Answer:
The answer to your question is below
Explanation:
A.
[H₃O⁺] = 2 x 10⁻¹⁴ M
pH = ?
Formula
pH = - log [H₃O⁺]
Substitution
pH = - log [2 x 10⁻¹⁴]
Result
pH = 13.7
B.
[H₃O⁺] = ?
pH = 3.12
Formula
pH = - log [H₃O⁺]
Substitution
3.12 = - log [H₃O⁺]
![10^{-3.12} = [H_{3} O^{+}]](https://tex.z-dn.net/?f=10%5E%7B-3.12%7D%20%3D%20%5BH_%7B3%7D%20O%5E%7B%2B%7D%5D)
Result
[H₃O⁺] = 7.59 M
This process is called meiosis! good luck!
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5