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Chapter 15: Problem 49

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 of the acid is \(Ka = 0.000814\).

Step by step solution

01

Converting Percentage Ionization into Concentration

Firstly, calculate the concentration of the ionized acid. Given that the 0.040 M solution of acid is 14 percent ionized, the concentration of ionized acid ([H+] = [A-]) would be 0.14 * 0.040 M = 0.0056 M.
02

Substitute Values into the Ka Formula

Secondly, substitute the values into the equilibrium constant expression for ionization of a monoprotic acid which is \(Ka = [H+][A-]/[HA]\). Given that [H+] = [A-] and the remaining concentration of HA is ([HA] initial - [H+]), the Ka can be calculated as follows: \(Ka = (0.0056)^2 / (0.040 - 0.0056)\).
03

Calculate the Ionization Constant

Finally, carrying out the computation from Step 2 yields Ka. This gives \(Ka = 0.000814\).

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Key Concepts

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

Monoprotic Acid
A monoprotic acid is a type of acid that donates just one hydrogen ion (H鈦�) per molecule to an aqueous solution. This characteristic is crucial because it defines how this acid will behave in reactions and helps in calculating its ionization in a solution.

Examples of monoprotic acids include:
  • Hydrochloric acid (HCl)
  • Acetic acid (CH鈧僀OOH)
  • Nitric acid (HNO鈧�)
Understanding that a monoprotic acid releases one H鈦� ion helps us set up equations for calculating its ionization constant, or Ka. It simplifies calculations because each molecule of acid that ionizes contributes equally to the concentration of H鈦� ions.
Percentage Ionization
Percentage ionization is the measure of how much of the acid dissociates into ions in a solution. This is depicted as a percentage and gives insight into the strength of an acid.

In the exercise, 14 percent ionization means that 14% of the original acid molecules ionize in water.

How to Calculate Percentage Ionization

  • Find the initial concentration of the acid (given as 0.040 M).
  • Multiply by the percentage ionization to find the concentration of ionized acid (0.14 * 0.040 M = 0.0056 M).
This calculation reveals how much the acid ionizes, which is essential when finding the equilibrium constant.
Equilibrium Constant
The equilibrium constant, known as Ka for acids, reflects the extent of ionization in an equilibrium state. A larger Ka value indicates a stronger acid, meaning it dissociates more in solution.

The equilibrium expression for a monoprotic acid (HA) is:\[Ka = \frac{[H^+][A^-]}{[HA]}\]In the formula:
  • [H鈦篯 is the concentration of hydrogen ions.
  • [A鈦籡 is the concentration of the conjugate base.
  • [HA] is the concentration of the undissociated acid.
Substituting the values from our exercise into this formula gives us the ionization constant. This expression is essential for understanding acid strength and behavior in reactions.
Concentration Calculation
Calculating concentrations is often necessary for understanding the behavior of solutions in chemistry. In the exercise, concentrations are derived from given initial values and percentages.

Steps for Calculating Concentration Changes

  • Establish the initial concentration of the acid (e.g., 0.040 M).
  • Determine the ionized concentration from percentage ionization (0.14 of 0.040 M = 0.0056 M).
  • Calculate the remaining concentration of the un-ionized acid using the formula: initial concentration minus ionized concentration (0.040 - 0.0056 M).
These steps help us find the concentrations required to determine the equilibrium constant, reflecting the acid's dissociation level.

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Most popular questions from this chapter

Classify each of the following species as a weak or strong acid: (a) \(\mathrm{HNO}_{3},\) (b) HF, (c) \(\mathrm{H}_{2} \mathrm{SO}_{4},\) (d) \(\mathrm{HSO}_{4}^{-}\), (e) \(\mathrm{H}_{2} \mathrm{CO}_{3},\) (f) \(\mathrm{HCO}_{3}^{-},(\mathrm{g}) \mathrm{HCl}\) (h) \(\mathrm{HCN}\), (i) \(\mathrm{HNO}_{2}\).

Malonic acid \(\left[\mathrm{CH}_{2}(\mathrm{COOH})_{2}\right]\) is a diprotic acid. Compare its two \(K_{\mathrm{a}}\) values with that of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\left(K_{\mathrm{a}}\right),\) and account for the differences in the three \(K_{\mathrm{a}}\) values.

Give an example of the following: (a) a weak acid that contains oxygen atoms, (b) a weak acid that does not contain oxygen atoms, (c) a neutral molecule that acts as a Lewis acid, (d) a neutral molecule that acts as a Lewis base, (e) a weak acid that contains two ionizable \(\mathrm{H}\) atoms, (f) a conjugate acidbase pair, both of which react with \(\mathrm{HCl}\) to give carbon dioxide gas.

Compare the strengths of the following pairs of acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SeO}_{4}\), (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{3} \mathrm{AsO}_{4}\).

A certain salt, MX (containing the \(\mathrm{M}^{+}\) and \(\mathrm{X}^{-}\) ions), is dissolved in water, and the \(\mathrm{pH}\) of the resulting solution is \(7.0 .\) Can you say anything about the strengths of the acid and the base from which the salt is derived?
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