Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t
he ionization constant of this acid?
2 answers:
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
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Answer:
The value of for this reaction at 1200 K is 4.066.
Explanation:
Partial pressure of water vapor at equilibrium =
Partial pressure of hydrogen gas at equilibrium =
Total pressure of the system at equilibrium P = 36.3 Torr
Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:
The expression of is given by:
The value of for this reaction at 1200 K is 4.066.