Novocaine, which is used by dentists as a local anesthetic, is a weak base with \(\mathrm{pK}_{\mathrm{b}}=5.05\). Blood has a pH of \(7.4\). What is the ratio of concentrations of Novocaine to its conjugate acid in the bloodstream?

Short Answer

Expert verified
The ratio of concentrations of Novocaine to its conjugate acid in the bloodstream is approximately 562.34 to 1.

Step by step solution

01

Understand the Relationship between pKa, pKb, and pH

Remember that for water at 25°C, \( pK_a + pK_b = pK_w \), which is 14. First, convert \( pK_b \) to \( pK_a \) for Novocaine by using the formula \( pK_a = 14 - pK_b \) to proceed with the calculation involving the Henderson-Hasselbalch equation.
02

Calculate the pKa of Novocaine

The \( pK_a \) of Novocaine can be calculated using the relationship \( pK_a = 14 - pK_b \). Substitute \( pK_b = 5.05 \) into the equation to find \( pK_a \) of Novocaine.
03

Apply the Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation \( pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{conjugate acid}]}\right) \) to find the ratio \( \frac{[\text{base}]}{[\text{conjugate acid}]} \). Rearrange the equation to solve for \( \log\left(\frac{[\text{base}]}{[\text{conjugate acid}]}\right) \) by isolating it on one side.
04

Solve for the Ratio

Plug in the known values of \( pH \) and \( pK_a \) into the rearranged Henderson-Hasselbalch equation and solve for \( \frac{[\text{base}]}{[\text{conjugate acid}]} \) by taking the antilogarithm of the result.

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Key Concepts

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

pKa and pKb relationship
When studying acid-base chemistry, it's essential to understand the relationship between the pKa and pKb values. They are indicators of the strengths of acids and bases, respectively. The pKa value measures the tendency of a compound to lose a proton (H+), whereas the pKb measures the tendency to gain a proton. For water at 25°C, the sum of pKa and pKb is constant and equal to 14, known as the pKw.

This is expressed by the equation: \[ pK_a + pK_b = pK_w \]
So, if you know either the pKa or pKb of a substance, you can easily calculate the other. For instance, with Novocaine—a weak base with a given pKb value—you can determine its pKa by subtracting pKb from 14. This conversion is crucial for using the Henderson-Hasselbalch equation, which requires the pKa value to calculate the pH of a solution or, as in our exercise, the ratio of base to conjugate acid concentrations.
Weak Base Equilibria
Weak bases do not completely dissociate in water. Instead, they establish an equilibrium between the unprotonated base (B) and their conjugate acid (BH+), which is represented in the equation:\[ B + H_2O \leftrightarrow BH^+ + OH^- \]
In this equilibrium, only a proportion of the base is converted into its conjugate acid. The position of this equilibrium, and thus the relative concentrations of the base and its conjugate acid, are determined by the base's pKb value: the lower the pKb, the stronger the base, and the more the equilibrium lies to the right. Understanding this equilibrium is key to many calculations and predictions in acid-base chemistry.

The Henderson-Hasselbalch equation often applies to these equilibria as it can calculate the pH of buffered solutions, or in our case, the ratio of concentrations of a base to its conjugate acid given their pKa and the pH of the solution.
pH Calculation
pH is a measure of the acidity or basicity of a solution. It is the negative logarithm of the concentration of hydrogen ions (\(H^+\)) in a solution:\[ pH = -\log[H^+] \]
The pH scale generally ranges from 0 to 14, with lower values being more acidic, higher values more basic, and 7 being neutral, like pure water at 25°C. For solutions of weak bases like Novocaine, the pH is not straightforward to measure directly because the base does not dissociate completely.

Instead, we use the Henderson-Hasselbalch equation, which relates the pH of the solution to the pKa and the ratio of concentrations of the base and its conjugate acid. As our problem shows, knowing the pH of blood and the pKa of Novocaine allows us to calculate the ratio of Novocaine to its conjugate acid in the bloodstream.
Conjugate Acids and Bases
Each base has a conjugate acid that forms when the base accepts a proton, just as every acid has a conjugate base that remains when the acid donates a proton. This concept of conjugate pairs is fundamental in acid-base chemistry. For example, when Novocaine – a weak base – accepts a proton, it becomes its conjugate acid.

The strength of a base is related to the tendency of its conjugate acid to donate a proton. The stronger the base, the weaker its conjugate acid. The Henderson-Hasselbalch equation, used in our textbook exercise, is based on the assumption that acid-base reactions reach an equilibrium state, involving weak acids or bases and their conjugate partners, and thus allows for the quantification of this equilibrium position in terms of pH, pKa, and concentration ratios.

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Most popular questions from this chapter

Determine the \(K_{\text {sp }}\) for the following sparingly soluble compounds, given their molar solubilities: (a) AgI, \(9.1 \times 10^{-9} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\), \(0.011 \mathrm{~mol}-\mathrm{L}^{-1}\); (c) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, 2.7 \times 10^{-6} \mathrm{~mol}^{-\mathrm{L}}^{-1}\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, 5.2 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\).

The narcotic cocaine is a weak base wirh \(\mathrm{p} K_{\mathrm{b}}=5.59\). Calculate the ratio of the concentration of cocaine and its conjugate acid in a solution of \(\mathrm{pH}=8.00\).

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of (a) a solution that is \(0.17 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4}\) (aq) and \(0.25 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) (aq); (b) a solution that is \(0.66 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq})\) and \(0.42 \mathrm{M}\) \(\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ;\) (c) a solution that is \(0.12 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq})\) and \(0.12 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})\).

A 20-mL sample of \(0.020 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) was titrated with \(0.035 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\). Calculate the \(\mathrm{pH}\) at the following points in the titration and sketch the \(\mathrm{pH}\) curve: (a) no \(\mathrm{KOH}\) added; (b) \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) added; (c) an additional \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) (for a total of \(10.0 \mathrm{~mL}\).) added; (d) another \(5.0 \mathrm{~mL}\) of \(\mathrm{KOH}(\mathrm{aq})\) added; (e) another \(5.00 \mathrm{~mL}\). \(\mathrm{KOH}(\mathrm{aq})\) added. (f) Determine the volume of \(\mathrm{KOH}\) (aq) required to reach the stoichiometric point.

An old bottle labcled "Standardized \(6.0 \mathrm{M}\) \(\mathrm{NaOH}^{\prime \prime}\) was found on the back of a shelf in the stockroom. Over time, some of the NaOH had reacted with the glass and the solution was no longer \(6.0 \mathrm{M}\). To determine its purity, \(5.0 \mathrm{~mL}\) of the solution was diluted to \(100 \mathrm{ml}\). and titrated to the stoichiometric point with 11.8 mL. of \(2.05 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\). What is the molarity of the sodium hydroxide solution?

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