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Chapter 14: Problem 75

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

Short Answer

Expert verified
The acid dissociation constant \(K_a\) for the given weak acid is \(1.39 \times 10^{-5}\).

Step by step solution

01

Write down the given information

We are given: 1. A 0.15 M solution of a weak acid. 2. 3.0% dissociated
02

Set up the reaction of the weak acid dissociation

Let the weak acid be represented by HA. The dissociation reaction of the weak acid (HA) can be written as: \(HA <=> H^+ + A^-\) We know that the initial concentration of the weak acid is 0.15 M, and 3.0% of it dissociates. Therefore, we can calculate the change in concentration of each of the species in the equilibrium.
03

Calculate the change in concentration

- HA dissociates by 3.0%, so \(0.15 \times 0.03 = 0.0045\, M\). - The change in concentration of \(H^+\) and \(A^-\) is equal to that of HA, so they increase by 0.0045 M.
04

Write the equilibrium concentrations of each species

In the equilibrium we have: - \(HA: 0.15\, M - 0.0045\, M = 0.1455\, M\) - \(H^+: 0.0045\, M\) - \(A^-: 0.0045\, M\)
05

Set up the Ka expression

The expression for the acid dissociation constant (Ka) is: \(K_a = \dfrac{[H^+][A^-]}{[HA]}\)
06

Plug in the equilibrium concentrations and calculate Ka

Using the calculated equilibrium concentrations from step 4, the Ka expression becomes: \(K_a = \dfrac{(0.0045)(0.0045)}{0.1455}\) Now, calculate the value of K_a: \(K_a = 1.39 \times 10^{-5}\)
07

Writing the final answer

The acid dissociation constant \(K_a\) for the given weak acid is \(1.39 \times 10^{-5}\).

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Most popular questions from this chapter

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\). a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) b. Calculate the \(\mathrm{pH}\) of this solution.

The following illustration displays the relative number of species when an acid, HA, is added to water. a. Is HA a weak or strong acid? How can you tell? b. Using the relative numbers given in the illustration, determine the value for \(K_{\mathrm{a}}\) and the percent dissociation of the acid. Assume the initial acid concentration is \(0.20 \mathrm{M}\).

A \(0.100-\mathrm{g}\) sample of the weak acid HA (molar mass \(=\) \(100.0 \mathrm{~g} / \mathrm{mol}\) ) is dissolved in \(500.0 \mathrm{~g}\) water. The freezing point of the resulting solution is \(-0.0056^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{a}}\) for this acid. Assume molality equals molarity in this solution.

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Brønsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10-M\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\).
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