$pH = pK_a + \log \frac{[A^-]}{[HA]}$
Since we are given $K_b$ for NH4OH, we can find $K_a$ for NH4+ using the relationship:
$K_a * K_b = K_w = 1.0 \times 10^{-14}$
$K_a = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10}$
Now, we can find
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