization of excess is $58.08 \text{mL}$.
2. The amount of NaOH required to get the first equivalence point is $7.82 \text{mL}$.
3. The value of pH at first equivalence point is $6.302$.

Step 1: Definition of Concept.

Heat Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

Step 2: Find the number of milliliters are needed to titrate the excess HCI.

(a)

Quinine is acidified by HCI because it contains two basic nitrogen atoms.

As a result, one mole of quinine requires two moles of HCl to acidify.

Considering the given information:

Molar mass of quinine is $324.41\mathrm{g}/\mathrm{mol}$.

Quinine present in solution is $33.85 \mathrm{mg}$.

Hence,

$\begin{array}{c}\text{No.ofmolesofquinine} =\frac{\mathrm{weighttaken}}{\mathrm{molarmass}}\\ = \frac{33.85\times {10}^{-3}\mathrm{g}}{324.41\mathrm{g}/\mathrm{mol}}\\ = 1.04\times {10}^{-4} \mathrm{moles}\end{array}$

$2\times$no. of moles of quinine = No. of moles of HCl reacted.

Hence,

$\begin{array}{c}\text{No. of moles of HCl No of moles of HCl fused for quinine} = 1.04\times {10}^{-4}\times 2\mathrm{moles}\\ = 2.08\times {10}^{-4} \mathrm{moles}\end{array}$

Now, the total number of HCl added moles = volume concentration.

As a result,

HCI has a volume of $6.55 \mathrm{mL}$.

HCI has a concentration of $0.150\mathrm{M}$.

$\begin{array}{c}\text{Hence total no. of HCl taken }= \frac{6.55\mathrm{mL}\times 0.150\mathrm{mol}}{1000\mathrm{mL}}\\ = 9.8\times {10}^{-4} \mathrm{moles}\end{array}$

$\begin{array}{l}\text{Excess HCl after quinine acidification,}\\ = \left(9.8\times {10}^{-4}\right)-\left(2.08\times {10}^{-4}\right.)\mathrm{moles}\\ = 7.72\times {10}^{-4}\mathrm{moles}\end{array}$

The excess amount of HCI must be neutralized with NaOH.
$\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{excess} \mathrm{HCl} = \frac{\left[7.72\times {10}^4\mathrm{moles}\times 1000\mathrm{mL}\right]}{[0.150\mathrm{moles}]}\\ = 5.15 \mathrm{mL}\end{array}$

                                              

Given that,

Strength of NaOH is  $0.0133\mathrm{M}$.

Hence,

$\begin{array}{c}\text{Volume of NaOH needed for titration} = \frac{[\mathrm{volumeofacid}\times \mathrm{strengthofacid}]}{\left[\mathrm{strengthofbase}\right]}\\ = \frac{[5.15\mathrm{mL}\times 0.150\mathrm{M}]}{[0.0133\mathrm{M}]}\\ = 58.08 \mathrm{mL}\end{array}$

                                                                    

Therefore, NaOH needed for the neutralization of excess HCI is $58.08 \mathrm{mL}$.

Step 3: Find the additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride.

(b)

The number of moles of quinine dihydrochloride and the number of moles of NaOH required for neutralization must be the same at the equivalence point.

Now from part (a) the no. of moles of quinine dihydrochloride is $1.04\times {10}^{-4} \mathrm{mole}$.

Considering the given information:

Molarity of NaOH is $0.0133\mathrm{M}$.

$\begin{array}{c}\text{Volume of NaOH at equivalence point} = \frac{[\mathrm{volumeofacid}\times \mathrm{strengthofacid}]}{\left[\mathrm{strengthofbase}\right]}\\ = \frac{\left[1.04\times {10}^{-4}\mathrm{mol}\times 1000\mathrm{mL}\right]}{[0.0133\mathrm{mol}]}\\ = 7.82\mathrm{mL}\end{array}$                             

Therefore, amount of NaOH required to get the first equivalence point is $7.82 \mathrm{mL}$.

Step 4: Find the pH at the first equivalence point.

(c)

When you treat quinine with HCI, you get quinine dihydrochloride, which is dispositive, meaning it has two protons. One of the protons is now neutralized at the first equivalence point, leaving one more positive charge to participate in the acid-base reaction.

$\begin{array}{c}\mathrm{pH} = -\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pOH} = -\log \left[{\mathrm{OH}}^{-}\right]\end{array}$

From ionization constant of water at ${25}^{\circ }{\mathrm{CK}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$.

After the first equivalence point, unipositive quinine can act as an acid or a base with water in the medium. As a result, the concentration of ${\mathrm{OH}}^{-}$at the equivalence point must be determined.

$\left[{\mathrm{OH}}^{-}\right]=\sqrt{{\mathrm{K}}_{\mathrm{b}1}\times {\mathrm{K}}_{\mathrm{b}2}}$

(If the concentration of quinine at the first equivalence point is greater than ${\mathrm{K}}_{\mathrm{bl}}$)

Considering the given values:

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}1}=4.0\times {10}^{-6}\\ {\mathrm{K}}_{\mathrm{b}2}=1.0\times {10}^{-10}\end{array}$

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

$\begin{array}{c}\left[{\mathrm{OH}}^{-}\right]=\sqrt{4.0\times {10}^{-6}\times 1.0\times {10}^{-10}} \mathrm{M}\\ =2.0\times {10}^{-8} \mathrm{M}\end{array}$

As a result, pH can be calculated as follows:

$\begin{array}{c}\mathrm{pOH}= -\log \left[{\mathrm{OH}}^{-}\right]\\ \mathrm{pOH}= -\log \left[2.0\times {10}^{-8}\right]\\ \mathrm{pOH}= 7.698\end{array}$

Therefore,

The value of pH is calculated as follows:

$\begin{array}{c}\mathrm{pH} = 14-\mathrm{pOH}\\ = 14-7.698\\ = 6.302\end{array}$

Therefore, the value of pH at first equivalence point is $6.302$.