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For each of the following cases, decide whether the \(\mathrm{pH}\) is less than \(7,\) equal to \(7,\) or greater than 7 (a) Equal volumes of 0.10 M acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) and \(0.10 \mathrm{M} \mathrm{KOH}\) are mixed. (b) \(25 \mathrm{mL}\) of \(0.015 \mathrm{M} \mathrm{NH}_{3}\) is mixed with \(25 \mathrm{mL}\) of 0.015 M HCl. (c) \(150 \mathrm{mL}\) of \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(75 \mathrm{mL}\) of 0.40 M NaOH.

Short Answer

Expert verified
(a) pH > 7, (b) pH < 7, (c) pH = 7.

Step by step solution

01

Determine Reaction Type (a)

In case (a), equal volumes of 0.10 M acetic acid (CH鈧僀O鈧侶) and 0.10 M KOH are mixed. Both are weak acid and strong base, respectively. The reaction is a neutralization leading to the formation of CH鈧僀O鈧傗伝 (acetate ions) and H鈧侽. Acetate ions create a basic solution, leading to a pH > 7.
02

Calculate Products and pH (a)

Since the base KOH completely neutralizes the weak acid CH鈧僀O鈧侶, the acetic acid is converted into its conjugate base, acetate (CH鈧僀O鈧傗伝). The resulting solution contains a weak base, causing the pH to be greater than 7.
03

Determine Reaction Type (b)

In case (b), 25 mL of 0.015 M NH鈧 is mixed with 25 mL of 0.015 M HCl. Both solutions are equal in volume and concentration, and NH鈧 is a weak base reacting with a strong acid HCl in a 1:1 molar proportion, leading to complete neutralization, forming NH鈧凜l, a weak acid. Thus, the pH is slightly below 7.
04

Calculate Products and pH (b)

The equal moles of NH鈧 and HCl completely react to form NH鈧勨伜, increasing the acidity as NH鈧勨伜 solutions are slightly acidic. Therefore, the resulting pH will be less than 7 due to H鈦 from NH鈧勨伜 dissociation.
05

Determine Reaction Type (c)

In case (c), 150 mL of 0.20 M HNO鈧 is mixed with 75 mL of 0.40 M NaOH. Since both substances are strong acid and strong base respectively, we calculate moles to understand the reaction's complete or incomplete state.
06

Calculate Moles and Compare (c)

Calculate moles: moles of HNO鈧 = 0.150 L 脳 0.20 M = 0.030 mol; moles of NaOH = 0.075 L 脳 0.40 M = 0.030 mol. Moles are equal, resulting in complete neutralization into NaNO鈧 and H鈧侽, making the pH neutral and equal to 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Reactions
Acid-base reactions involve the transfer of protons from the acid to the base. This fundamental chemical interaction determines the acidity or basicity of solutions. In these reactions, acids, which are proton donors, react with bases, which are proton acceptors, to form water and a salt. The ability of an acid or base to donate or accept protons varies depending on the strength of the acid or base involved. The products of an acid-base reaction are neutralization, which could shift the pH depending on the resultant products.
  • Acids are hydrogen ion (H鈦) donors.
  • Bases are hydrogen ion (H鈦) acceptors.
  • The reaction typically yields water and a salt.
Learning to identify whether the resultant mixture after such reactions is acidic, basic, or neutral plays a crucial role in understanding the behavior of different substances in a solution.
Neutralization
Neutralization is a crucial concept in understanding acid-base reactions. It occurs when an acid and a base fully react to form water and a salt. The end product is neither as strongly acidic nor as strongly basic as the reactants were before mixing. In the exercise, we see examples of different neutralization reactions:
  • Case (a) mixes acetic acid, a weak acid, with KOH, a strong base. The base completely neutralizes the acid, leading to a basic solution with pH greater than 7.
  • Case (b) involves NH鈧, a weak base, and HCl, a strong acid. They neutralize each other, but because HCl is stronger, the resultant solution is slightly acidic, with pH less than 7.
  • Case (c) has equal moles of HNO鈧 and NaOH, both strong acid and strong base respectively, perfectly neutralizing each other to achieve a pH equal to 7.
Understanding neutralization helps predict the pH changes during chemical reactions involving acids and bases.
Strong and Weak Acids and Bases
The strength of an acid or base influences the extent of dissociation of its molecules in water, which in turn affects the solution's pH. Strong acids and bases disassociate completely in water, while weak acids and bases only disassociate partially. This distinction plays a significant role in determining the pH of solutions and the outcome of acid-base reactions:
  • Strong acids like HCl and HNO鈧 provide more H鈦 ions, leading to lower pH values.
  • Weak acids like acetic acid release fewer H鈦 ions.
  • Strong bases like NaOH fully dissociate, releasing more OH鈦 ions and raising the pH.
  • Weak bases like NH鈧 only partially dissociate in solution.
Comprehending these properties allows us to predict and calculate the pH levels in solutions formed from these acids and bases, as seen in the exercise cases. Choosing strong or weak acids or bases in reactions influences the strength of the neutralization and whether the resulting solution is more acidic or basic.
Buffer Solutions
Buffer solutions play a vital role in maintaining the pH level of a solution despite the addition of an acid or base. A buffer consists of a weak acid and its conjugate base or a weak base and its conjugate acid. This combination stabilizes the pH by neutralizing added acids or bases through reversible reactions. While the exercise provided does not directly involve buffer solutions, understanding them is crucial:
  • Buffers resist significant pH changes.
  • They work best within a certain pH range related to their pKa values.
  • The presence of both weak acid and its conjugate base allows it to neutralize small amounts of additional acid or base.
These solutions maintain stability in biochemical and industrial processes where maintaining a consistent pH level is essential. A deep understanding of buffers enriches one's comprehension of complex reactions and biological systems.

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Most popular questions from this chapter

Which of the following common food additives would give a basic solution when dissolved in water? (a) \(\mathrm{NaNO}_{3}\) (used as a meat preservative) (b) \(\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\) (sodium benzoate; used as a softdrink preservative) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) (used as an emulsifier in the manufacture of pasteurized cheese)

A weak base has \(K_{\mathrm{b}}=1.5 \times 10^{-9} .\) What is the value of \(K_{\mathrm{a}}\) for the conjugate acid?

The base ethylamine \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)\) has a \(K_{\mathrm{b}}\) of \(4.3 \times 10^{-4} .\) A closely related base, ethanolamine \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right),\) has a \(K_{\mathrm{b}}\) of \(3.2 \times 10^{-5}\) (a) Which of the two bases is stronger? (b) Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of the stronger base.

In each of the following acid-base reactions, identify the Bronsted acid and base on the left and their conjugate partners on the right. (a) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightleftarrows\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightleftarrows\) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Decide whether each of the following substances should be classified as a Lewis acid or a Lewis base. (a) \(\mathrm{H}_{2} \mathrm{NOH}\) in the reaction \(\mathrm{H}_{2} \mathrm{NOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\left[\mathrm{H}_{3} \mathrm{NOH}\right] \mathrm{Cl}(\mathrm{aq})\) (b) \(\mathrm{Fe}^{2+}\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (Hint: Draw the electron dot structure.

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