What is the percent ionization of a 1.38 mol/L weak acid if its Ka = 2.7 x 10-6? Answer to two (2) decimal places.


I know what steps to use but have difficulty applying them to the quesiton

Can someone please complete the question for me since there are about 10 other questions just like this one on my practice sheet.

Step one is to use an ICE grid.

Step two is to use an ionization equation.

Step three is to use the equilibrium constant expression.

Step four is to solve the equation.

Step five is to use the quadratic.

Step six is to determine percent ionization by dividing ionized over initial concentration

1.38 x 100 = 1.4%

Is this correct?

No it isn't. I gave step by step instructions and worked most of it. You didn't follow any of them except to multiply by 100 to change to percent.

To find the percent ionization of a weak acid, you need to follow these steps:

Step 1: Use an ICE (Initial-Change-Equilibrium) grid.
The ICE grid helps you keep track of the initial concentration (I), the change in concentration (C), and the equilibrium concentration (E) of species involved in the acid's ionization. For this problem, you will only have one species in your grid, which is the weak acid itself.

Step 2: Use an ionization equation.
The ionization equation for a weak acid is written as:
HA ⇌ H+ + A-

Step 3: Use the equilibrium constant expression.
In this question, the equilibrium constant expression is given as Ka = [H+][A-]/[HA]. Since we're looking for the percent ionization, we don't need to use the exact values of the concentrations. Instead, we can express them in terms of x, where x represents the extent of ionization.

Step 4: Solve the equation.
Using the expression from step 3, plug in the corresponding expressions for each concentration (in terms of x) and the given value of the Ka. Rearrange the equation to solve for x, which represents the extent of ionization.

Step 5: Use the quadratic.
If the concentration of the weak acid is relatively high, the assumption x << [HA] can be made, and you can solve for x straightforwardly. However, if the concentration is low or the Ka value is small, you may need to use the quadratic equation to solve for x.

Step 6: Determine percent ionization.
The percent ionization is calculated by dividing the ionized concentration (in terms of x) by the initial concentration (in this case, [HA]) and multiplying by 100.

Now, let's apply these steps to the given problem:

Step 1: Initialize your ICE grid.
I: [HA] = 1.38 mol/L
C: Let "x" represent the extent of ionization.
E: [HA] - x, [H+] = x, [A-] = x

Step 2: Ionization equation.
HA ⇌ H+ + A-

Step 3: Equilibrium constant expression.
Ka = [H+][A-]/[HA]
Ka = (x)(x)/([HA] - x)

Step 4: Solve the equation.
Since this is a weak acid and the concentration is relatively high, we can assume x << [HA]. Therefore, we can approximate [HA] - x as [HA] and solve the equation directly.
Ka = (x)(x)/[HA]
Ka = (x^2)/[HA]
2.7 x 10^-6 = (x^2)/(1.38)

Step 5: Use the quadratic.
To solve for x, rearrange the equation:
(x^2) = (2.7 x 10^-6)(1.38)
x ≈ √(2.7 x 10^-6)(1.38)

Step 6: Determine percent ionization.
To find the percent ionization, divide x by the initial concentration [HA] and multiply by 100.
Percent ionization = (x/[HA]) * 100

Now, you can substitute the value of x from step 5 and [HA] from the initial information into this equation to calculate the percent ionization. Round the answer to two decimal places as instructed.

Can someone please complete the question for me since there are about 10 other questions just like this one on my practice sheet.

First you write the ionization equation. For a weak acid, which I will call HA, it ionizes as follows:
HA <==> H^+ + A^-

Next you want to write the Ka expression.
Ka = (H^+)(A^-)/(HA) = 2.7 x 10^-6


Step one is to use an ICE grid.
For spacing reasons, this board doesn't handle multiple spaces at all; therefore, I will need to covert this a little differently. Basically, I do it vertically, like this.
I
C
E
Initial concn:
(HA) = 1.38 M
(H^+) = 0
(A^-) = 0

Change in concn:
(H^+) = +y
(A^-) = +y
(HA) = -y

Equilibrium concn:
(HA) = 1.38 - y
(H^+) = 0 + y = y
(A^-) = 0 + y = y


Step two is to use an ionization equation.

Step three is to use the equilibrium constant expression.

Step four is to solve the equation.
I will leave the equation for you to solve. You may want to try making the simplifying assumption that 1.38 - y = 1.38. That will avoid solving a quadratic IF it is almost true.

Step five is to use the quadratic.

Step six is to determine percent ionization by dividing ionized over initial concentration
percent ionization = (amount ionized/amount to start)*100.

The amount ionized is just (H^+) so you can use that for the numerator. The amount unionized will be 1.38. Then multiply by 100 to convert to percent.
Post your work if you get stuck.