Enough of a monoprotic acid is dissolved in water to produce a 0.0165 M solution. The pH of the resulting solution is 2.38. Calculate the Ka for the acid.

Nevermind I figured it out,

With the equation
Ka = x^2 / 0.0165 - x

x = [H+]
I forgot to do the 0.0165-x in my original calculation, but after doing that I got the correct answer.

That's right. 0.0165-x can't be assumed = 0.0165

To calculate the Ka (acid dissociation constant) for the monoprotic acid, we need to use the pH of the solution and the concentration of the acid.

The pH of a solution is related to the concentration of the hydrogen ions (H+) in the solution. In this case, the pH of the solution is given as 2.38, which means the concentration of H+ is 10^(-2.38) M.

The acid is monoprotic, meaning it donates one proton (H+) when it ionizes. So, if we denote the initial concentration of the acid as x M, it will produce an equal concentration of H+ when it ionizes.

The equation for the dissociation of the acid is:
HA(aq) ⇌ H+(aq) + A-(aq)

The equilibrium expression for this reaction is:
Ka = [H+][A-] / [HA]

Since the concentration of H+ is the same as the concentration of A-, we can simplify the expression to:
Ka = [H+]^2 / [HA]

Given that the concentration of H+ is 10^(-2.38) M, and the concentration of the acid is 0.0165 M, we can plug these values into the expression:

Ka = (10^(-2.38))^2 / 0.0165

Calculating this, we find that the Ka value for the acid is approximately 1.19 x 10^(-5).

To calculate the Ka for the acid, we need to use the pH of the solution.

Ka, or acid dissociation constant, is a measure of the strength of an acid. It can be calculated using the equation:

Ka = [H3O+][A-] / [HA]

Where [H3O+] is the concentration of hydronium ions, [A-] is the concentration of the acid's conjugate base, and [HA] is the concentration of the acid itself.

In this case, the pH of the solution is given as 2.38. Since pH is a measure of the concentration of hydronium ions, we can use the formula:

pH = -log[H3O+]

To find the concentration of hydronium ions ([H3O+]), we need to rearrange the formula as follows:

[H3O+] = 10^(-pH)

Plugging in the given pH value of 2.38, we get:

[H3O+] = 10^(-2.38) = 4.02 x 10^(-3) M

Now that we have the concentration of hydronium ions, we can use it to calculate the concentration of the acid's conjugate base ([A-]). In a monoprotic acid, the concentration of the acid ([HA]) is equal to the concentration of the acid's conjugate base ([A-]). Therefore, we can substitute [A-] with [HA] in the equation.

Now, let's look at the information given in the question. It states that the acid is dissolved to produce a 0.0165 M solution. This means that [HA] = 0.0165 M.

Plugging the values into the equation for Ka:

Ka = (4.02 x 10^(-3) M)(0.0165 M) / (0.0165 M)

Simplifying the equation:

Ka = 4.02 x 10^(-3)

Therefore, the Ka for the acid is 4.02 x 10^(-3).