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= [tex]\frac{ [H^{+}] [A^{-} ]}{ [HA]}[/tex]
Ka= [tex]\frac{x x}{0.200 M -x}[/tex]
Ka= [tex]\frac{x^{2} }{0.200 M - x}[/tex]
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= [tex]\frac{(10^{-3})^{2} }{0.200 - (10^{-3}) }[/tex]= [tex]\frac{10^{-6} }{0.199}[/tex]= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
