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Chapter 16: Problem 47

A \(0.040 M\) solution of a monoprotic acid is 14 percent ionized. Calculate the ionization constant of the acid.

Short Answer

Expert verified
The Ionization constant (Ka) of the acid is approximately \(8.8 * 10^{-3}\).

Step by step solution

01

Understand what is given and what is asked

We know that the molar concentration of acid is \(0.040 M\) and the percentage of ionization is \(14\%\). We're asked to determine the Ionization constant (Ka) of the acid.
02

Correlate the percentage ionization with the amount of hydronium ions

The 14 percent ionized means 14 percent of the acid molecules have donated their protons and therefore became hydronium ions. So, we have \(0.040 M * 0.14 = 0.0056 M\) hydronium ions. Note that for every one acid molecule that ionizes, it donates one proton (H+) and becomes one hydronium ion (H3O+). So, the concentration of protons is equal to the concentration of hydronium ions.
03

Determine the Ionization Constant

The ionization constant for a weak acid (Ka) is given by: \[Ka = [H3O+][A–] / [HA]\] Where \([HA]\) is the concentration of the un-ionized acid, \([H3O+]\) is the concentration of the hydronium ion, \([A–]\) is the concentration of the conjugate base. For a monoprotic weak acid, \([H3O+]\) = \([A–]\). Therefore, \[Ka = [H3O+]^2 / (0.040 - [H3O+])\] Substituting the value we have for \([H3O+]\) (which is 0.0056), we find \[Ka= [(0.0056)]^2 / (0.040-0.0056)\] After calculation, we find that the Ionization Constant, Ka, is approximately \(8.8 * 10^{-3}\).

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Most popular questions from this chapter

What are the strongest acid and strongest base that can exist in water?

Predict the acid strengths of the following compounds: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{Se} .\)

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) has the following structure: An oxalic acid solution contains these species in varying concentrations: \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{HC}_{2} \mathrm{O}_{4}^{-}, \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\), and \(\mathrm{H}^{+}\). (a) Draw Lewis structures of \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\). (b) Which of the four species listed here can act only as acids, which can act only as bases, and which can act as both acids and bases?

Both the amide ion \(\left(\mathrm{NH}_{2}^{-}\right)\) and the nitride ion \(\left(\mathrm{N}^{3-}\right)\) are stronger bases than the hydroxide ion and hence do not exist in aqueous solutions. (a) Write the equations showing the reactions of these ions with water, and identify the Bronsted acid and base in each case. (b) Which of the two is the stronger base?

Most of the hydrides of Group \(1 \mathrm{~A}\) and Group \(2 \mathrm{~A}\) metals are ionic (the exceptions are \(\mathrm{BeH}_{2}\) and \(\mathrm{MgH}_{2}\), which are covalent compounds). (a) Describe the reaction between the hydride ion \(\left(\mathrm{H}^{-}\right)\) and water in terms of a Bronsted acid-base reaction. (b) The same reaction can also be classified as a redox reaction. Identify the oxidizing and reducing agents.
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