Answer
The percent of $NH_3$ that is $NH_4^+$ in this solution is 1.5%.
Work Step by Step
Equations needed:
$Kb = \frac{[OH^-][B^+]}{[BOH]}$**
**All these concentrations are the values in the equilibrium.
$[OH^-] = [B^+] = x$
$[BOH] = [BOH]_{initial} - x$
Kb ammonia = $1.8 \times 10^{-5}$
-Remember, for ammonia: $BOH = NH_3$ (The base) and $B^+ = NH_4^+$ (The conjugate acid)
1. Find $[NH_4^+]$.
It is the conjugate acid: $[NH_4^+] = [B^+]$
- Since 0.08M is a large number, compared to the Kb value, we can consider:
$[BOH] \approx [BOH]_{initial}$
Therefore:
$1.8 \times 10^{-5} = \frac{x * x}{0.08}$
$1.44 \times 10^{-6} = x^2$
$x = 1.2 \times 10^{-3}M = [OH^-] = [B^+]$
2. Now we know that:
$[NH_4^+] = 1.2 \times 10^{-3}M$
and
$[NH_3]_{initial} = 0.08M$
So the percent between these 2 is:
$\frac{1.2 \times 10^{-3}}{0.08} \times 100\% = 1.5\%$
