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Chapter 16: Problem 24

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ?

Short Answer

Expert verified
The stronger Brønsted-Lowry acid is \(\mathrm{HClO}_{3}\) and the stronger Brønsted-Lowry base is \(\mathrm{HS}^{-}\).

Step by step solution

01

Identify their conjugate bases

The conjugate base of an acid is formed by subtracting a proton (H+) from the acid. So, for \(\mathrm{HClO}_{3}\) and \(\mathrm{HClO}_{2}\), we will find their respective conjugate bases. \[ \mathrm{HClO}_{3} \rightarrow \mathrm{ClO}_{3}^{-} + \mathrm{H}^{+} \] \[ \mathrm{HClO}_{2} \rightarrow \mathrm{ClO}_{2}^{-} + \mathrm{H}^{+} \] Conjugate bases of these acids are \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{ClO}_{2}^{-}\), respectively.
02

Evaluate the stability of the conjugate bases

The stability of the conjugate bases is affected by the presence of oxygen atoms in these species. Oxygen atoms have a higher electronegativity, and therefore they are better at stabilizing negative charges. More oxygen atoms in the conjugate base will increase its stability. \(\mathrm{ClO}_{3}^{-}\) has more oxygen atoms (3) than \(\mathrm{ClO}_{2}^{-}\) (2). Thus, the increased oxygen presence in \(\mathrm{ClO}_{3}^{-}\) allows it to better stabilize the negative charge, making it a more stable conjugate base.
03

Determine the stronger acid

A stronger acid will have a more stable conjugate base. Since \(\mathrm{ClO}_{3}^{-}\) is more stable than \(\mathrm{ClO}_{2}^{-}\), the stronger Brønsted-Lowry acid is \(\mathrm{HClO}_{3}\). (b) Comparing the strength of \(\mathrm{HS}^{-}\) and \(\mathrm{HSO}_{4}^{-}\) as Bronsted-Lowry bases:
04

Identify their conjugate acids

The conjugate acid of a base is formed by adding a proton (H+) to the base. So, for \(\mathrm{HS}^{-}\) and \(\mathrm{HSO}_{4}^{-}\), we will find their respective conjugate acids. \[ \mathrm{HS}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{S} \] \[ \mathrm{HSO}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{SO}_{4} \] Conjugate acids of these bases are \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{H}_{2}\mathrm{SO}_{4}\), respectively.
05

Evaluate the stability of the conjugate acids

The presence of oxygen atoms in the conjugate acids affects their stability due to the electron-withdrawing inductive effect. A conjugate acid with more oxygen atoms will destabilize the positive charge on the acidic hydrogen more effectively. \(\mathrm{H}_{2}\mathrm{SO}_{4}\) has more oxygen atoms (4) than \(\mathrm{H}_{2}\mathrm{S}\) (0). Thus, the increased oxygen presence in \(\mathrm{H}_{2}\mathrm{SO}_{4}\) allows it to better stabilize the positive charge, making it a more stable conjugate acid.
06

Determine the stronger base

A stronger base will have a less stable conjugate acid. Since \(\mathrm{H}_{2}\mathrm{S}\) is less stable than \(\mathrm{H}_{2}\mathrm{SO}_{4}\), the stronger Brønsted-Lowry base is \(\mathrm{HS}^{-}\). So, the answers are (a) \(\mathrm{HClO}_{3}\) is a stronger Bronsted-Lowry acid, and (b) \(\mathrm{HS}^{-}\) is a stronger Bronsted-Lowry base.

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Most popular questions from this chapter

A 0.100\(M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is 11.0\(\%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right],\) and \(K_{a}\) for chloroacetic acid.

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{pK}_{a}=2.32\) at \(25^{\circ} \mathrm{C} .\) It ionizes in aqueous solution as follows: $$\mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \Longrightarrow \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q)$$ What is the pH of a 0.10\(M\) solution of this substance?

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 M \mathrm{HNO}_{3},(\mathbf{b}) 0.225 \mathrm{g}\) of \(\mathrm{HClO}_{3}\) in 2.00 \(\mathrm{L}\) of solution, \((\mathbf{c}) 15.00 \mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{L},\) (d) a mixture formed by adding 50.0 \(\mathrm{mL}\) of 0.020 \(\mathrm{MHCl}\) to 125 \(\mathrm{mL}\) of 0.010 \(\mathrm{M} \mathrm{HI} .\)

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (B) \(\mathrm{HCO}_{3}^{-}\)

The fluoride ion reacts with water to produce HF. (a) Write out the chemical equation for this reaction. (b) Will a concentrated solution of NaF in water be acidic, basic, or neutral? (c) Is fluoride acting as a Lewis acid or as a Lewis base when reacting with water?
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