An ecobotanist 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 nitrogens \(\left(K_{\mathrm{b} 1}=4.0 \times 10^{-6}\right.\) and \(\left.K_{\mathrm{b} 2}=1.0 \times 10^{-\mathrm{i} 0}\right) .\) To measure the concentration of quinidine, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens with excess \(\mathrm{HCl}\) and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine \((\mathscr{M}=324.41 \mathrm{~g} / \mathrm{mol})\) is acidified with \(6.55 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (a) How many milliliters of \(0.0133 \mathrm{M} \mathrm{NaOH}\) are needed to titrate the excess HCl? (b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride? (c) What is the \(\mathrm{pH}\) at the first equivalence point?

Step by step solution

01

- Calculate moles of quinidine

Use the mass of quinidine and its molar mass to find the number of moles: \[ \text{moles of quinidine} = \frac{33.85 \text{ mg}}{324.41 \text{ g/mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 1.044 \times 10^{-4} \text{ mol} \]

02

- Calculate moles of HCl added

Use the volume and concentration of HCl to calculate the number of moles: \[ \text{moles of HCl} = 6.55 \text{ mL} \times 0.150 \text{ M} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 9.825 \times 10^{-4} \text{ mol} \]

03

- Calculate excess moles of HCl

Subtract the moles of HCl that reacted with quinidine from the total moles of HCl added: \[ \text{moles of excess HCl} = 9.825 \times 10^{-4} \text{ mol} - 2 \times 1.044 \times 10^{-4} \text{ mol} = 7.737 \times 10^{-4} \text{ mol} \]

04

- Calculate volume of NaOH needed to titrate excess HCl

Use the moles of excess HCl and the concentration of NaOH to find the volume of NaOH needed: \[ \text{volume of NaOH} = \frac{7.737 \times 10^{-4} \text{ mol}}{0.0133 \text{ M}} = 58.2 \text{ mL} \]

05

- Calculate NaOH needed to reach the first equivalence point

Use the moles of quinidine that has one protonated nitrogen: \[ \text{moles of quinidine} = 1.044 \times 10^{-4} \text{ mol} \] This means same amount of NaOH needed: \[ \text{volume of NaOH} = \frac{1.044 \times 10^{-4} \text{ mol}}{0.0133 \text{ M}} = 7.85 \text{ mL} \]

06

- Calculate additional NaOH for the first equivalence point

Only additional NaOH needed since it already titrated excess HCl: \[ 7.85 \text{ mL} \]

07

- Calculate pH at the first equivalence point

Use pH to pKa conversion:\[ \text{pOH} = \frac{pK_{\text{b}} 1}{\text{2}} - \frac{\text{0.301}}{\text{2}} = 5.70 \] Therefore: \[ \text{pH} = 14 - 5.70 = 8.30 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

chromatography

Chromatography is a powerful technique used to separate components of a mixture. In this exercise, an ecobotanist uses chromatography to separate the components of a tropical bark extract. Chromatography works based on the different affinities of substances for a stationary phase and a mobile phase. Those substances that have a higher affinity for the stationary phase will move slower, while those with a higher affinity for the mobile phase will move faster. This separation technique is essential in various fields such as chemistry, biology, and environmental science. By separating these components, the ecobotanist identifies quinidine, a compound used for controlling arrhythmic heartbeats.

equivalence point

The equivalence point in titration is the point at which the quantity of titrant added is sufficient to completely neutralize the analyte in solution. In the given exercise, you observe two equivalence points related to the protonation of quinidine's two basic nitrogens. When reaching each equivalence point, the number of moles of titrant (NaOH in this case) is stoichiometrically equal to the number of moles of analyte (quinidine), or the excess HCl that hadn't neutralized initially.
This concept is crucial to understand because it indicates when the reaction is completed, allowing accurate determination of concentration and understanding of reaction dynamics. In our calculations, we find not only the volume of NaOH needed to titrate the excess HCl but also the volume required to reach the first equivalence point of quinidine dihydrochloride.

pH calculation

Knowing how to calculate the pH at various points during a titration is vital for understanding the acid-base properties of the solution. pH calculation involves determining the concentration of hydrogen ions \(\text{H}^{+}\) in the solution.
In our exercise, to find the pH at the first equivalence point, we initially find the pOH by taking the negative log of the concentration of hydroxide ions \(\text{OH}^{-}\). The formula is:
\[ \text{pOH} = \frac{pK_{\text{b}} 1}{\text{2}} \]
We then convert the pOH to pH using the relationship:
\[ \text{pH} = 14 - \text{pOH} \]
Understanding these calculations allows you to determine the acidity or basicity of the solution at any stage of the titration, providing deeper insight into the chemical behavior of the substances involved.

moles and molarity

Moles and molarity are fundamental concepts in chemistry that relate to the quantity of a substance. The number of moles represents the quantity of a substance based on Avogadro’s number, with one mole being equal to approximately \({6.022} \times 10^{23} \) particles, atoms, or molecules.
Molarity (M) is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula is:
\[ \text{Molarity} \ (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the exercise, we calculate the moles of quinidine using its mass and molar mass, and we use the molarity of HCl and NaOH to determine the volumes needed for the titration. Understanding moles and molarity is crucial for performing accurate stoichiometric calculations in any chemical reaction.

quinidine

Quinidine is an important compound derived from the tropical bark extract, identified using chromatography. It is a dextrorotatory isomer of quinine and is used primarily to treat arrhythmic heartbeat. Quinidine has two basic nitrogens that can react with acids and bases, making it an ideal subject for acid-base titration experiments.
In our exercise, the ecobotanist needed to determine the concentration of quinidine by first protonating both nitrogens with excess HCl. It is crucial to understand that quinidine's basic nature and its interaction with acids are what make titration possible. By understanding these interactions, scientists can accurately measure and use quinidine in medical applications. Through the titration process and the calculations of equivalence points, we derive crucial information about quinidine's concentration and its behavior in the solution, showcasing its practical relevance in pharmacology.