The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
Given the concentration of aniline hydrochloride is 
Aniline hydrochloride is the conjugate acid of aniline a weak base.
pH can be calculated from 
anilinium ion the conjugate acid of aniline.
Answer:
92%
Explanation:
I believe. Hope this was helpful.
