Answer to Question #136936 in Chemistry for Melo

Question

What is the pH of 0.15M of hydroxylamine of kb=1.1*10^-8

Answer 1

Solution:

Hydroxylamine, NH₂OH, is a weak base.

The equilibrium equation for its reaction with water:

NH₂OH(aq) + H₂O(l) ⇌ NH₃OH⁺(aq) + OH^–(aq);  K_b = 1.1×10^-8

From this, we can construct an ICE table:

The K_b expression for NH₂OH is:

K_b = [ x ] × [ x ] / [ 0.15 - x ]

C(NH₂OH) / K_b = (0.15) / (1.1×10^-8) = 1.364×10⁷ > 500

Since the ratio is greater than 500, we can remove (–x) from the equation.

The equation simplifies to:

K_b = [ x ] × [ x ] / [ 0.15 ]

K_b = x² / 0.15

Solving for x:

x² = 1.1×10^-8 × 0.15 = 0.165×10^-8

x = (0.165×10^-8)^0.5 = 4.062×10^-5

This means [OH^–] = [NH₃OH⁺] = x = 4.062×10^-5 M. 

From [OH^–], we can calculate pOH:

pOH = -log[OH^-] = -log(4.062×10^-5) = 4.39

From pOH, we can calculate pH:

pH + pOH = 14;

pH = 14 - pOH = 14 - 4.39 = 9.61

pH = 9.61

Answer: pH = 9.61