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>
pH of 0.048 M HClO is 4.35.
<u>Explanation:</u>
HClO is a weak acid and it is dissociated as,
HClO ⇄ H⁺ + ClO⁻
We can write the equilibrium expression as,
Ka = ![$\frac{[H^{+}] [ClO^{-}] }{[HClO]}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BClO%5E%7B-%7D%5D%20%20%7D%7B%5BHClO%5D%7D)
Ka = 4.0 × 10⁻⁸ M
4.0 × 10⁻⁸ M = 
Now we can find x by rewriting the equation as,
x² = 4.0 × 10⁻⁸ × 0.048
= 1.92 × 10⁻⁹
Taking sqrt on both sides, we will get,
x = [H⁺] = 4.38 × 10⁻⁵
pH = -log₁₀[H⁺]
= - log₁₀[ 4.38 × 10⁻⁵]
= 4.35
Answer:
Weak acid and base solutions contain multiple charged and uncharged species in dynamic equilibrium. Strong acids and strong bases refer to species that completely dissociate to form ions in solution.
Explanation:
2+ charge electrons of course
