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Chapter 16: Problem 22

A \(0.500 \mathrm{M}\) solution of a weak acid, HA, is \(0.68 \%\) ionized. Calculate the ionization constant, \(K_{\mathrm{a}}\), for the acid.

Short Answer

Expert verified
The ionization constant, \(K_{\text{a}}\), is \(2.33 \times 10^{-5}\).

Step by step solution

01

Identify Given Data

The initial concentration of the weak acid, HA, is given as 0.500 M and the percentage of ionization is 0.68%.
02

Calculate Ionized Concentration

Determine the concentration of HA that ionizes by using the percentage ionization: \[ \text{Ionized concentration} = 0.500 \text{ M} \times \frac{0.68}{100} = 0.0034 \text{ M} \]
03

Calculate Equilibrium Concentrations

Since the ionized form, HA, dissociates as follows: \[ HA \rightleftharpoons H^+ + A^- \]the concentrations of both \(H^+\) and \(A^-\) ions will be equal to the ionized concentration, \(0.0034 \text{ M}\). The concentration of the undissociated acid at equilibrium will be: \[ [HA] = 0.500 \text{ M} - 0.0034 \text{ M} = 0.4966 \text{ M} \]
04

Set up the Ka Expression

The ionization constant \(K_{\text{a}}\) is expressed as: \[ K_{\text{a}} = \frac{[H^+][A^-]}{[HA]} \]
05

Substitute Values and Solve for Ka

Substitute the equilibrium concentrations into the expression: \[ K_{\text{a}} = \frac{(0.0034 \text{ M}) (0.0034 \text{ M})}{0.4966 \text{ M}} \] \[ K_{\text{a}} = \frac{0.00001156 \text{ M}^2}{0.4966 \text{ M}} \] \[ K_{\text{a}} = 2.33 \times 10^{-5} \]

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Key Concepts

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

Weak Acid Ionization
When a weak acid, such as HA, dissolves in water, it partially ionizes into its ions. Unlike strong acids, which completely ionize, weak acids only dissociate slightly. Let's use the exercise example where a 0.500 M solution of HA is 0.68% ionized. This tells us only a small fraction of HA (0.68%) breaks apart into ions.
Acid Dissociation Constant
The acid dissociation constant, denoted as Ka, helps us understand how much a weak acid dissociates in water. In the example, we calculated the concentration of ionized HA as 0.0034 M. Next, we found the equilibrium concentrations of the ions formed: H^+ and A^-. To calculate Ka, use the expression:

Ka = ( [H^+] [A^-] ) / [HA]
The initial concentration of HA was given as 0.500 M, and the percentage of ionization was 0.68%, so our next step is calculating the ionization constant Ka.
Equilibrium Concentrations
Equilibrium concentrations are crucial for solving problems involving weak acids. Once we know how much of the weak acid has ionized, we can determine the equilibrium concentrations of all species in the solution. Using the example:
  • For HA (undissociated): 0.4966 M
  • For H^+: 0.0034 M
  • For A^-: 0.0034 M
    Substituting these values into the ionization constant expression gives us the value of Ka, providing insight into the strength of the acid.

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Most popular questions from this chapter

If it takes \(0.048 \mathrm{~g} \mathrm{BaF}_{2}\) to saturate \(15.0 \mathrm{~mL}\) of water, what is the \(K_{\mathrm{sp}}\) of \(\mathrm{BaF}_{2}\) ?

A \(0.37 M\) solution of a weak acid (HA) has a pH of \(3.7\). What is the \(K_{\mathrm{a}}\) for this acid?

If \(\mathrm{BaCl}_{2}\) is added to a saturated \(\mathrm{BaSO}_{4}\) solution until the \(\left[\mathrm{Ba}^{2+}\right]\) is \(0.050 \mathrm{M}\), (a) what concentration of \(\mathrm{SO}_{4}^{2-}\) remains in solution? (b) how much \(\mathrm{BaSO}_{4}\) remains dissolved in 100 . \(\mathrm{mL}\) of the solution? \(\left(K_{\mathrm{sp}}=1.5 \times 10^{-9}\right.\) for \(\left.\mathrm{BaSO}_{4}\right)\)

Calculate the \(\left[\mathrm{H}^{+}\right]\)in each of the following solutions: (a) \(\left[\mathrm{OH}^{-}\right]=4.5 \times 10^{-2}\) (b) \(\left[\mathrm{OH}^{-}\right]=5.2 \times 10^{-9}\)

One day in a laboratory, some water spilled on a table. In just a few minutes the water had evaporated. Some days later, a similar amount of water spilled again. This time, the water remained on the table after 7 or 8 hours. Name three conditions that could have changed in the lab to cause this difference.
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