Calculate the pH of a solution of 1.5 × 10-5 M NH4OH. What is the result?

Calculate the pH of a solution of 1.5 × 10-5 M NH4OH. What is the result?

• A. 4.2
• B. 7.2
• C. 9.2
• D. 11.2

Answer: (C)

To determine the pH of a solution of ammonium hydroxide (NH4OH), which is a weak base, we first need to recognize that it partially dissociates in water to form NH4+ and OH-. The concentration of OH- produced through this dissociation can be calculated using the base dissociation constant (Kb) of NH4OH. Given that ammonium hydroxide is a weak base, we can use the Kb value for NH4OH (around 1.8 × 10^-5) to set up a dissociation equilibrium expression: \[ K_b = \frac{[NH4^+][OH^-]}{[NH4OH]} \] Assuming that the concentration of NH4OH before dissociation is 1.5 × 10^-5 M, we can let x be the concentration of OH- produced after dissociation. The expression then simplifies to: \[ K_b = \frac{x^2}{1.5 \times 10^{-5} - x} \] At the equilibrium for weak bases, x (the concentration of OH-) is usually small compared to the initial concentration of NH4OH, so we can approximate: \[ K_b = \frac{x^2

To determine the pH of a solution of ammonium hydroxide (NH4OH), which is a weak base, we first need to recognize that it partially dissociates in water to form NH4+ and OH-. The concentration of OH- produced through this dissociation can be calculated using the base dissociation constant (Kb) of NH4OH.

Given that ammonium hydroxide is a weak base, we can use the Kb value for NH4OH (around 1.8 × 10^-5) to set up a dissociation equilibrium expression:

[ K_b = \frac{[NH4^+][OH^-]}{[NH4OH]} ]

Assuming that the concentration of NH4OH before dissociation is 1.5 × 10^-5 M, we can let x be the concentration of OH- produced after dissociation. The expression then simplifies to:

[ K_b = \frac{x^2}{1.5 \times 10^{-5} - x} ]

At the equilibrium for weak bases, x (the concentration of OH-) is usually small compared to the initial concentration of NH4OH, so we can approximate:

[ K_b = \frac{x^2