Question

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

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

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

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 \(\mathrm{pH} = -\log{[\mathrm{H}^+]}\).

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

Key concepts

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.