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Chapter 14: Problem 20

A buffer is prepared by dissolving \(\mathrm{HONH}_{2}\) and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
The buffer solution prepared using HONH2 (a weak base) and HONH3NO3 (its conjugate acid) neutralizes added H+ ions through the following reaction: HONH2 (aq) + H+ (aq) -> HONH3+ (aq) It neutralizes added OH- ions via this reaction: HONH3NO3 (aq) + OH- (aq) -> HONH2 (aq) + H2O (l) These reactions help to maintain the overall pH of the buffer solution and resist significant changes in acidity or alkalinity.

Step by step solution

01

1. Neutralization of added H+ ions

To demonstrate how this buffer neutralizes added H+ ions, we'll focus on the weak base HONH2 present in the mixture. When H+ ions are added to the buffer solution, the weak base reacts with them to form its conjugate acid. Here's the equation representing this process: HONH2 (aq) + H+ (aq) -> HONH3+ (aq) This reaction shows the buffer neutralizing the added H+ ions by forming HONH3+ ions, as the weak base HONH2 consumes the added H+ ions.
02

2. Neutralization of added OH- ions

Now let's analyze how the buffer neutralizes added OH- ions. To do this, we'll focus on the conjugate acid HONH3NO3 present in the buffer solution. When OH- ions are added to the buffer, the conjugate acid reacts with the added OH- ions to form its conjugate base, HONH2, and water. The equation for this reaction is as follows: HONH3NO3 (aq) + OH- (aq) -> HONH2 (aq) + H2O (l) This equation illustrates the process of neutralizing added OH- ions, as the conjugate acid HONH3NO3 reacts with OH- ions to produce the weak base HONH2 and water. In summary, the buffer solution composed of HONH2 and HONH3NO3 neutralizes added H+ ions by allowing the weak base HONH2 to react with them, and added OH- ions by allowing the conjugate acid HONH3NO3 to react with them. This maintains the overall pH of the buffer solution and prevents significant changes in its acidity or alkalinity.

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Most popular questions from this chapter

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=\) \(7.4 \times 10^{-5} .\) Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0\) \(\mathrm{M} ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0\) \(\mathrm {M} ?\)

A \(10.00-g\) sample of the ionic compound \(\mathrm{NaA}\), where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 \(M\) HCl. After 500.0 mL HCl was added, the pH was \(5.00 .\) The experimenter found that 1.00 L of \(0.100 M\) HCl was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(p\) H of the solution at the stoichiometric point of the titration.

A student dissolves 0.0100 mole of an unknown weak base in \(100.0 \mathrm{mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) After \(40.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100 \(M\) lactic acid (HC \(_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), \(\mathrm{p} K_{\mathrm{a}}=3.86\) is titrated with \(0.100 \mathrm{M}\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL}\) \(20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\) \(26.0 \mathrm{mL}, 28.0 \mathrm{mL},\) and \(30.0 \mathrm{mL}\) of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Could a buffered solution be made by mixing aqueous solutions of HCl and NaOH? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?
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