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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

A 0.210 -g sample of an acid (molar mass \(=192 \mathrm{g} / \mathrm{mol}\) ) is titrated with \(30.5 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

The active ingredient in aspirin is acetylsalicylic acid. A \(2.51-\mathrm{g}\) sample of acetylsalicylic acid required \(27.36 \mathrm{mL}\) of \(0.5106 M\) NaOH for complete reaction. Addition of \(13.68 \mathrm{mL}\) of \(0.5106 \mathrm{M}\) HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with \(\mathrm{pH}=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{\mathrm{a}}\) value. State any assumptions you must make to reach your answer.

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

Consider the following four titrations (i-iv): i. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) by \(0.2 \mathrm{M} \mathrm{HCl}\) ii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M}\) HCl by \(0.2 \mathrm{M} \mathrm{NaOH}\) iii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) iv. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) a. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the halfway point to equivalence (lowest to highest \(\mathrm{pH}\) ). b. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or \(\mathrm{NaOH}\) ) to reach the equivalence point?

Consider the titration of \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) NaOH. a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)
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