The pH of a 0.095 M solution of an unknown monoprotic acid is 5.42. Calculate the Ka of the acid.

A) 3.8 x 10-6 B) 3.6 x 10-7 C) 1.5 x 10-10 D) 2.6 x 10-9 E) 2.8 x 10-8

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

I....0.095.....0.....0
C......-x......x.....x
E...0.095-x....x.....x

The problem tells you that the pH is 5.42 which means you can calculate (H^+) from pH = -log(H^+) and that is x in the above. Then (Ac^-) is the same and
you can calculate 0.095-x. Substitute each of those into the Ka expression and solve for Ka.

To calculate the Ka of the acid, we can use the equation for the pH of a weak acid:

pH = -log[H+]

To find the concentration of [H+], we can solve for it by taking the antilog of the negative pH value:

[H+] = 10^(-pH)

Given that the pH of the solution is 5.42, we can calculate the concentration of [H+] as follows:

[H+] = 10^(-5.42)

[H+] = 2.51 x 10^(-6) M

Since the acid is monoprotic, the initial concentration of [HA] (undissociated acid) is the same as the concentration of the acid itself, which is 0.095 M.

The equilibrium expression for the dissociation of a monoprotic acid is:

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

Assuming the concentration of [A-] is equal to [H+], we can substitute the values into the equation:

Ka = (2.51 x 10^(-6))^2 / 0.095

Ka = 6.30 x 10^(-12) / 0.095

Ka ≈ 6.63 x 10^(-10)

Therefore, the answer is: C) 1.5 x 10^(-10)

To calculate the Ka of the acid, we need to use the formula for pH. The pH of a solution is given by the equation:

pH = -log[H+],

where [H+] represents the concentration of hydrogen ions in the solution. In this case, since the solution is acidic, the [H+] concentration is equal to the concentration of the acid.

First, convert the pH value to [H+] concentration. To do this, we take the antilog of the pH value:

[H+] = 10^(-pH).

In this case, the pH is 5.42, so:

[H+] = 10^(-5.42).

Next, we need to calculate the concentration of the acid. Since the acid is monoprotic, its concentration is the same as the concentration of hydrogen ions.

The concentration of the acid is 0.095 M, so the [H+] concentration is also 0.095 M.

Now, we can calculate the Ka using the formula for Ka:

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

where [A-] represents the concentration of the conjugate base of the acid and [HA] represents the concentration of the acid.

Since the acid is monoprotic, the [A-] concentration is the same as the [H+] concentration, which is 0.095 M.

Plugging in the values:

Ka = (0.095)(0.095) / 0.095 = 0.095.

Therefore, the Ka of the acid is 0.095.

Looking at the answer choices, we can see that none of them match the calculated value. Therefore, the answer is not listed among the options given.

However, the correct calculated value for Ka is 0.095.

-log[H⁺] = 5.42

... [H⁺] = 10^-5.42 = 3.80E-6

Ka = [H⁺][A⁻] / [HA]
... = 3.80E-6 * 3.80E-6 /
... (.095 - 3.80E-6)
... = ?