Consider the following data on some weak acids and weak bases

Consider the following data on some weak acids and weak bases

HNO2 Ka 4.5×10^-4

HCH3CO2 Ka 1.8×10^-5

C6H5NH2 Kb 4.3×10^-10

C2H5NH2 Kb 6.4×10^-4

Use the data to rank the following solutions in order of increasing pH

0.1 M C6H5NH3Cl
0.1 M KNO3
0.1 M NaNO2
0.1 M KCH3CO2

The correct answer and explanation is :

To rank the solutions in order of increasing pH, we must analyze the effect of each substance on the solution’s pH. Solutions of salts, acids, and bases influence pH depending on whether they produce acidic, neutral, or basic conditions when dissolved in water.

Solution Analysis:

1. 0.1 M C6H5NH3Cl:

• C6H5NH3Cl is the salt of aniline (C6H5NH2), a weak base, and HCl, a strong acid. When dissolved, C6H5NH3Cl dissociates into C6H5NH3⁺ (the conjugate acid of aniline) and Cl⁻ (a neutral ion).
• The conjugate acid C6H5NH3⁺ will undergo hydrolysis, reacting with water to produce H⁺ ions and thus lower the pH. The hydrolysis reaction is: $$
  C6H5NH3^+ + H_2O \leftrightarrow C6H5NH2 + H_3O^+
  $$
• Since the Ka of C6H5NH3⁺ is determined by the relationship $K_a = \frac{K_w}{K_b}$, where $K_b = 4.3 \times 10^{-10}$ for aniline, the pH will be acidic but not extremely low.

1. 0.1 M KNO3:

• KNO3 is a neutral salt, derived from a strong acid (HNO3) and a strong base (KOH). Neither K⁺ nor NO3⁻ undergo hydrolysis, so the solution remains neutral. The pH of this solution will be approximately 7.

1. 0.1 M NaNO2:

• NaNO2 is a salt of a weak acid (HNO2, Ka = 4.5 x 10⁻⁴) and a strong base (NaOH). The NO2⁻ ion is the conjugate base of HNO2 and will hydrolyze to produce OH⁻ ions, making the solution basic: $$
  NO_2^- + H_2O \leftrightarrow HNO_2 + OH^-
  $$
• Since the Kb for NO2⁻ can be calculated as $K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{4.5 \times 10^{-4}} \approx 2.2 \times 10^{-11}$, the solution will be weakly basic.

1. 0.1 M KCH3CO2:

• KCH3CO2 is a salt of acetic acid (CH3COOH, Ka = 1.8 x 10⁻⁵) and KOH. The CH3CO2⁻ ion is the conjugate base of acetic acid and will hydrolyze in water to produce OH⁻ ions: $$
  CH_3CO_2^- + H_2O \leftrightarrow CH_3COOH + OH^-
  $$
• The Kb for CH3CO2⁻ is $K_b = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.6 \times 10^{-10}$, which means the solution will also be weakly basic but not as much as NaNO2.

Ranking the Solutions by pH (Increasing pH):

1. 0.1 M C6H5NH3Cl: This solution is acidic due to the presence of the conjugate acid of aniline.
2. 0.1 M NaNO2: This solution is weakly basic, with a higher pH than the acidic C6H5NH3Cl solution.
3. 0.1 M KCH3CO2: This solution is weakly basic but slightly less so than NaNO2.
4. 0.1 M KNO3: This is a neutral solution, so it has the highest pH of around 7.

Conclusion:

The correct order of increasing pH is:

0.1 M C6H5NH3Cl < 0.1 M NaNO2 < 0.1 M KCH3CO2 < 0.1 M KNO3