Answer:
C₁₂H₂₂O₁₁ and CH₃OH
Explanation:
Sucrose and methyl alcohol are nonelectrolytes. They do not ionize or conduct a current in aqueous solution.
HC₂H₃O₂ is a weak electrolyte. It produces only a few ions and is a poor conductor of electricity in aqueous solution.
HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻
H₂SO₄ is a strong electrolyte. Its first ionization is complete, so it is a good conductor of electricity in aqueous solution.
H₂SO₄ + H₂O ⟶ H₃O⁺ + HSO₄⁻
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka=
Ka=
Ka=
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= =
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶