Hydroxylamine (HONH2) is a weak base with Kp = 1.1 x 10-8.
(a) Write the chemical formula for the conjugate acid of hydroxylamine.
(b) What is the value of the acid ionization constant, Ka, for the conjugate acid?
(c) Write the equation for the reaction between hydroxylamine and water.
(d) Calculate the concentrations of all species in solution, and the pOH and pH of a 1.0 M
aqueous solution of hydroxylamine.
The Correct Answer and Explanation is :
(a) Chemical formula for the conjugate acid of hydroxylamine:
The conjugate acid of hydroxylamine (HONH₂) is formed when it accepts a proton (H⁺) from water. Therefore, the conjugate acid of hydroxylamine has the formula HONH₃⁺.
(b) Value of the acid ionization constant (Ka) for the conjugate acid:
The acid ionization constant (Ka) is related to the base ionization constant (Kb) by the equation: Ka×Kb=KwK_a \times K_b = K_w
where KwK_w is the ionization constant for water, equal to 1.0×10−141.0 \times 10^{-14} at 25°C. Given that the base ionization constant for hydroxylamine is Kb=1.1×10−8K_b = 1.1 \times 10^{-8}, we can calculate KaK_a as follows: Ka=KwKb=1.0×10−141.1×10−8=9.1×10−7K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.1 \times 10^{-7}
Thus, the Ka for the conjugate acid of hydroxylamine is 9.1×10−79.1 \times 10^{-7}.
(c) Equation for the reaction between hydroxylamine and water:
When hydroxylamine dissolves in water, it reacts as a weak base, partially ionizing to produce hydroxide ions (OH⁻) and its conjugate acid (HONH₃⁺). The equilibrium reaction can be written as: HONH2(aq)+H2O(l)⇌HONH3+(aq)+OH−(aq)\text{HONH}_2 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{HONH}_3^+ (aq) + \text{OH}^- (aq)
This reaction shows hydroxylamine accepting a proton from water to form its conjugate acid, HONH₃⁺, and hydroxide ions, OH⁻.
(d) Concentrations of all species in a 1.0 M aqueous solution of hydroxylamine, pOH, and pH:
To calculate the concentrations of all species, we first set up the equilibrium expression for the ionization of hydroxylamine: Kb=[HONH3+][OH−][HONH2]K_b = \frac{[\text{HONH}_3^+][\text{OH}^-]}{[\text{HONH}_2]}
Let the concentration of hydroxylamine that ionizes be represented as xx. At equilibrium, the concentrations of the species will be:
- [HONH₃⁺] = xx
- [OH⁻] = xx
- [HONH₂] = 1.0 – xx (since the initial concentration of hydroxylamine is 1.0 M)
Substituting into the equilibrium expression: 1.1×10−8=x⋅x1.0−x1.1 \times 10^{-8} = \frac{x \cdot x}{1.0 – x}
For weak bases, we can approximate that xx is small compared to 1.0 M, so 1.0−x≈1.01.0 – x \approx 1.0. Therefore, the equation simplifies to: 1.1×10−8=x21.1 \times 10^{-8} = x^2
Solving for xx: x=1.1×10−8≈1.05×10−4x = \sqrt{1.1 \times 10^{-8}} \approx 1.05 \times 10^{-4}
Thus, the concentrations at equilibrium are:
- [OH⁻] = 1.05 × 10⁻⁴ M
- [HONH₃⁺] = 1.05 × 10⁻⁴ M
- [HONH₂] ≈ 1.0 M
To find the pOH: pOH=−log[OH−]=−log(1.05×10−4)≈3.98\text{pOH} = -\log [\text{OH}^-] = -\log(1.05 \times 10^{-4}) \approx 3.98
Now, to find the pH: pH=14−pOH=14−3.98=10.02\text{pH} = 14 – \text{pOH} = 14 – 3.98 = 10.02
Summary of results:
- Conjugate acid: HONH₃⁺
- Ka for the conjugate acid: 9.1×10−79.1 \times 10^{-7}
- Equilibrium reaction: HONH₂ + H₂O ⇌ HONH₃⁺ + OH⁻
- Concentrations at equilibrium: [OH⁻] = 1.05 × 10⁻⁴ M, [HONH₃⁺] = 1.05 × 10⁻⁴ M, [HONH₂] ≈ 1.0 M
- pOH: 3.98
- pH: 10.02