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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 \(\begin{array}{lll}0.150 & M \text { benzenesulfonic acid and } 0.125 \mathrm{M} \text { sodium }\end{array}\) benzensulfonate.

Short Answer

Expert verified
The pH of the buffer solution composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzenesulfonate is approximately 2.17.

Step by step solution

01

Identify the given information

In this exercise, we are given the following information: - pKa of benzenesulfonic acid: 2.25 - Concentration of benzenesulfonic acid: 0.150 M - Concentration of sodium benzenesulfonate: 0.125 M
02

Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a simplified form of the Ka expression for a weak acid and its conjugate base in a buffer solution. The equation is as follows: \( pH = pKa + \log \frac{[A^-]}{[HA]} \) 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 problem, we are given values for pKa, [A^-], and [HA], and we are asked to find the pH.
03

Plug in the given values into the equation

Using the given values, we can plug them into the Henderson-Hasselbalch equation: \( pH = 2.25 + \log \frac{0.125}{0.150} \)
04

Calculate the pH value

Now, we need to perform the calculations within the equation to find the pH value: \( pH = 2.25 + \log \frac{0.125}{0.150} = 2.25 + \log(0.833) \approx 2.25 + (-0.08) \) \( pH \approx 2.17 \) The pH of the buffer solution composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzenesulfonate is approximately 2.17.

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Key Concepts

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

pKa
The term 'pKa' refers to the acid dissociation constant, but it is expressed in a logarithmic scale for better manageability. It quantifies the strength of an acid in a solution, defining how easily the acid can donate a proton (H+). The lower the pKa value, the stronger the acid, because it more readily gives up its proton. This means that benzenesulfonic acid with a pKa of 2.25 is a relatively strong acid as it has a low pKa value, indicating a high tendency to dissociate in solution. Understanding this concept is essential, as it is used in the Henderson-Hasselbalch equation to calculate pH.

Understanding pKa also helps predict the degree of dissociation of an acid at different pH levels. For instance, when the pH of a solution equals the pKa of the solute acid, there is an equal concentration of the acid and its conjugate base, which is a key point in the buffer region.
pH of a buffer solution
The pH of a buffer solution is its measure of acidity or basicity. A buffer solution is composed of a weak acid and its conjugate base and is designed to maintain a stable pH level despite additions of acids or bases. This stability is a direct result of the buffer's ability to neutralize added acid or base, with the weak acid donating protons (H+) when the pH increases (basic substance added), and the conjugate base capturing protons when the pH decreases (acidic substance added).

The Henderson-Hasselbalch equation is particularly useful for calculating the pH of buffer solutions. Due to the relationship between pKa and the ratio of the species in the buffer (acid and conjugate base), we can determine the expected pH when we know the concentrations of these components. In the given exercise, the equation facilitates the prediction of the buffer's pH with known concentrations of benzenesulfonic acid and sodium benzenesulfonate.
Acid dissociation constant
The acid dissociation constant, represented as Ka, is a quantitative measure of the strength of an acid in solution. It specifically describes the equilibrium state wherein an acid, HA, dissociates into its conjugate base, A-, and a proton, H+. The formula for this expression is:\[\begin{equation}Ka = \frac{[A^-][H^+]}{[HA]}\end{equation}\]Where [HA] is the concentration of the acid, [A-] is the concentration of the conjugate base, and [H+] is the concentration of the free protons, or hydrogen ions. A higher Ka value indicates a stronger acid because it implies that more acid has dissociated into its constituents at equilibrium. By taking the negative logarithm of Ka, we get the pKa value, which offers a clearer comparison between acids due to its smaller and more workable numbers.
Conjugate base
A conjugate base is what remains after an acid has donated a proton during a chemical reaction. This concept is pivotal in understanding how buffer solutions work. The conjugate base, usually denoted as A-, is able to react with excess hydrogen ions (H+), in the solution, thus resisting changes in pH. In our exercise, the sodium benzenesulfonate acts as the conjugate base of benzenesulfonic acid.

In buffer equations, the concentration of the conjugate base relative to the concentration of its corresponding acid determines the pH of the solution. If the ratio of conjugate base to acid is higher, the pH will be above the pKa, indicating a more basic solution, and vice versa. Hence, the balance between the acid and its conjugate base is vital for the buffer's ability to mitigate changes in pH upon the addition of other acids or bases.

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Most popular questions from this chapter

Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppb}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is 8 ppb. Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

(a) Precipitation of the group 4 cations (Figure \(17.22)\) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

Show that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{pK}_{a}\) for the acid.

What is the pH of a solution made by mixing \(0.30 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.20 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(1.00 \mathrm{~L}\) ?

Seawatercontains \(0.13 \%\) magnesium by mass, and has a density of \(1.025 \mathrm{~g} / \mathrm{mL}\). What fraction of the magnesium can be removed by adding a stoichiometric quantity of \(\mathrm{CaO}\) (that is, one mole of \(\mathrm{CaO}\) for each mole of \(\mathrm{Mg}^{2+}\) )?

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