Answer: 16.32 g of as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of as excess reagent are left.
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