Question

A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.

Select one:

a.
9.4 × 10−3


b.
1.8 × 10−3


c.
1.9 × 10−2


d.
3.8 × 10−3

Answer 1

According to the problem, we have a 0.200 M solution of a weak acid HA that is 9.4% ionized. To find the Ka (acid dissociation constant) for HA, we can use the percent ionization formula:

• 
  `\sf \% \: ionization = ([H+])/([HA]initial * 100)`

From this formula, we know that [H+] (the concentration of hydrogen ions) is equal to 9.4% of [HA]initial, and [HA]initial is equal to the initial concentration of the weak acid, which is 0.200 M. Solving for [H+], we get:

• 
  `\sf [H+] = 0.094 * 0.200 \: M = 0.0188\: M`

Now we can use the equation for Ka:

• 
  `\sf Ka = ([H+][A-])/([HA])`

We don't know the concentration of the conjugate base (A-) at this point, but we can assume that it is equal to [H+] because the weak acid is only slightly ionized. Therefore, we can substitute [A-] = [H+] = 0.0188 M into the equation and solve for Ka:

`\sf Ka = ((0.0188)^2)/((0.200 - 0.0188)) = \bold{1.8 * 10^(-3)}`

So the answer is
`\bold{ B.\: 1.8 * 10^(-3)}`.

I hope this helps!

Answer 2

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant (
`{ \alpha}`) is :

`\qquad \sf  \dashrightarrow \: \alpha = (dissociation \: \: percentage)/(100)`

`\qquad \sf  \dashrightarrow \: \alpha = (9.4)/(100) = 0.094`

Now, at initial stage :

`\textsf{ Conc of HA = 0.200 M}`

`\textsf{Conc of H+ = 0 M}`

`\textsf{Conc of A - = 0 M}`

At equilibrium :

`\textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}`

`\textsf{Conc of H+ = 0.094(0.200) = 0.0188 M}`

`\textsf{Conc of A - = 0.094(0.200) = 0.0188 M}`

Now, we know :

`\qquad \sf  \dashrightarrow \: { K_a = ([H+] [A-])/([HA])}`

( big brackets represents concentration )

`\qquad \sf  \dashrightarrow \: { K_a = (0.0188×0.0188)/(0.1812)}`

`\qquad \sf  \dashrightarrow \: { K_a = (0.00035344)/(0.1812)}`

`\qquad \sf  \dashrightarrow \: { K_a \approx 0.00195 }`

`\qquad \sf  \dashrightarrow \: {K_a \approx 1.9 × {10}^(-3) }`

And the closest value to that is :

`\qquad \sf \dashrightarrow \: { 1.8 × 10^(-3)}`