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Benzenesulfonic acid is a monoprotic acid with \(\mathrm{p} K_{a}=2.25 .\) Calculate the \(\mathrm{pH}\) of a buffer composed of \(0.150 \mathrm{M}\) benzenesulfonic acid and \(0.125 \mathrm{M}\) sodium benzensulfonate.

Short Answer

Expert verified
The pH of the buffer solution composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzensulfonate can be calculated using the Henderson-Hasselbalch equation: \(pH = pKa + log_{10} \frac{[A^-]}{[HA]}\). Substituting the given values, we get \(pH = 2.25 + log_{10} \frac{0.125}{0.150} \approx 2.03\). Therefore, the pH of the buffer solution is approximately 2.03.

Step by step solution

01

Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution. The equation is given as: \(pH = pKa + log_{10} \frac{[A^-]}{[HA]}\) where pH is the pH of the buffer solution, pKa is the acid dissociation constant, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In this case, the weak acid is benzenesulfonic acid and its conjugate base is benzensulfonate.
02

Substitute the given values into the equation

In this exercise, we are given the pKa value (2.25), the concentration of the weak acid [HA] (0.150 M), and the concentration of the conjugate base [A^-] (0.125 M). We can substitute these values in the Henderson-Hasselbalch equation: \(pH = 2.25 + log_{10} \frac{0.125}{0.150}\)
03

Calculate the pH of the buffer

By solving the equation, we can calculate the pH of the buffer solution: \(pH = 2.25 + log_{10} (\frac{0.125}{0.150}) \approx 2.25 - 0.22 = 2.03\) The pH of the buffer solution composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzensulfonate is approximately 2.03.

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a fundamental tool in chemistry for estimating the pH of buffer solutions. It relates the pH of a solution to the pKa of the weak acid and the ratio of the concentrations of the conjugate base and weak acid. The equation is expressed as: \[ pH = pKa + \log_{10} \left( \frac{[A^-]}{[HA]} \right) \] In this formula, - **pH** represents the acidity or basicity level of the solution.- **pKa** is the negative logarithm of the acid dissociation constant \(K_a\), a measure of the strength of the acid. - **[A^-]** denotes the concentration of the conjugate base.- **[HA]** represents the concentration of the weak acid.Understanding this equation is crucial when analyzing buffers, as it allows for easy determination of the pH based on known concentrations, revealing how acids and bases can maintain pH stability.
Acid Dissociation Constant
The acid dissociation constant, denoted as \(K_a\), is a quantitative measure of the strength of an acid in solution. It specifically describes the degree to which an acid dissociates into its ions in water. The larger the \(K_a\), the stronger the acid, because it implies a greater tendency to donate protons.The formula for \(K_a\) is: \[ K_a = \frac{[A^-][H^+]}{[HA]} \] This describes the ratio of the concentrations of protons \([H^+]\) and the conjugate base \([A^-]\) to the concentration of the undissociated acid \([HA]\). In practical terms:- A **high \(K_a\)** implies more dissociation, meaning a stronger acid.- A **low \(K_a\)** signifies less dissociation, indicating a weaker acid. pKa is simply the negative logarithm of \(K_a\): \[ pKa = -\log_{10} (K_a) \] It is often more convenient to use in calculations and discussions, as it translates large variations into more manageable numbers.
Conjugate Base
In the context of acid-base chemistry, a conjugate base results when an acid donates a proton during a reaction. It is essentially the remaining portion of the acid, minus one hydrogen ion \(H^+\). Conjugate bases play a pivotal role in maintaining buffer solutions.- For a given acid \([HA]\), the removal of a proton forms its conjugate base \([A^-]\).- Conjugate bases can accept protons back, functioning as bases in the reverse reaction.In buffers, the conjugate base helps neutralize added acids, while the original weak acid can neutralize added bases. This symbiotic relationship helps stabilize the pH within a narrow range, which is crucial in many biological and chemical processes. In the original exercise, sodium benzensulfonate acts as the conjugate base of benzenesulfonic acid.
Weak Acid
Weak acids are acids that do not completely dissociate in solution. This incompleteness in dissociation results in an equilibrium mixture of non-ionized acid molecules and a small concentration of ions.Why are weak acids special? - Weak acids have **smaller \(K_a\) values**, indicating limited ionization.- They serve effectively in buffer solutions because they can release or accept protons when small amounts of acids or bases are added, minimizing pH changes.An example is benzenesulfonic acid in the original exercise, which mildly dissociates to form benzensulfonate and hydronium ions in solution. Understanding weak acids and their properties helps in comprehending how buffer solutions function and resist changes in pH. These characteristics make them invaluable in both laboratory and real-world applications where pH stability is key.

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Most popular questions from this chapter

How many grams of sodium lactate \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\right.\) or \(\left.\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) should be added to \(1.00 \mathrm{~L}\) of \(0.150 \mathrm{M}\) lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to form a buffer solution with \(\mathrm{pH} 4.00\) ? Assume that no volume change occurs when the sodium lactate is added.

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?

Toothenamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right){ }_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box in Section \(17.5\), fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose \(K_{s p}=1.0 \times 10^{-60}\). (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

A buffer contains a weak acid, \(\mathrm{HX}\), and its conjugate base. The weak acid has a \(\mathrm{pK}_{a}\) of \(4.5\), and the buffer solution has a pH of 4.3. Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right],[\mathrm{HX}]>\left[\mathrm{X}^{-}\right]\), or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right]\) Explain. [Section 17.2]

The solubility products of \(\mathrm{PbSO}_{4}\) and \(\mathrm{Sr} \mathrm{SO}_{4}\) are \(6.3 \times 10^{-7}\) and \(3.2 \times 10^{-7}\), respectively. What are the values of \(\left[\mathrm{SO}_{4}{ }^{2-}\right],\left[\mathrm{Pb}^{2+}\right]\), and \(\left[\mathrm{Sr}^{2+}\right]\) in a solution at equilibrium with both substances?

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