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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

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?

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH - to form A Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right)\). Thus how can we claim that a buffered solution resists changes in the pH of the solution?" How would you explain buffering to this friend?

A buffer is made using \(45.0 \mathrm{mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=\right.\) \(1.3 \times 10^{-5}\) ) and \(55.0 \mathrm{mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} .\) What volume of 0.10 \(M\) NaOH must be added to change the pH of the original buffer solution by \(2.5 \% ?\)

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125 \(M\) NaOH, requiring \(16.00 \mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding \(2.00 \mathrm{mL}\) NaOH was \(6.912 .\) Calculate \(K_{\mathrm{a}}\) for the weak acid.

Consider a solution formed by mixing \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HOCl}, 25.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\), \(25.0 \mathrm{mL}\) of \(0.100 M \mathrm{Ba}(\mathrm{OH})_{2},\) and \(10.0 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\) Calculate the pH of this solution.
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