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>
Answer:
alkali metals- Group 1
Explanation:
they have less valence electrons and therefore are more reactive
Answer:
24m/s
Explanation:
a=change of v/change of t
6m/s^2=v/4s
multiply both sides by 4s
v=24m/s
Yes because look in the book dh