Question

what is the ph of a 0.15 m nh₄oh solution with a kᵦ of 1.78 x 10⁻⁵?
a) 12.09
b) 11.21
c) 11.93
d) 10.85

Explanation:

Step1: Calculate $[OH^-]$

For weak bases, $[OH^-] = \sqrt{K_b \times c}$, where $K_b=1.78 \times 10^{-5}$ and $c=0.15\ \text{M}$.
$$[OH^-] = \sqrt{1.78 \times 10^{-5} \times 0.15}$$
$$[OH^-] = \sqrt{2.67 \times 10^{-6}} \approx 1.634 \times 10^{-3}\ \text{M}$$

Step2: Calculate pOH

$\text{pOH} = -\log[OH^-]$
$$\text{pOH} = -\log(1.634 \times 10^{-3}) \approx 2.787$$

Step3: Calculate pH

Use $\text{pH} + \text{pOH} = 14$
$$\text{pH} = 14 - 2.787 \approx 11.21$$

Answer:

b) 11.21