Classify NH4NO3 as an acidic, basic, or neutral salt, and then answer the following question. What is the pH of 0.03 M NH,NO, (aq)?
The Correct Answer and Explanation is :
To classify NH₄NO₃ as an acidic, basic, or neutral salt, let’s break it down into its constituent ions: NH₄⁺ (ammonium ion) and NO₃⁻ (nitrate ion).
- NH₄⁺ (Ammonium ion): The ammonium ion is derived from the weak base ammonia (NH₃). When dissolved in water, ammonium ions undergo hydrolysis, meaning they react with water to form ammonia (NH₃) and hydronium ions (H₃O⁺), which lowers the pH of the solution. This makes NH₄⁺ a weak acid. The reaction is as follows:
[
NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+
] - NO₃⁻ (Nitrate ion): The nitrate ion is the conjugate base of a strong acid, nitric acid (HNO₃). Conjugate bases of strong acids, such as NO₃⁻, do not hydrolyze in water and have a neutral effect on pH.
Given that NH₄⁺ is acidic and NO₃⁻ is neutral, the salt NH₄NO₃ is acidic overall. This is because the contribution of the ammonium ion to lowering the pH outweighs any neutral effect of the nitrate ion.
pH of a 0.03 M NH₄NO₃ Solution
To find the pH of a 0.03 M NH₄NO₃ solution, we need to consider the hydrolysis of the ammonium ion (NH₄⁺). The relevant equilibrium for NH₄⁺ is:
[
NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+
]
The equilibrium constant for this reaction, ( K_b ), can be found using the relationship between the base dissociation constant of ammonia (NH₃) and the acid dissociation constant of NH₄⁺. The value for ( K_b ) for ammonia is approximately ( 1.8 \times 10^{-5} ), and the value for ( K_a ) of NH₄⁺ can be calculated using:
[
K_a \cdot K_b = K_w = 1.0 \times 10^{-14}
]
Thus, the acid dissociation constant ( K_a ) of NH₄⁺ is:
[
K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}
]
Now, for the reaction:
[
NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+
]
We set up an ICE (Initial, Change, Equilibrium) table for NH₄⁺ dissociation:
- Initial concentration of NH₄⁺ = 0.03 M
- Change: Let the concentration of NH₄⁺ that dissociates be ( x ).
- At equilibrium, the concentrations will be: [NH₄⁺] = 0.03 – x, [NH₃] = x, and [H₃O⁺] = x.
The equilibrium expression for ( K_a ) is:
[
K_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} = \frac{x \cdot x}{0.03 – x}
]
Assuming ( x ) is small compared to 0.03, we can approximate the denominator as 0.03:
[
K_a = \frac{x^2}{0.03}
]
Now, solve for ( x ):
[
x^2 = (5.56 \times 10^{-10})(0.03)
]
[
x^2 = 1.668 \times 10^{-11}
]
[
x = \sqrt{1.668 \times 10^{-11}} = 4.09 \times 10^{-6} \text{ M}
]
This value for ( x ) is the concentration of hydronium ions ([H_3O^+]). To find the pH:
[
\text{pH} = -\log [H_3O^+] = -\log (4.09 \times 10^{-6})
]
[
\text{pH} \approx 5.39
]
Thus, the pH of a 0.03 M NH₄NO₃ solution is approximately 5.39, indicating that the solution is slightly acidic due to the presence of NH₄⁺.