Question

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 the ionization constant of this acid?

Answer 1

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= $\frac{ [H^{+}] [A^{-} ]}{ [HA]}$

Ka= $\frac{x x}{0.200 M -x}$

Ka= $\frac{x^{2} }{0.200 M - x}$

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= $\frac{(10^{-3})^{2} }{0.200 - (10^{-3}) }$= $\frac{10^{-6} }{0.199}$= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

Answer 2

Answer:5.0251 × 10^-6M

Explanation: Please see the attachments below