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Chapter 14: Problem 73

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

Short Answer

Expert verified
The dissociation constant of the weak acid, \(K_a\), is approximately \(1.39 × 10^{-4}\).

Step by step solution

01

Determine the initial concentration of the weak acid (HA)

The initial concentration of the weak acid is given as \(0.15 M\).
02

Calculate the dissociated concentration of HA

The percent dissociation is given as \(3.0\%\). To find the dissociated concentration, we will multiply the initial concentration by the percent dissociation: Dissociated concentration of HA = Initial concentration × Percent dissociation Dissociated concentration of HA = \(0.15 M × 0.03 = 0.0045 M\)
03

Calculate the concentrations of A⁻ and H⁺ ions after dissociation

As the dissociation of HA produces equal concentrations of A⁻ and H⁺ ions, the concentration of both ions after dissociation will be equal to the dissociated concentration of HA. [𝐴⁻] = [𝐻⁺] = 0.0045 M
04

Calculate the remaining concentration of HA after dissociation

Since we know the initial concentration of HA and the dissociated concentration, we can calculate the remaining concentration after dissociation: Remaining concentration of HA = Initial concentration - Dissociated concentration Remaining concentration of HA = 0.15 M - 0.0045 M = 0.1455 M
05

Calculate the dissociation constant (Ka)

Now that we know the concentrations of A⁻, H⁺, and HA after dissociation, we can use the formula for Ka: \[K_a = \frac{[\mathrm{A^-}][\mathrm{H^+}]}{[\mathrm{HA}]}\] Plug in the concentrations obtained in Steps 3 and 4: \[K_a = \frac{(0.0045 M)(0.0045 M)}{(0.1455 M)}\]
06

Compute the result

Finally, calculate the Ka value: \[K_a = \frac{(0.0045 M)(0.0045 M)}{(0.1455 M)} = 1.39 × 10^{-4}\] The dissociation constant of the weak acid, \(K_a\), is approximately \(1.39 × 10^{-4}\).

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Most popular questions from this chapter

The \(\mathrm{pH}\) of a sample of gastric juice in a person's stomach is \(2.1\). Calculate the \(\mathrm{pOH},\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is gastric juice acidic or basic?

A typical sample of vinegar has a pH of \(3.0\). Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in vinegar.

Arrange the following \(0.10 M\) solutions in order of most acidic to most basic. \(\mathrm{KOH}, \quad \mathrm{KNO}_{3}, \quad \mathrm{KCN}, \quad \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{HCl}\).

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5 \times 10^{-3}\) \(K_{\mathrm{a}_{2}}=8 \times 10^{-8}\), and \(K_{\mathrm{a}_{3}}=6 \times 10^{-10} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}^{3-}\right]\) in a \(0.20 \mathrm{M}\) arsenic acid solution.

A \(0.20 \mathrm{M}\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a \(\mathrm{pH}\) of \(8.65 .\) Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{M}\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.
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