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Chapter 14: Problem 172

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. \(\mathrm{NH}_{3}\) b. KOH c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KCl}\) e. HCl

Short Answer

Expert verified
The solutions can be ranked in order of increasing pH as follows: HCl (strong acid), \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (weak acid), \(\mathrm{KCl}\) (neutral), \(\mathrm{NH}_{3}\) (weak base), and KOH (strong base).

Step by step solution

01

Identify the nature of each compound

We have five compounds: \(\mathrm{NH}_{3}\), KOH, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), \(\mathrm{KCl}\), and HCl. Let's identify their nature: a. \(\mathrm{NH}_{3}\) is a weak base b. KOH is a strong base c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is a weak acid d. \(\mathrm{KCl}\) is a salt of a strong acid and strong base, thus neutral e. HCl is a strong acid
02

Determine the pH for each solution

Now we will calculate the pH for each solution, considering their nature and concentration. a. Ammonia (\(\mathrm{NH}_{3}\)): As a weak base, we need to use a simplified version of the Henderson-Hasselbalch equation: \(\mathrm{pOH} = \mathrm{pK}_{b} - \log{\frac{[\mathrm{NH}_{3}]}{[\mathrm{OH}^-]}}\). However, for ranking purpose, we don't need the exact pH value. b. Potassium hydroxide (KOH): As a strong base, it fully dissociates in water: \(\mathrm{KOH} \rightarrow \mathrm{K}^+ + \mathrm{OH}^-\). Hence, \([\mathrm{OH}^-]\) will be \(0.10\,\mathrm{M}\). We can calculate the pOH as \(\mathrm{pOH} = -\log{[\mathrm{OH}^-]}\), and then find the pH using \(\mathrm{pH} = 14 - \mathrm{pOH}\). c. Acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)): As a weak acid, we can also use the simplified version of the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{pK}_{a} - \log{\frac{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]}{[\mathrm{H}^+]}}\). Again, we don't need the exact pH value for ranking purposes. d. Potassium chloride (\(\mathrm{KCl}\)): As a neutral salt, the pH will be 7. e. Hydrochloric acid (HCl): As a strong acid, it fully dissociates in water: \(\mathrm{HCl} \rightarrow \mathrm{H}^+ + \mathrm{Cl}^-\). Hence, \([\mathrm{H}^+]\) will be \(0.10\,\mathrm{M}\). We can calculate the pH as \(\mathrm{pH} = -\log{[\mathrm{H}^+]}\).
03

Rank the solutions in order of increasing pH

Based on the pH calculations and the nature of each compound, we can rank the solutions as follows: 1. Hydrochloric acid (HCl): strong acid 2. Acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)): weak acid 3. Potassium chloride (\(\mathrm{KCl}\)): neutral 4. Ammonia (\(\mathrm{NH}_{3}\)): weak base 5. Potassium hydroxide (KOH): strong base

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Key Concepts

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

Weak and Strong Acids
Acids are substances that donate protons (\( ext{H}^+ \)) in a solution. They can be classified into two categories: weak and strong acids.

**Strong Acids**
  • Completely dissociate in water, releasing all their protons.
  • Have a high concentration of \( ext{H}^+ \), resulting in a low pH value.
  • Example: Hydrochloric acid (\( ext{HCl} \)) fully dissociates to \( ext{H}^+ \) and \( ext{Cl}^- \).
**Weak Acids**
  • Partially dissociate in water, releasing fewer protons.
  • Have a higher pH compared to strong acids at the same concentration.
  • Example: Acetic acid (\( ext{HC}_2 ext{H}_3 ext{O}_2 \)) only partially dissociates.
Understanding the dissociation level helps predict the pH level and strength of the acid in a solution.
Weak and Strong Bases
Bases are substances that accept protons or donate hydroxide ions (\( ext{OH}^- \)). Like acids, they can be classified as weak or strong.

**Strong Bases**
  • Completely dissociate in water, releasing hydroxide ions.
  • Result in high pH values due to a high concentration of \( ext{OH}^- \).
  • Example: Potassium hydroxide (KOH) dissociates into \( ext{K}^+ \) and \( ext{OH}^- \).
**Weak Bases**
  • Partially dissociate in water, producing fewer hydroxide ions.
  • Have a lower pH compared to strong bases at the same concentration.
  • Example: Ammonia (\( ext{NH}_3 \)) forms \( ext{NH}_4^+ \) and \( ext{OH}^- \) to a limited extent.
Knowing the strength of a base helps determine how it will influence the pH of a solution.
Neutral Salts
Neutral salts are formed from the neutralization reaction of a strong acid and a strong base. They do not affect the pH of the solution significantly.

**Characteristics of Neutral Salts**
  • Composed of the cation of a base and the anion of an acid.
  • Have a pH close to 7, indicating neutrality.
  • Example: Potassium chloride (\( ext{KCl} \)) is the result of neutralizing \( ext{HCl} \) and KOH.
Neutral salts like \( ext{KCl} \) do not hydrolyze in water, maintaining a neutral pH in solution. This neutrality is important for many industrial and laboratory applications.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is crucial for calculating the pH of buffer solutions, which contain weak acids or bases and their salts.

**For Weak Acids**
  • The equation is: \( \text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \)
  • Useful for weak acids partially dissociated in solution.
  • Helps find the relationship between pH, pKa, and concentrations.
**For Weak Bases**
  • Similar form: \( \text{pOH} = \text{pKb} + \log \left(\frac{[\text{B}^+]}{[\text{BOH}]}\right) \)
  • Applicable to weak bases like \( ext{NH}_3 \)
This equation is a key tool for chemists, offering a way to estimate the pH of solutions containing weak acids and bases, thus facilitating the preparation of desired pH levels.

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},\) 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH. Calculate the pH of each solution. $$\begin{aligned} \text { Group I: } & 0.20 M \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MCl}, 0.20 M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \\ & 0.20 M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl} \end{aligned}$$ $$\begin{aligned} \text { Group II: } 0.20 \quad M\quad \mathrm{KOI}, 0.20 \quad\mathrm{M} \quad\mathrm{NaCN}, 0.20\quad \mathrm{M}\quad \mathrm{KOCl}, 0.20 \\ \mathrm{M}\quad \mathrm{NaNO}_{2} \end{aligned}$$ a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the pH of a \(1.0 \times 10^{-6}-\mathrm{M}\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

The pH of a \(0.063-M\) solution of hypobromous acid (HOBr but usually written \(\mathrm{HBrO}\) ) is \(4.95 .\) Calculate \(K_{\mathrm{a}} .\)

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make 250.0 \(\mathrm{mL}\) of solution having a pH of 10.00\(\left(K_{\mathrm{b}}\right.\) \(=1.1 \times 10^{-8} )\)
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