A 0.185 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid.

..........HA ==> H^+ + A^-

initial..0.185...0.....0
change....-x.....x.....x
equil...0.185-x..x.....x

x = 0185(0.0155) = ?
Then substitute into the Ka expression and solve for Ka.

0.0185*0.0155=2.8675*10-4

2.8675*10-4/0.0185M=1.55*10-3

To determine the acid ionization constant (Ka), we need to first calculate the concentration of the dissociated (ionized) acid and the undissociated (unionized) acid.

Let's assume the initial concentration of the monoprotic acid is x M.

The percent ionization is given as 1.55%, which means that 1.55% of the initial concentration of the acid has dissociated. Therefore, the concentration of the dissociated acid is (1.55/100) * x = (0.0155) * x.

The remaining concentration of the undissociated acid is the initial concentration minus the concentration of the dissociated acid. So, the concentration of the undissociated acid is x - (0.0155) * x = (1 - 0.0155) * x = (0.9845) * x.

Given that the overall concentration of the acid solution is 0.185 M, we can write the equation:

x + (0.0155) * x = 0.185

Simplifying the equation, we get:

1.0155x = 0.185

Solving for x, we find:

x = 0.185 / 1.0155

x ≈ 0.182 M

The concentration of the undissociated acid (0.9845 * x) is approximately:

0.9845 * 0.182 ≈ 0.179 M

Now, to calculate the acid ionization constant (Ka), we can use the equation:

Ka = [H+][A-] / [HA]

In this case, [H+] is the concentration of the dissociated acid, [A-] is 0.182 M, and [HA] is 0.179 M.

Substituting these values into the equation, we have:

Ka = (0.0155 * 0.182) / 0.179

Calculating further, we find:

Ka ≈ 0.0157

To determine the acid ionization constant (Ka), we need to know the concentration of the acid and the percent ionization.

Given:
Concentration of the acid (CH3COOH): 0.185 M
Percent ionization: 1.55%

Step 1: Convert percent ionization to decimal form.
Percent ionization = 1.55%
1.55% = 1.55/100 = 0.0155

Step 2: Calculate the concentration of the ionized acid (H+).
Since the acid is monoprotic, the concentration of H+ is the same as the percent ionization.

[H+] = 0.0155

Step 3: Write the balanced chemical equation for the ionization of the acid.
The monoprotic acid (HA) ionizes into its conjugate base (A-) and releases one H+ ion.
HA → H+ + A-

Step 4: Write the expression for the acid ionization constant (Ka).
Ka = [H+][A-]/[HA]

Step 5: Substitute the known values into the expression for Ka.
Ka = (0.0155)(0.0155) / (0.185 - 0.0155)

Step 6: Calculate Ka.
Ka = 0.000240025 / 0.1695

Ka ≈ 0.001416

Therefore, the acid ionization constant (Ka) for the given acid is approximately 0.001416.