Answer:
Ka = 
Explanation:
Initial concentration of weak acid =
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:

(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)

0.000000135 <<< 0.00045

It is "LICI" I believe. I could be wrong please let me know if I am wrong.
Answer:
C. Precipitation of white bismuth hydroxide.
Explanation:
When aqueous ammonia is added to a solution that has Bi3+ and CU2+ cations, what we would have as a chemical result is the precipitation of white Bismuth hydroxide.
The chemical reaction for this can be written as,
Bi3+ + 2NH3 + 3H2O ⇌ Bi(OH)3(s) + 3NH4
Thank you!