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Chapter 13: Problem 63

What are the major species present in 0.250 \(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Short Answer

Expert verified
In a 0.250 M solution of HNO2, the major species present are HNO2, its conjugate base NO2-, and water (H2O) molecules. The pH of this solution is approximately 1.35. In a 0.250 M solution of CH3CO2H (HC2H3O2), the major species present are CH3CO2H (HC2H3O2), its conjugate base CH3CO2- (C2H3O2-), and water (H2O) molecules. The pH of this solution is approximately 2.63.

Step by step solution

01

Identifying Major Species of HNO2

HNO2, known as nitrous acid, is a weak acid. In a 0.250 M solution of HNO2, the major species present will be HNO2 itself, along with its conjugate base NO2-, and water (H2O) molecules.
02

Identifying Major Species of CH3CO2H (HC2H3O2)

CH3CO2H (HC2H3O2), known as acetic acid, is also a weak acid. In a 0.250 M solution, the major species present will be CH3CO2H (HC2H3O2) itself, along with its conjugate base CH3CO2- (C2H3O2-), and water (H2O) molecules. Now, let's find the pH for each solution.
03

Calculate pH for HNO2 solution

First, we need the Ka value for HNO2. The Ka value for HNO2 is \(4.50 \times 10^{-4}\). Now, set up the equilibrium expression: \(Ka = \frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]}\) Since the problem states the initial concentration of HNO2 is 0.250 M, we can use the following concentrations changes at equilibrium: - HNO2: 0.250 - x - H+: x - NO2-: x Plugging in the values, we get: \(4.50 \times 10^{-4} = \frac{x^2}{0.250-x}\) Solve for x to find the hydrogen ion concentration, [H+]. Then, calculate the pH as follows: \(pH = -\log[H^{+}]\)
04

Calculate pH for CH3CO2H (HC2H3O2) solution

For acetic acid, the Ka value is \(1.76 \times 10^{-5}\). Set up the equilibrium expression: \(Ka = \frac{[H^{+}][CH_{3}CO_{2}^{-}]}{[CH_{3}CO_{2}H]}\) Since the initial concentration of CH3CO2H (HC2H3O2) is 0.250 M, we can use the following concentrations changes at equilibrium: - CH3CO2H (HC2H3O2): 0.250 - x - H+: x - CH3CO2- (C2H3O2-): x Plugging in the values, we get: \(1.76 \times 10^{-5} = \frac{x^2}{0.250-x}\) Solve for x to find the hydrogen ions concentration, [H+]. Then, calculate the pH as follows: \(pH = -\log[H^{+}]\) Once you've calculated the [H+] for each solution and found the pH values, the problem is solved.

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Most popular questions from this chapter

The pOH of a sample of baking soda dissolved in water is 5.74 at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is the solution acidic or basic?

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