Question

What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 2.43 × 10 − 12 .

Answer 1

The percent ionization of a monoprotic weak acid solution can be calculated using the formula: Percent ionization = (concentration of ionized acid / initial acid concentration) x 100. To find the concentration of ionized acid, consider the equilibrium equation for the ionization of the weak acid and use the given Ka value to calculate the concentration of H+ ions.

Step-by-step explanation:

The percent ionization of a monoprotic weak acid solution can be calculated using the formula:

Percent ionization = (concentration of ionized acid / initial acid concentration) x 100

In this case, the initial acid concentration is 0.188 M. To find the concentration of ionized acid, we need to consider the equilibrium equation for the ionization of the weak acid:

HA(aq) <--> H+(aq) + A-(aq)

The equilibrium constant for this reaction is given as Ka = 2.43 × 10-12. Since the acid is monoprotic, the concentration of H+ ions in the solution is equal to the concentration of ionized acid. Therefore, we can use the given Ka value to calculate the concentration of H+ ions, and then use it to determine the percent ionization.

Note: Please provide the specific weak acid for a more accurate calculation.

Answer 2

Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is
`3.59* 10^(-4)\%`

Step-by-step explanation:

Dissociation of weak acid is represented as:

`HA\rightleftharpoons H^+A^-`

cM 0 0

`c-c\alpha`
`c\alpha`
`c\alpha`

So dissociation constant will be:

`K_a=((c\alpha)^(2))/(c-c\alpha)`

Give c= 0.188 M and
`\alpha` = ?

`K_a=2.43* 10^(-12)`

Putting in the values we get:

`2.43* 10^(-12)=((0.188* \alpha)^2)/((0.188-0.188* \alpha))`

`(\alpha)=3.59* 10^(-6)=3.59* 10^(-4)\%`

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is
`3.59* 10^(-4)\%`