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>
Realize that pH + pOH = 14
so, 9 + pOH = 14 -> pOH = 5
pOH = -log[OH-]
5 = -log[OH-]
plug it into a calculator and you get 1.0 x 10^-5
alternatively, use [OH-] = 10^-pOH to get the same answer
[OH-] = 1.0 x 10^-5
Answer:
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Answer:
All chemical equations goes to the law of conservation of mass which says that matter can not be destroyed nor created which means there has to be an equal amount of atoms of each element on both sides of the equation. I hope that helps, I'm learning this as well.
Answer:
Oxybenzone contains ether, phenol and ketone (-CO) functional group alonq with two aromatic rings.
Explanation:
